Arihant CBSE Class 11 Book

CBSE Term II

2022

Applied
Mathematics

Class XII



CBSE Term II

2022

Applied
Mathematics

Class XII

Complete Theory Covering NCERT
Case Based Questions
Short/Long Answer Questions
3 Practice Papers

Authors
Raju Regar
Sagar Verma

ARIHANT PRAKASHAN (School Division Series)

ARIHANT PRAKASHAN (School Division Series)

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CBSE Term II

2022

Contents -
-
CHAPTER -
Indefinite Integrals -
-
CHAPTER -
Definite Integrals -
-
CHAPTER -
Application of Integrals -
-
CHAPTER
Differential Equations

CHAPTER
Inferential Statistics
CHAPTER
Index Number and Time Based Data
CHAPTER
Perpetuity, Sinking Funds, Bonds EMI

CHAPTER
Stock, Shares and Debentures
CHAPTER
Return, Growth and Depreciation
CHAPTER
Linear Programming

Practice Papers -

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Syllabus

CBSE Term II Class XII

One Paper Time : Hrs Max Marks:

No. Units Marks
III. Calculus Continued
V. Inferential Statistics
VI. Index Numbers and Time Series Continued
VII. Financial Mathematics
VIII. Linear Programming
Total

SI.No. Contents Learning Outcomes Notes Explanations
Students Will Be Able To
l Integration as a reverse process of
UNIT - CALCULUS differentiation

Integration and its Applications l Vocabulary and Notations related to
Integration
. Integration l Understand and determine indefinite
integrals of simple functions as anti- l Simple integrals based on each method
derivative non-trigonometric function

. Indefinite l Evaluate indefinite integrals of simple l Evaluation of definite integrals using
properties
Integrals as algebraic functions by method of:
Problems based on finding
family of curves i substitution l Total cost when Marginal Cost is given
l Total Revenue when Marginal Revenue
ii partial fraction
iii by parts is given
l Equilibrium price and equilibrium
. Definite l Define definite integral as area under the
quantity and hence consumer and
Integrals as curve producer surplus

area under the l Understand fundamental theorem of

curve Integral calculus and apply it to evaluate

the definite integral

l Apply properties of definite integrals to
solve the problems

. Applications of l Identify the region representing C.S. and

Integrations P.S. graphically

l Apply the definite integral to find
consumer surplus-producer surplus

Differential Equations and Modeling

. Differential l Recognize a differential equation l Definition, order, degree and examples
Equations
l Find the order and degree of a l Growth and Decay Model in Biological
differential equation sciences, Economics and business, etc.

. Applications l Define Growth and Decay Model

of Differential l Apply the differential equations to solve
Equations Growth and Decay Models

UNIT - INFERENTIAL STATISTICS

. Population l Define Population and Sample l Population data from census, economic
and Sample surveys and other contexts from
l Differentiate between population and practical life
sample
l Examples of drawing more than one
l Define a representative sample from a sample set from the same population
population
l Examples of representative and non-
l Differentiate between a representative representative sample
and non-representative sample
l Unbiased and biased sampling
l Draw a representative sample using
simple random sampling l Problems based on random sampling
using simple random sampling and
l Draw a representative sample using and systematic random sampling sample
systematic random sampling size less than

. Parameter l Define Parameter with reference to l Conceptual understanding of Parameter
and Statistics Population and Statistics
and Statistical
Interferences l Define Statistics with reference to Sample l Examples of Parameter and Statistic
limited to Mean and Standard deviation
l Explain the relation between Parameter only

and Statistic l Examples to highlight limitations of

l Explain the limitation of Statistic to generalizing results from sample to

generalize the estimation for population population

l Interpret the concept of Statistical l Only conceptual understanding of
Significance and Statistical Inferences Statistical Significance Statistical

l State Central Limit Theorem Inferences

l Explain the relation between Population- l Only conceptual understanding of
Sampling Distribution-Sample Sampling Distribution through

simulation and graphs

. t-test l Define a hypothesis l Examples and non-examples of Null and

one sample l Differentiate between Null and Alternate Alternate hypothesis only non-
t-test and two hypothesis directional alternate hypothesis
independent
groups t-test l Define and calculate degree of freedom l Framing of Null and Alternate hypothesis

l Test Null hypothesis and make inferences l Testing a Null Hypothesis to make
using t-test statistic for one group two Statistical Inferences for small sample size

independent groups l for small sample size: t- test for one group

and two independent groups

l Use of t-table

UNIT - INDEX NUMBERS AND TIME-BASED DATA

. Time Series l Identify time series as chronological data l Meaning and Definition

. Components l Distinguish between different l Secular trend
l Seasonal variation
of Time Series components of time series l Cyclical variation
l Irregular variation
. Time Series l Solve practical problems based on
l Fitting a straight line trend and
analysis for statistical data and Interpret the result estimating the value

univariate data

. Secular Trend l Understand the long term tendency l The tendency of the variable to increase
or decrease over a long period of time

. Methods of l Demonstrate the techniques of finding l Moving Average method
Measuring trend by different methods l Method of Least Squares
trend

UNIT - FINANCIAL MATHEMATICS

. Perpetuity, l Explain the concept of perpetuity and l Meaning of Perpetuity and Sinking Fund
l Real life examples of sinking fund
Sinking Funds sinking fund l Advantages of Sinking Fund
l Sinking Fund vs. Savings account
l Calculate perpetuity
l Meaning of Bond Valuation
l Differentiate between sinking fund and l Terms related to valuation of bond:
saving account
Coupon rate, Maturity rate and
. Valuation l Define the concept of valuation of bond Current price
of Bonds and related terms l Bond Valuation Methods:
i Present Value Approach
l Calculate value of bond using present ii Relative Price Approach
value approach
l Methods to calculate EMI:
. Calculation l Explain the concept of EMI i Flat-Rate Method
of EMI l Calculate EMI using various methods ii Reducing-Balance Method

l Real life examples to calculate EMI of
various types of loans, purchase of
assets, etc.

. Calculation l Explain the concept of rate of return and l Formula for calculation of Rate of Return,

of Returns, nominal rate of return Nominal Rate of Return

Nominal Rate l Calculate rate of return and nominal rate
of Return of return

. Compound l Understand the concept of Compound l Meaning and use of Compound Annual

Annual Growth Annual Growth Rate Growth Rate

Rate l Differentiate between Compound Annual l Formula for Compound Annual

Growth Rate and Annual Growth Rate Growth Rate

l Calculate Compound Annual
Growth Rate

. Linear method l Define the concept of linear method of l Meaning and formula for Linear Method
of Depreciation Depreciation of Depreciation

l Interpret cost, residual value and useful l Advantages and disadvantages of Linear
life of an asset from the given Method
information

l Calculate depreciation

UNIT - LINEAR PROGRAMMING

. Introduction l Familiarize with terms related to Linear l Need for framing linear programming
problem
and related Programming Problem
l Definition of Decision Variable,
terminology Constraints, Objective function,
Optimization and Non-negative
conditions

. Mathematical l Formulate Linear Programming Problem l Set the problem in terms of decision

formulation of variables, identify the objective function,

Linear identify the set of problem constraints,

Programming express the problem in terms of

Problem inequations

. Different types l Identify and formulate different l Formulate various types of LPP s like
Manufacturing Problem, Diet Problem,
of Linear types of LPP Transportation Problem, etc.

Programming

Problems

. Graphical l Draw the Graph for a system of linear l Corner Point Method for the Optimal
method of inequalities involving two variables and solution of LPP
solution for to find its solution graphically
problems in

two variables

. Feasible and l Identify feasible, infeasible and bounded l Definition and Examples to explain the

Infeasible regions terms

Regions

. Feasible and l Understand feasible and infeasible l Problems based on optimization

infeasible solutions l Examples of finding the solutions by
graphical method
solutions, l Find optimal feasible solution

optimal feasible

solution

CBSE Circular

Acad - 51/2021, 05 July 2021

Exam Scheme Term I & II

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CBSE Term II Applied Mathematics XII 1

CHAPTER 01

Indefinite
Integrals

In this Chapter...

l Indefinite Integral
l Integration by Substitution
l Integration by Partial Fractions
l Integration by Parts

Indefinite Integral where, C is any arbitrary constant.

Let F( x) and f ( x) be two functions connected together, such (ii) ò { f (x) ± g (x)} dx = ò f (x) dx ± ò g (x) dx
that d [ F( x)] = f ( x), then F( x) is called integral of f ( x) or (iii) ò k f (x) dx = k ò f (x) dx

dx (iv) In general, if f1 , f 2 , ..., f n are functions and
k1 , k2 , ..., kn are numbers, then
indefinite integral or anti-derivative.
ò [ k1 f1 ( x) + k2 f 2 ( x) + ... + kn f n ( x)] dx
Thus, ò f ( x) dx = F( x) + C ò ò ò= k1 f1 ( x) dx + k2 f 2 ( x) dx + ... + kn f n ( x) dx + C

where, C is any arbitrary constant. where, C is the constant of integration.

Some Standard Formulae Integration by Substitution

ò(i) xn dx = x n +1 + C, n ¹ - 1 The method of reducing a given integral into one of the
n+1 standard integrals by a proper substitution is called method of
substitution. To evaluate an integral of the type
(ii) ò dx = x + C
ò f {g( x)}× g¢ ( x) dx, we substitute g( x) = t, so that g¢ ( x) dx = dt
(iii) ò 1 dx = log|x| + C, x ¹ 0
x

ò(iv) ex dx = ex + C Integrals of Some Particular Functions

ò(v) a x dx = a x + C, a > 0, a ¹ 1 ò(i) dx = 1 log x-a +C
( x2 - a 2 ) 2a x+a
log e a

Properties of Indefinite Integral ò(ii) dx = 1 log a+x
( a 2 - x2 ) 2a a -x
(i) The process of differentiation and integration are inverse +C

of each other. ò(iii) dx = log|x + x2 - a 2| + C

i.e. d ò f ( x) dx = f ( x) and ò f ¢( x) dx = f (x) + C (x2 - a2 )
dx

2 CBSE Term II Applied Mathematics XII

ò(iv) dx = log x + x2 + a 2 + C px2 ± qx ± r A+ B+ C
4. ( x ± a) ( x ± b)2 ( x ± a) ( x ± b) ( x ± b)2
x2 + a2
px2 ± qx ± r A+ B +C
ò(v) x2 - a 2 dx = x x2 - a 2 - a 2 log|x + x2 - a 2|+ C 5. ( x ± a)2 ( x ± b) ( x ± a) ( x ± a)2 ( x ± b)

22 px2 ± qx ± r (x A a) + (x B + (x C
6. ( x ± a)3 ± ± a)2 ± a)3
ò(vi) x2 + a 2 dx = x x2 + a 2 + a 2 log|x + x2 + a 2|+ C
px2 ± qx ± r A + Bx + C
22 7. ( x ± a) ( x2 ± bx ± c) ± a) x2 ± bx ±

Some Standard Integrals and Substitutions for them

(x c where,

Integral Substitution x2 ± bx ± c cannot be
Put px + q = t
ò( ax + b) px + q dx factorised further.

or ò ( ax + dx + q Method of Solving Integral by Partial Fractions
b) px

or ò ( ax2 + dx px + q Suppose given integral is of the form ò P( x) dx, where P( x)
+ c) Q( x)
bx

and Q( x) are polynomials in x and Q( x) ¹ 0. Then, to evaluate

ò ( px + q)( dx Put px + q = 1 such integrals by partial fraction, we firstly take the given
ax2 + bx + c ) t
integrand P( x) and decompose it into suitable partial fraction
Q( x)

ò dx Put x = 1 and then put a2 + bt 2 = u form by above method and then integrate each term by using
t suitable method to get the required answer.
( px2 + q) ax2 + b

Integration by Partial Fraction Integration by Parts

Sometimes, an integral of the form ò P( x) dx, where P( x) and Let u and v be two differentiable functions of a single variable
Q( x) x, then the integral of the product of these two functions

Q( x) are polynomials in x and Q( x) ¹ 0, also Q( x) has only = Ist function ´ Integral of the IInd function

linear and quadratic factors given to us, if we cannot integrate -Integral of [Derivatives of Ist function

it directly or by previous methods, then we use the partial ´ Integral of the IInd function]

fractions. For this, firstly we have to know partial fraction i.e. ò u×v dx = uò v dx - ò çæ d ( u)ò v dx ö÷ dx
è dx ø
decomposition which is given below I II

Partial Fraction Decomposition If in the product, two functions are of different types, then
take that function as first function (i.e. u) which comes first in
According to nature of factors of Q( x), corresponding form of word ILATE, where
the partial fraction is given below.
I : Inverse trigonometric function. e.g. sin -1 x
S.No. Form of the rational Form of the partial
1. function fraction L : Logarithmic function. e.g. log x
A : Algebraic function. e.g. 1, x, x2
(x px ± q , a ¹ b x A + x B
± a) ( x ± b) ±a ±b T : Trigonometric function. e.g. sin x, cos x
E : Exponential function. e.g. ex

px ± q (x A a) + (x B Integral of the type ò e x[f( x) + f ¢( x)] dx
2. ( x ± a)2 ± ± a)2
If the given integrand is of the form ex [ f ( x) + f ¢( x)], then we
px2 ± qx ± r (x A a) + (x B b) + C c)
3. ( x ± a) ( x ± b) ( x ± c) ± ± (x ± can directly write the integral as

ò ex [ f ( x) + f ¢( x)] dx = ex f ( x) + C

CBSE Term II Applied Mathematics XII 3

Solved Examples

Example 1. Evaluate the following integrals. Example 3. Evaluate ò 3ax dx.
+ c2x2
(i) ò x6 dx ò(ii) 1 dx (iii) ò 5 x dx b2
x 3/4
òSol. Let I = b2 3ax dx
Sol. (i) Let I = ò x6dx + c2x2

= x6 +1 + òé xndx = xn+1 + C, ¹ -1 úù Now, put b2 + c2x2 = t Þ 2c2x dx = dt Þ x dx = dt
6+1 n+1 û 2c2
C êQ if n
ë
ò\ I = 3a dt = 3a log|t| + C
= x7 + C 2c2 t 2c2
7
= 3a log|b 2 + c2x2| + C
é 1 ù 2c2
ò ò(ii) Let I = êë x 3/ ûú dx = x-3/ 4 dx
4 = 3a log ( b2 + c2x2 ) + [Q b2 + c2x2 > 0]
2c2 C

- 3+1 òé xn +1 ù
x4 n+ 1 Cú
= -3+ + C êQ xndx = + Example 4. Evaluate ò dx .
ë û x(x3 +8)
4 1

= x1/ 4 + C = 4x1/ 4 + C dx x2
1/4 x( x3 + x3(x3 +
ò òSol. Let I = = dx
8) 8)

where, C is a constant of integration. [multiply numerator and denominator by x2]
Now, put x3 + 8 = t
ò(iii) Let I = 5xdx = 5x + C òé axdx = ax + ù
log a Cú Þ 3x2 dx = dt Þ x2 dx = dt
log5 êQ 3
ë û

where, C is a constant of integration. 1 ò dt
3 - 8)
Example 2. Evaluate ò (1 + x)2 dx. \ I= (t t

x = 1 ò dt
3 -
(1 + x)2 t 2 8t
x
Sol. Let I = ò dx ò= 1 dt

= ò (1 + x2 + 2x) dx 3 t 2 - 8t + ( 4)2 - ( 4)2
x
[adding and subtracting by ( 4)2 from denominator]

=ò 1 dx + ò x2 dx + 2ò x dx [by property (iv)] ò= 1 dt
x x x ( t - 4)2 - ( 4)2
3

-1 2-1 1- 1 = 1 ´ 1 t -4-4 +C
3 2´ t -4+ 4
x 2 dx x 2 dx x 2 dx
ò ò ò= + + 2 4 log

ò ò ò= x–1/ 2dx + x3/ 2dx + 2 x1/ 2dx = 1 log t - 8 + C
24 t
-1 +1 3 +1 1 +1 é xn +1 ù
êQ n+1 Cú
x2 x2 2x 2 1 ë
-1 + 1 3+1 1+ û
ò= + + + C xndx = + = 1 log x3 +C [Q t = x3 + 8]
24 x3 + 8
222

= x1/ 2 + x5 / 2 + 2x3/ 2 + C Example 5. Evaluate ò dx .
1/ 2 5/ 2 3/ 2
x ax - x 2
= 2x1/ 2 + 2 x5 / 2 + 4 x3/ 2 + C
53 Sol. Let I = ò dx
ax - x2
= 2 x + 2 x5 / 2 + 4 x3/ 2 + C x
53
Now, put x = 1 Þ dx = - 1 dt
where, C is a constant of integration. t t2

4 CBSE Term II Applied Mathematics XII

èçæ - 1 öø÷ dt - dt = - dt ò ( x2 + 1) e x
t2 t 2 × 1 at - 1 at - 1 (x + 1 )2
Example 7. Evaluate dx.
tt
ò ò\
òI = 1 a 1 =
t t t2
- x 2 +1
x + 1)2
- 1 +1 òSol. Let I = ex ( dx
2
ò= - ( at - 1 ) - 1 = - ( at - 1 ) +
2 dt C
a × çæ - 1 + 1ö÷ ex ççæè1 2x ÷øö÷
è2 ø òÞ I= - (x + 1)2 dx

= - 2 ( at 1 +C =- 2 a× 1 -1 +C é t = 1ù = ò ex dx - 2ò ex (x x dx
êëQ x úû + 1)2
- 1)2
a ax
+1 -1
òÞ ex x x + 1)2
= -2 a - x + C I = ex - 2 ( dx
ax
exíì ü
ò x òÞ îx 1 -1 ý
Example 6. Evaluate (x2 + 1) (x - 1) dx. I = ex - 2 + 1 + (x + 1)2 þ dx

Sol. Let I = ò (x2 x dx ìü
+ 1) (x - 1) I = ex - 2íï 1 1 dxïý
îï + + 1)2 ïþ
ò òÞ ex× ex
x P( x) x 1 dx - (x
1) (x Q( x) II
Here, integrand = = (say)
( x2 + - 1) I

ò òÞ =ex ì 1 ex 1 ex 1 ü
where, degree of P( x) is less than degree of Q( x), so it is a I - 2í x + 1 - - x + 1)2 exdx - (x + 1)2 dxý
î (
proper fraction. þ

Now, let x = A + Bx + C ...(i) [using integration by part]
1) (x - 1) x2 + 1
( x2 + - 1) (x ò òÞ ì 1 ex + e x × 1 dx ü
2ïïí x +1 x + 1) ïýï
On multiplying both sides of Eq. (i) by ( x2 + 1) ( x - 1), we get = ex - ïîï- ( 2 ïïþ +
ex
I 1 C
+ 1)2
x = A ( x2 + 1) + ( Bx + C) ( x - 1) ...(ii) × ( x dx

On putting x = 1 in Eq. (ii), we get Þ I = ex - 2 x ex 1 + C
1 = A (1 + 1) + ( B + C) ( 0) Þ 1 = 2 A Þ A = 1 +
2

On equating the coefficients of x2 and x from both sides of Example 8. Evaluate ò ( 2x - 5) e2x dx.
Eq. (ii), we get (2x - 3 )3

0 = A + B and 1 = - B + C

Þ 0=1 + BÞ B=-1 é A = 1ù òSol. Let I = ( 2x - 5) e2x dx
22 ëêQ 2 ûú ( 2x - 3)3

and 1 = - æçè - 1 öø÷ + C Þ C =1 - 1 = 1 ò= ( 2x - 3 - 2)e2x dx
2 2 2 ( 2x - 3)3

From Eq. (i), we get -1 x+1 e2x e2x
+2 2 ( 2x - 3)2 ( 2x - 3)3
x = 1 ò ò= dx -2 dx

(x2 + 1) (x - 1) 2 (x - 1) x2 + 1 …(iii)

Now, putting partial fraction form of integrand from Eq. (iii) ò ò= e2x( 2x - 3)- 2 dx - 2 e2x( 2x - 3)- 3 dx
II I

in given integral, we get ò ò ò= éëê( 2x - 3)-2 ìîí d 3)-2 dxþüý ù
dx ûú
= ò 1 + 1 ò ( - x + 1 ) e2xdx - ( 2x - e 2 x dx
2(x - 1) 2 x + 1
I dx 2 dx

= 1 log|x - 1| + 1 é - ò x ´ 2 dx + ò 1 ù ò- 2 e2x ( 2x - 3)- 3 dx
2 2 ê + 2 x2 + dxú
ë x2 1 1 [using integration by parts]
û
ò= ( 2x - 3)- 2 e2x - - 2( 2x - 3)-3 ´ 2 ´ e2x dx
= 1 log|x - 1| - 1 ò 2x 1 dx + 1 ò 1 1 dx
2 4 x2 + 2 x2 + 22

= 1 log|x - 1| - 1 log|x2 + 1| + 1 tan-1 x + C ò- 2 e2x ( 2x - 3)- 3 dx
24 2 ò= e2x( 2x - 3)-2 + 2 e2x( 2x - 3) -3 dx

òé for 2x dx, put x2 + 1 = t Þ 2x dx = dt, ù ò2 - 2 e2x( 2x - 3)-3 dx
x2 +1 ú
ê = e2x( 2x - 3)-2 + C
evaluating 2

ê ú
ê 2x 1 ú
ëê then ò x2 + 1 dx = ò t dt = log|t| = log|x2 + 1| + C úû

CBSE Term II Applied Mathematics XII 5

ò9. Evaluate çæ + - 1 ö÷ex + 1 A =1
x
Example 1 x dx. Again, putting x = - 1, we get
è xø
B = -1
1
òSol. Let I = çæ1 + x - 1 ÷ö x+ x dx \ I1 = ò çæçè 1 - 1 1 ø÷÷ö dx
x +
è xø e x

x+ 1 é çæè1 1 ÷øö x+ 1ù = ò 1 dx - ò dx
ê x2 xú x x+1
ex ê e ú
ò ò= dx + x êë - úû dx
I

II = log|x| - log|x + 1| + C ...(ii)

x+ 1 x+ 1 x + 1
x x
ex xe e
ò ò= dx + - 1 ´ dx Now, from Eqs. (i) and (ii), we get

é ò u× v dx = u ò v dx - ò íîì d ( u) òv dxýüþ ù I = - log x + log|x| - log|x + 1| + C
êQ dx = dxú x+1
òê dt
I1 = æç1 - 1 ö÷ e x+ 1 dx ú = - log x + log x +C é log m - log n = log mù
ê put è x2 ø x - ú x+1 x+1 êëQ n úû
ê + 1 ú
ê x x = t Þ æçè1 1 ÷øö dx ú Example 11. Evaluate ò xex dx.
ê x2 ú (1 + x)2
ê ú
òê ú
et dt = et = x + 1 + C úû
ëê\ x
e xex ( x + 1 -1) e x
ò òSol. Let I = + x)2 (1 + x)2
x+ 1 (1 dx = dx

= xe x + C é 1+x 1 ù
ëê (1 + x)2 + x)2 ûú
Example 10. Evaluate ò log x dx. ò= - (1 exdx
x + 1)
( 2

= ò log x dx = ò log x× ( x 1 òÞ I = e x é 1 - 1 ù
x + 1)2 + 1)2 êë + + x)2 ûú dx
Sol. Let I dx (1 x) (1

(

I II Let f ( x) = 1 .
+ x
On applying integration by parts, we get 1

I = log x × ò ( x dx - ò d (log x) × çèæç ò ( x dx ) 2 ö÷ø÷ dx Then, f ¢( x) = - 1
+ 1)2 dx +1 + x)2
(1

é ò u× v dx = uò v dx - ò æèç d ( u)ò v dxø÷ö dx ù \Given integral is of the form
ëêQ I II dx ûú
I = ò ex[ f ( x) + f ¢( x)],
(- 1) ò 1
= log x× x+1 + x( x + 1) dx where f ( x) = 1
+
é 1 1 ù 1 x
êQ + 1)2 + Cú
ë ò ( x dx =- 1) + òWe know that, ex[ f ( x) + f ¢( x)]dx = ex f ( x) + C
(x û

- log x dx \ I = ex f(x) + C
x+1 x+
ò= + I1, = (say) …(i)
where I1 x( 1) 1
+
= ex × x) + C
(1
Now, using partial fraction method, consider
ex
1 = A + x B Þ I = (1 + x) + C
x( x + 1) x +1

Þ 1 = A( x + 1) + Bx

On putting x = 0, we get

6 CBSE Term II Applied Mathematics XII

Chapter
Practice

PART 1 ò6. The value of (3x 6 + 5x 4 + 1) d(x 2 ) is
Objective Questions
(a) 3x7 + x5 + x + C (b) 3x8 + 5x6 + x2 + C
7 43

(c) 3x8 + 5x6 + x2 + C (d) None of these
4
l Multiple Choice Questions é x ù
êe2 log x + çæ e ÷ö ú
ò1. The value of a 3loga x dx is ò7. The value of dx is
êë è 3 ø úû
(b) x4 + C æç e ö÷ x
(a) x4 + C 4 (a) x3 + 1 è 3ø + C (b) x3 + æçè e öø÷ x - log3) + C
(c) 4x4 + C 3 - log3 3 3
(d) a3loga x + C (1
loga x
æç e ö÷ x
2. The anti-derivative of çæ x - 1 ö÷ is equal to (c) x3 - è 3ø + C (d) None of these
è 2x ø 3 loge/ 3

(a) 2 x3/ 2 + 1 log|x| + C (b) 2 x3/ 2 + log x + C 8. The value of ò (2x + 1) (9x - 1) dx is
32 3
(a) 4 ( 9x - 1)5 / 2 - 22 ( 9x - 1)3/ 2 + C
(c) 2 x3/ 2 - log x + C (d) None of these 405 243
3
(b) 4 ( 9x - 1)5 / 2 + 22 ( 9x - 1)3/ 2 + C
3. The value of ò 5 x dx is 405 243
5x -
1 (c) 4 ( 9x - 1)5 / 2 + 22( 9x - 1)3/ 2 + C
45 27
(a) x - 1 log|5x - 1| + C (b) x + 1 log|5x - 1| + C
5 5 (d) None of the above

(c) x - log|5x - 1| + C (d) None of these 9. The anti-derivative of function f (x) = e3x is
(2 + e3x )4
ò4. The simplified form of (x3 - x2 + x - 1) dx is
(x - 1)
(a) 1 + C (b) - 1 + C
(b) x3 - x + C 9( 2 + e3x )3 9( 2 + e3x )3
(a) 3x3 + x + C 3
(c) x3 + x + C (c) ( 2 + e3x )3 + C (d) None of these
(d) None of these 9
3
ò10. The value of e3x - e-3x
ò5. The value of 3x + 5 x dx is e3x + e-3x dx is

7x (a) 1 log|e3x - e-3x| + C (b) 3log|e3x + e-3x| + C
3 (d) None of these
çæè 3 ÷öø x 1 çæè 5 öø÷ x 1
7 7 (c) 1 log|e3x + e-3x| + C
(a) ´ æçè 3 öø÷ + ´ çæè 5 ÷øö + C 3
7 7
loge loge

çæè 3 ÷øö x çæè 3 ÷øö æçè 5 ÷øö x èæç 5 ø÷ö 11. The value of ò 1 dx is
7 7 7 7 + log x)
(b) loge + loge + C x (5

(c) 3x + 5x + 7x + C (a) 1 log|5 + logx| + C (b) 5 log|5 + logx| + C
log3 log5 log7 5 (d) log|5 + logx| + C

(d) None of the above (c) 5 + logx + C

CBSE Term II Applied Mathematics XII 7

ò12. The value of 6x 2 + 4x dx is 18. The simplified value of ò 1 dx is

x3 + x2 +1 x 2 - 4x

(a) 4log|x3 + x2 + 1| + C (b) 2log|x3 + x2 + 1| + C (a) log|( x - 2) + x2 - 4x| + C

(c) 1 log|x3 + x2 + 1| + C (d) None of these (b) log|( x - 2) - x2 - 4x| + C
2 (c) 1 log|( x - 2) + x2 - 4x| + C

13. òThe value of x 3 + x + 1 dx is 2
(d) None of the above
x2 -1

(a) x2 - log|x2 - 1| + 1 log x -1 +C l Case Based MCQs
2 2 x +1
19. If f(x) is a continuously differentiable function,
(b) x2 + log|x2 - 1| + 1 log x -1 +C then to evaluate integrals of the form
2 2 x +1
ò f F(f(x))f¢ (x) dx, we substitute
x2
(c) 2 + log|x2 - 1| - 1 log x -1 +C f(x) = t and f¢ (x) dx = dt
2 x +1
This substitution reduces the above integral to
(d) None of the above
ò f (t) dt. Further evaluating this integral and
ò dx
14. The value of 1) 2x is substitute back the value of t.

( x - +3 Based on the above information, answer the

(a) log 2x + 3 - 5 +C (b) 1 log 2x + 3 - 5 +C following questions.
2x + 3 + 5 5 2x + 3 + 5
(i) The integral value of e3x + 5 is

(c) 5 log 2x + 3 - 5 + C (d) None of these (a) ( e3x + 5 )( 3x) + C (b) 3( e3x + 5 ) + C
2x + 3 + 5
(c) 1 ( e3x + 5 ) + C (d) None of these
3
1
15. The value of ò dx is ò(ii) x2 dx is equal to
7
x (x + 1) ex3

(a) 1 log x7 + 1 +C (b) 7log x7 +C (a) -3 + C (b) 1 +C
7 x7 x7 + 1 ex3 3e x3

(c) 1 log x7 +C (d) None of these (c) - 1 +C (d) None of these
7 x7 + 1 3e x3

16. The value of ò x 4 2x dx ò(iii) e4 x dx is equal to
+ 2x 2
-5 5 + e2x

(a) 1 log x2 + 1 + 6 +C (b) 1 log x2 +1- 6 +C (a) 1 [ e2x - 5 log|e2x + 5|] + C
26 x2 + 1 - 6 2 x2 +1+ 6 2

(b) e2x - 5 log|e2x + 5| + C

(c) 1 log x2 + 1 - 6 + C (d) None of these (c) 1 [ e2x + 5 log|e2x + 5|] + C
26 x2 + 1 + 6 2

ò dx (d) None of the above
8x -
17. The value of 2 is (iv) Evaluate ò 1 dx.
+
5 - x x x

(a) 1 log 21 +x+ 4 +C (a) log|1 - x | + C (b) 2log|1 + x | + C
2 21 21 -x+ 4
(c) 1 log|1 + x | + C
(b) 1 log 21 + x + 4 2 (d) None of these
2 21 21 - x - 4
+ C (v) Find the value of ò 5 x 3 + x2 dx

(c) 1 log 21 +x + 4 +C (a) 5 ( 3 + x2 )1/ 2 + C (b) 5 ( 3 + x2 )3/ 2 + C
21 21 -x - 4 3 3

(d) None of the above (c) 5 ( 3 + x2 )3/ 2 + C (d) None of these
6

8 CBSE Term II Applied Mathematics XII

20. The integral of the product of two functions ò3. Find æç ex loga + 1 - 1 + x5 ö÷ dx.
= (first function) ´ (integral of second function) è x4 x ø

- Integral of [(differential coefficient of first function) ò4. Evaluate x 4 + x 2 + 1 dx.
´ integral of second function)]
x2 - x +1
i.e. ò f (x) g(x) dx = f (x)ò g(x) dx
-ò [ f ¢ (x)ò g(x) dx] dx 5. Evaluate ò (4x + 2) x 2 + x + 1 dx.

Based on above information, answer the following ò6. Evaluate x 2 dx.
questions. (2 + 3x 3 )3

ò(i) The value of x2 ex dx is 7. If f (x) = x - 1 dx, then find the value of ò f (x) dx.

(a) ex( x2 - 2x + 2) + C (b) ex( x2 - 2) + C x2 -1
(c) ex( x - 2) + C (d) ex( x + 2) + C
8. Evaluate ò 1 dx.
(ii) The value of ò x log 2x dx is (x - 1) (x - 2)

(a) x2 logx - x2 + C (b) x2 log2x - x2 + C ò9. Evaluate 333x × 33x × 3x dx.
24 24

(c) x logx - x2 + C (d) None of these 10. Find the value of ò dx .
24 1 + 4x 2

(iii) The value of ò x2 log x dx is òEvaluate x 2 dx.

(a) x3 logx - x3 + C (b) x logx + C 11. 1- x6
39 3
12. Find the value of ò 1 dx.
(c) x3 logx + C (d) x logx + C
3 2 x 2 + 2x + 2

(iv) ò ex ( x + 1)dx is equal to 13. Evaluate ò (x 2 + 1) log x dx.

(a) ex + C (b) xe2x + C 14. òSimplify xex dx.
(c) xex + C (d) xe-x + C
(1 + x)2
(v) ò ex (log x)dx - ò ex dx is equal to
x 15. Solveò log5 x 3dx.

(a) ex æç 1 ö÷ + C (b) ex(logx) + C 16. Simplify ò x 2 + 4x + 6 dx.
èxø

(c) - ex + C (d) ex + C 17. Evaluate ò (x x - 2) dx.
x x - 1) (x

18. Evaluate ò dx .
(x2 + 1)
PART 2 x
Subjective Questions
19. Evaluate ò ex × x dx.

20. Evaluate ò log x dx.

l Short Answer Type Questions 21. òEvaluate e 2x dx.

1. Evaluate ò x 3 + 3x + 4 dx. 22. Evaluate ò 1 + x 2 dx.
9
x

2. Evaluate ò 1 dx. 23. Evaluate ò x 2 + 4x + 6 dx.
x+a + x+b

CBSE Term II Applied Mathematics XII 9

l Long Answer Type Questions òII. ex dx = ex + C

24. Evaluate the following : III. ò 1 dx = log|x| + C
x
ò ( x 2 + 2) ò(ii) e6 log x - e5 log x dx
(i) x+1 dx òIV. a x dx = a x log a + C
e4 log x - e 3 log x

ò25. Evaluate x 2e2x dx. On the basis of the above information, solve the
following questions.
ò26. Evaluate ex dx. (i) If f ( x) = x3 + 1, find the integral value of f ( x).
(1 + ex ) (2 + ex )
x

27. Find the value of ò 1 dx. ò(ii) Evaluate e3 log x dx.

x( x n + 1) é ù
ê3 ú
28. Evaluate ò x + 2 dx. ò(iii) Find x + 1 úû dx.
+ 2)
x 2 + 2x + 3 êë ( x 2

29. Solve ò x2 x+3 dx. 33. Suppose a function g(x) = 1 , where p(x) is a
- 2x - 5 p( x )

quadratic polynomial of the form ax 2 + bx + c.

30. Evaluate ò x dx. While solving such type of integrals, we use the
2(x
( x - 1) + 2) result such as

ò31. Evaluate x 2 - 1 dx. ò dx = 1 log x -a +C
2a x +a
x4 + x2 +1 (x2 - a2)

l Case Based Questions òand dx = 1 log a + x +C
a2 - x2 2a a - x

32. Suppose f (x) is continuous function and the On the basis of the above information, solve the

anti-derivative of f (x) is defined as F(x) + C. following questions

i.e. ò f (x) dx = F(x) + C, (i) Find the value of ò x dx.
x4 - 4

where C is a constant of integration. ò(ii) Evaluate the value of ex dx.

Some standard integrals are defined below. ( ex ) 2 - 16

òI. xn dx = xn +1 +C (iii) Solve ò dx .
n+1 3 - x2 + 2x

10 CBSE Term II Applied Mathematics XII

SOLUTIONS

Objective Questions ò6. (b) Let I = ( 3x6 + 5x4 + 1) d( x2 )

ò1. (b) Let I = a3loga x dx [Q aloga b = b] Consider x2 = t, then
ò ò= aloga x3 dx = x3dx
òI = ( 3t 3 + 5t 2 + 1) d( t )
= x4 + C òé n+ 1 ù
4 xndx = x + Cú = 3ò t 3dt + 5ò t 2d( t ) + ò1 dt
êQ n + 1
ë û = 3t4 + 5t3 + t + C
43
2. (c) The anti-derivative of èæç x - 1 øö÷ is
2x
= 3( x2 )4 + 5( x2 )3 + x2 + C
ò æèç x - 1 ÷öø dx = ò x dx - 1 ò 1 dx 43 [Q put t = x2 ]
2x 2 x
= 3 x8 + 5 x6 + x2 + C
1 +1 43
x2 1 log|x| +
= 1+ - 2 C é èæç e øö÷ x ù
ê e2 log x 3 ú
2 1 ò7. (a) Let I = êë + ûú dx

= 2 x3/ 2 - log x + C ò= é x2 æèç e ÷öø x ù
3 ê 3 ú
e log + ûú dx

3. (b) Let I = ò 5x dx ëê
(5x -
1) ò x2dx ò æçè e ÷øö x
3
= ò 5x - 1 +1 = + dx
(5x - 1)
dx

=ò çèæç1 + 1 1) ÷÷øö dx = x3 + æç e ÷ö x + C
(5x - 3 è 3ø
log e
1 dx 3
= ò1 dx + 5 ò -1
çèæ x ö÷ø æç e ÷ö x
5 = x3 + è 3ø + C
3 loge - log3
=x+ 1 log x-1 + C1
5 5 æç e ö÷ x
è 3ø
= x + 1 log|5 x - 1| - 1 log5 + C1 = x3 + - log3 + C
5 5 3
1
= x + 1 log|5x - 1| + C é 1 ù
5 êëQ C = - 5 log5 + C1 úû 8. (b) Let I = ò( 2x + 1) 9x - 1 dx

ò4. (c) Let I = x3 - x2 + x -1 dx = ò é 2( 9x - 1) + 11 ù 9x - 1 dx
(x - 1) ëê 9 9 ûú

= ò x2( x - 1) + (x - 1) dx = 2 ò( 9x - 1) 9x - 1 dx + ò 11 9x - 1 dx
(x - 1) 9 9

= ò( x2 + 1) dx ò ò= 2 ( 9x - 1)3/ 2dx + 11 ( 9x - 1)1/ 2 dx

99

= ò x2dx + ò1 dx = 2 ( 9x -1)5 / 2 + 11 ( 9x - 1)3/ 2 +C
9 5/ 2´9 9 3/ 2´9
= x3 + x + c
3 òé ( ax + b ) n dx = ( ax + b) n + 1 + ù
(n + 1)a Cú
3x + 5x êQ û
ò5. (a) Let I = 7x dx ë

ò= é æèç 3 öø÷ x çæè 5 ö÷ø x ù = 4 ( 9x - 1)5 / 2 + 22 ( 9x - 1)3/ 2 + C
ê 7 7 ú 405 243
êë + ûú
dx 9. (b) The anti-derivative of function is

çæ 3÷ö x æç 5 ÷ö x ò òf ( x) dx = e 3x dx
( 2 + e3x )4
òé ax ù
= è 7ø + è 7ø + C axdx = log a + Cú Put 2 + e3x = t
êQ Þ 3e3x dx = dt
loge æçè 3 öø÷ loge çæè 5 ÷öø ë û
7 7

CBSE Term II Applied Mathematics XII 11

\ I = 1 ò dt = 1 ò t -4dt 14. (b) Let I = ò dx +
3 t4 3 - 1) 2x
(x 3

= 1 t -4 + 1 + C Now, put 2x + 3 = t
3 -4+1
Þ 2x + 3 = t 2 Þ x = t 2 - 3
= 1 ´ 1 3) + C 2
3 t 3( -
Then, dx = ( 2t - 0) dt = t dt
= - 1 + C [Q put t = 2 + e3x] 2
+ e3x )3
9( 2 \ I=ò t dt = ò dt
-3 - t2 -3-2
ò10. (c) Let I = e 3x - e -3x dx èæçç t2 2 1 ø÷÷ö × t
e 3x + e -3x 2

Put e3x + e-3x = t Þ ( 3e3x - 3e-3x ) dx = dt = 2ò dt = 2ò dt
t2 - t2 -( 5 )2
Þ ( e3x - e-3x ) dx = dt 5
3
= 2 ´ 1 log t- 5 +C
\ = ò 1 ´ dt 25 t+ 5
I t 3

= 1 log|t| + C òé dx = 1 log x-a + ù
3 x2 - a2 2a x+a C úû
ëêQ

= 1 log|e3x + e-3x| + C [Q put t = e3x + e-3x] = 1 log 2x + 3 - 5 +C [Q t = 2x + 3]
3 5 2x + 3 + 5

11. (d) Let I = ò 1 dx 15. (c) Let I = ò 1 dx
+ logx) x( x7 +
x(5 1)

Put 5 + logx = t Þ 1 dx = dt ò= 1 ´ x6 dx
x x( x7 + 1) x6

\ I = ò 1 dt = log|t| + C ò= x6
t x7(x7 +
dx
= log|5 + logx| + C [Q put t = 5 + logx] 1)

ò12. (b) Let I = 6x2 + 4x dx Put x7 = t Þ 7x6dx = dt
x3 + x2 +1
\ I = ò dt = 1 ò dt
Put x3 + x2 + 1 = t Þ ( 3x2 + 2x) dx = dt 7t( t + 1) 7 t2 + t

\ I = ò 2 dt = 2 log|t| + C = 1 ò dt 1 1
t 7 t+
t2 + -
= 2log|x3 + x2 + 1| + C [Q put t = x3 + x2 + 1] 44

x3 + x+ 1 ò= 1 dt
x2 -1
ò13. (b) Let I = dx 7 æçè t 1 ÷øö 2 çæè 1 ö÷ø 2
2 2
+ -

= ò é x + 2x + 1 ù dx = òx dx + ò 2x +1 dx t+1 -1
ê x2 - 1 ú x2 -1 22
ë û 1 1
= 7 ´ ´ log t+1+1 +C
òx ò 2x ò 1 1
= dx + x2 - 1 dx + x2 - 1 dx 2
2 22

= x2 + log|x2 - 1| + 1 log x - 1 +C òé x2 1 dx = 1 log x-a + ù
2 2 ´1 x + 1 - a2 2a x+a Cú
êQ û
ë

ò[Q integral 2x dx, put x2 -1 = tÞ 2x dx = dt = 1 log t +C
x2 - 1 7 +1
t

\ ò dt = log|t| + C = log|x2 - 1| + C = 1 log x7 +C [Q put t = x7 ]
t 7 x7 + 1

òand dx = 1 log x-a + C ù ò16. (c) Let I = 2x dx
x2 - a2 2a x+a ûú 2x2
x4 + -5
= x2 1 log x -1
2 + log|x2 - 1| + 2 x+1 +C Put x2 = t Þ 2x dx = dt

12 CBSE Term II Applied Mathematics XII

\ = ò dt e4x e2x × e2x
+ 2t + e2x 5 + e2x
I t2 -5 ò ò(iii) (a) Let I = dx = dx

5

= ò t2 + dt - - Substitute e2x = t Þ 2e2xdx = dt
+1
2t 5 1 \ = ò t
+
ò= dt I (5 t )2 dt
(t + 1)2 - (
6)2 = 1 ò t+ 5 -5
2 (t + 5)
1 t + 1 - 6 +C dt
6) t + 1 + 6
= 2( log ò æèçç1 ø÷÷ö

òé 1 1 log x- a ù = 1 - t 5 5 dt
- a2 2a x+ a Cú 2 +
êQ dx = +
ë x2 û = 1 [ t - 5 log|t + 5|] + C
2
= 1 x2 +1 - 6 +C [Q put t = x2]
26 log x2 +1 + 6 = 1 [ e2x - 5 log|e2x + 5|] + C [Q put t = e2x]
2
dx dx
- 8x - x2 5 - 2× 4× x - x2 - ( 4)2 + ( 4)2 1 1
ò ò17. (b) Let I =5 = (iv) (b) Let I = ò + x dx = ò x + 1) dx

x x(

ò= dx Substitute 1 + x = t Þ 0 + 1 dx = dt
2x
5 + 16 - [ x2 + ( 4)2 + 2× 4× x]

= ò 21 - dx 4)2 \ I = ò 2dt = 2 log|t|
(x + t

ò= dx = 2log|1 + x| + C

( 21 )2 - ( x + 4)2 (v) (b) Let I = ò5x 3 + x2 dx

= 1 log 21 + x + 4 +C Substitute 3 + x2 = t 2 Þ 0 + 2x dx = 2t dt
2 21 21 - x - 4 Þ x dx = t dt

òé dx = 1 log a+ x + ù \ I = ò5 ´ t ´ òt 2 dt = 5 t 2dt
a2 - x2 2a a-x Cú
êQ û
ë = 5t 3 + C = 5 ( 3 + x2 )3/ 2 + C
33
dx = dx
x2 - 4x x2 - 2× 2× x + 22 - 22 ò20. (i) (a) Let I = x2exdx
ò ò18. (a) Let I =

ò= dx On taking x2 as first function and ex as second function

( x - 2)2 - 22 and integrating by parts, we get

= log|( x - 2) + ( x - 2)2 - 22| + C I = x2 ò exdx - ò é d (x2) ò e xdx ù dx
êë dx úû
= log|( x - 2) + x2 - 4x| + C
ò= x2ex - ( 2xex ) dx
ò19. (i) (c) Let I = ( e3x + 5) dx
Again, integrating by parts, we get
Substitute 3x + 5 = t Þ 3dx = dt
x2ex ìíî2x ò exdx 2ò é d x)ò xdx ù dxþýü
ò\ I = et dt = 1 et + C I = - - ëê dx ( e ûú

33 ò= x2ex - 2xex + 2 exdx

= 1 e(3x + 5 ) + C [Q put t = 3x + 5] = x2ex - 2xex + 2ex + C
3 [Q put t = x3] Þ I = ex( x2 - 2x + 2) + C

ò(ii) (c) Let I = x2 dx
ex3
(ii) (b) On taking log 2x as first function and x as second
Substitute x3 = t Þ 3x2dx = dt
function and integrating by parts, we get
\ I = ò dt = ò1 e-tdt
3et I = ò x log2x dx
3
é d (log x dxùúû
= 1 ( - e-t) + C = log 2xòx dx - ò êë dx 2x) ò dx
3
é ù
= - 1 e-x3 + C ò= x2 log2x - ê 1 ´ 2 ´ x2 dxú
3 ë 2x 2
2 û

=- 1 +C ò= x2 log2x - 1 x dx Þ I = x2 log 2x - 1 x2 + C
3e x3
22 24

CBSE Term II Applied Mathematics XII 13

(iii) (a) On taking logx as first function and x2 as second ò= 1 {( x + a)1/ 2 - ( x + b)1/ 2}dx
-
function and integrating by parts, we get a b

I = ò x2 logx dx ìü
ïí( x + a)(1/ 2) +1 + b)(1/ 2) + 1 ï
1 bï 1 +1 (x 1 +1 ý
éd ù = - 2 - 2 + C
= log x ò x 2 dx - ò ëê dx (log x ) ò x 2 dx ûú dx a î
ï
þ
ò= x3 logx - é x3 ù
ê 1 × 3 ú dx = 1 ìí( x + a)3/ 2 - (x + b)3/ 2 ü +
3 ë x û a- bî 3/ 2 3/ 2 ý
þ C

ò= x3 logx - 1 x2dx = x3 logx - x3 + C = 2 [( x + a)3/ 2 -(x + b)3/ 2 ] + C
33 39 3( a - b)

(iv) (c) Let I = ò ex( x + 1) dx ò3. Let I = çæ ex log a + 1 -1 + x5 ö÷ dx
II I è x4 x ø

Using integration by parts, we get çæ elog ax 1 -1 x5 ÷ö
è x4 x ø
= (x + 1) ò exdx - ò é d (x + 1) ò e xdx ù dx ò= + + dx
êë dx ûú
çæ ax 1 -1 x5 ö÷
ò= ( x + 1) ex - (1 + 0)ex dx ò= è + x4 x + ø dx

ò= ( x + 1)ex - ex dx = ax loga + x-4 + 1 - log|x|+ x6 + C
-4 + 1 6
= ( x + 1)ex - ex + C = xex + C

ò ò(v) (b) Let I = ex( logx dx) + ex dx x = ax loga - 1 - log|x|+ x6 +C
3x3 6
II I

Using integration by parts, we get ò4. Let I = x4 + x2 + 1 dx
x2 -x + 1
= log x ò exdx - ò é d (log x ) ò e x dxúùû dx +ò ex dx
ëê dx x
ò= (x2 )2 + 2x2 + (1)2 - x2 dx
(x2 - x + 1)
= (logx)ex - ò 1 ex dx + ò ex dx
x x (x2 + 1)2 - ( x2 )
ò= ( x2 -x + 1) dx
= (logx)ex + C

Subjective Questions ò= ( x2 + 1- x)( x2 +1 + x) dx
( x2 -x+ 1)
1. Let I = ò x3 + 3x + 4 dx = ò x3 dx + 3ò x dx + 4 ò 1 dx
x x x x
= ò( x2 + 1 + x) dx
3 ´ x-1/ 2 x-1/ 2 dx
ò ò ò=
x dx + 3 x1/ 2 dx + 4 = x3 + x + x2 + C
32
ò ò ò= x(6-1)/ 2 dx + 3 x1/ 2 dx + 4 x-1/ 2 dx
5. Let I = ò ( 4x + 2) x2 + x + 1 dx.
ò ò ò= x5 / 2 dx + 3 x1/ 2 dx + 4 x-1/ 2 dx
Let x2 + x + 1 = t
x(5 / 2) + 1 3x(1/ 2)+1 4x(-1/ 2) + 1
= (5 / 2) + 1 + (1 / 2) + 1 + ( -1 / 2) + 1 + C On differentiating both sides w.r.t. x, we get

òé xn dx = xn +1 + ù 2x + 1 = dt Þ dx = dt 1)
n+1 Cú dx ( 2x +
êQ
ë û \ I = ò( 4x + 2) t dt
( 2x + 1)
= x7/ 2 + 3x3/ 2 + 4x1/ 2 + C
7/ 2 3/ 2 1/ 2 = ò2( 2x + 1) t dt = 2ò t dt
( 2x +
= 2 x7/ 2 + 2 × x3/ 2 + 8x1/ 2 + C 1)
7
= t (1/ 2) +1 + = 4( x2 + x+ 1 )3/ 2 +
2. Let I = ò 1 x + b dx 2 ( 1 / 2) 1 C 3 C
x+a+ +

=ò 1 x+b ´ x+a- x + b dx ò6. Let I = ( 2 x2 3 )3 dx
x+a+ x+a- x + b + 3x

x+ a - x+b Let 2 + 3x3 = t
x+ a) - x + b)
= ò ( ( dx Þ 9x2 = dt Þ dx = dt
dx 9x2

14 CBSE Term II Applied Mathematics XII

x2 x2 dt = 1 1 = log x - 3 + çæ x - 3÷ö 2 - 1 + C
( 2+ 3x3)3 t 3 9x2 9 t3 2 è 2ø 4
ò ò ò\ dx = dt

ò= 1 t - 3dt = 1 çæèç t - 3+1 ÷öø÷ + C = log æç x - 3ö÷ + x2 - 3x + 2 + C
9 -3 + 1 è 2ø
9

=1 ´ 1 t - 2 +C = -1 +C ò9. Let I = 333x × 33x × 3x dx
9 -2 18t 2

= -1 +C Put 333x = t Þ 333x log3 d æçè 33x ÷øö = dt
+ 3x3)2 dx
18( 2
Þ 333x (log3) ´ 33x (log3) ´ 3x(log3)dx = dt
7. Let I = ò x - 1 dx
x2 - 1 Þ 333x 33x 3x (log 3)3 dx = dt

x dx - 1
x2 - 1 x2 - 1
ò ò= dx = I1 - I2 ...(i) Þ 333x 33x 3x dx = dt
(log 3) 3
x dx,
òNow, I1 = x2 - 1 ò\ I= dt = t + C
(log 3) 3 (log 3) 3

Let x2 - 1 = t Þ 2x = dt Þ dx = dt = 333x + C é put t = 333x ù
dx 2x (log 3) 3 ëêQ úû

-1 é 1ù
x dt = 1 dt = 1 1 ê t2 ú
t 2x 2 t2 t 2 dt 2
ò ò ò\ I1 = = ê ú + C1 10. Let I = ò dx =ò dx
êë ûú 1 + 4x2
1 / 2 çæèç æèç 1 ÷øö 2 x2 öø÷÷
2
4 +

= t + C1 = x2 - 1 + C1 [Q t = x2 - 1]

Now, òI2 = 1 dx 1 ò dx = 1 log x + çèæ 1 öø÷ 2
x2 -1 2 x2 + æç 1 ö÷ 2 2 2
= x2 + +C
è 2ø
= log|x + x2 -1|+ C2

òé dx = log|x + ù òé 1 ù
x2 - a2 x2 - a2|+ Cú dx = log|x + x2 + a2|+ Cú
êêëQ êêëQ x2 + a2 ûú
úû

On putting the values of I1 and I2 in Eq. (i), we get = 1 log x + 4x2 + 1 + C
ò x -1 dx = x2 -1 - log|x + x2 -1|+ C 22

x2 -1
[where, C =C1 - C2]
= 1 log|2x + 4x2 + 1| - 1 log 2 + C
8. Let I = ò 1 (x- dx = ò 1 dx 22
- 1) x2 - 3x + 2
(x 2) é log çèæ m ø÷ö = log m - log nûúù
êëQ n
ò= 1 dx
çèæ 3 ÷öø 2 æçè 3 ÷øö 2 = 1 log|2x + 4x2 + 1| + C
2 2 2
x2 - 3x + - + 2

=ò 1 dx = ò 1 dx [Q 1 log 2 is constant and consider as C]
2
èçæ x 3 ö÷ø 2 - 9+ 8 æçè x 3 ø÷ö 2 çèæ 1 ö÷ø 2
2 4 2 2 x2 x2
- + - - ò ò11. Let I = - x6 dx = - (x3)2 dx

Let x - 3 = t Þ dx = dt 1 1
2
Let x3 = t Þ 3x2 = dt Þ dx = dt
dx 3x2
ò\ I = 1 dt
x2 dt = 1 dt
1 - t 2 3x2 3 1 - t2
t 2 - æç 1 ÷ö 2 ò ò\ I =

è 2ø 1+t é a+ x ù
1-t êQ a-x ú
= log t + t2 - 1 + C ò= 1 × 1 log + C ë dx = 1 log + C û
4 a2 - x2 2a
32

òé dx = log|x + ù = 1 log 1 + x3 +C [Q t = x3]
x2 - a2 x2 - a2|+ Cú 6 1 - x3
êêëQ
ûú

CBSE Term II Applied Mathematics XII 15

12. Let I = ò 1 dx = ò 1 dx 16. Let I = ò x2 + 4x + 6 dx = ò x2 + 4x + 22 + 6 - 4 dx
x2 + 2x + 2 ( x2 + 2x + 1) + 1
ò= ( x + 2)2 + ( 2)2 dx
=ò 1 dx
(x+ 1)2 + 1 = x + 2 x2 + 4x + 6 + 2 log|( x + 2) + x2 + 4x + 6 | + C
22
Let x + 1 = t Þ dx = dt
òé x2 + a2 dx = x x2 + a2 + a2 log|x + x2 a2|+ ù
\ I=ò 1 dt = log|t + t 2 + 1| + C 2 2 + Cú
t2 + 1 êQ û
ë

òé dx ù Þ I = x + 2 x2 + 4x + 6 + log|( x + 2) + x2 + 4x + 6| + C
= log|x + x2 + a2|+ Cú 2
ëêêQ
x2 + a2 ûú

= log|( x + 1) + ( x + 1)2 + 1| + C [Q t = x + 1] 17. Let I = ò - x - dx
1) (
( x x 2)

= log|( x + 1) + x2 + 2x + 2| + C By using partial fraction,

13. Let I = ò( x2 + 1) logx dx Let (x x - 2) = (x A + (x B 2)
- 1) (x - 1) -
On taking log x as first function and ( x2 + 1) as second
Þ x = A( x - 2) + B( x - 1)
function and integrating by parts, we get
Putting x = 1, we get
logxò( x2 ò é d ò ù
I = + 1 ) dx - êë dx (log x) ( x 2 + 1) dx úû dx A = -1

èæçç x3 x÷÷øö 1 æççè x3 xø÷ö÷ dx and putting x = 2, we get
3 x 3
òÞ I = log x + - × + B=2

èæçç x3 xø÷ö÷ logx èççæ x2 + 1ö÷÷ø dx Thus, ò (x x - 2) = ò -1 dx + 2ò dx
3 3 - 1) (x x -1 x-2
òÞ I = + -
= - log|x - 1| + 2log|x - 2| + C

= çèæç x3 + x÷÷öø log x - x3 - x + C = log ( x - 2)2 +C
3 9 x -1

xex ( x + 1) -1 18. Let I = ò dx
(1 + x)2 (1 + x )2 x(x2 +
ò ò14. Let I = dx = e x dx 1)

ò= é 1 1 ù By using partial fraction,
êë + + x)2 úû
e x - dx 1 = A + Bx + C
(1 x( x2 + x x2 + 1
( 1 x) Let 1)

Let f ( x) = 1 1 x Þ 1 = A( x2 + 1) + ( Bx + C)x
+
Þ A + B = 0, C = 0 and A = 1
Þ f ¢( x) = - 1
(1 + x)2 On solving these equations, we get

We know that, A = 1, B = - 1 and C = 0
\ 1 =1 + -x
ò ex[ f ( x) + f ¢ ( x)] dx = ex f ( x) + C
x( x2 + 1) x x2 + 1

òÞ I= exíîì1 1 x - (1 1 ü = ex + C Þ ò 1 dx = ò ìíî1x - x ü
+ + x)2 ý dx 1+x x( x2 + + dxý
þ 1) x2 1
þ
log x 3
15. Let I = ò log5 x3 dx = ò log 5 dx = log|x| - 1 log ( x2 + 1) + C
2

ò= 3 1 × logx dx 19. Let I = ò ex× x dx
II I
log 5 II I
Using integration by parts
Using integration by parts, we get
I = xò exdx - òíîìddx ( x)üýþ ´ ò exdx
= 3 5 é log xò 1 dx - ò îíì d (log x) ò1 dxüþý dxúûù
log ëê dx
ò= xex - 1 ´ exdx
= 3 éêë(log x ) - ò 1 ´ dxûùú
log 5 x x x ò= xex - exdx

= 3 5 [x log x - ò1 dx] = 3 5 [ x logx - x] + C = xex - ex + C
log log = (x - 1) ex + C

16 CBSE Term II Applied Mathematics XII

20. Let I = ò 1× logx dx ò(ii) Let I = çæçè e6 log x - e5 log x ÷÷öø dx
II I e4 log x - e3log x

Using integration by parts, we get çæ elog x6 ÷ö
èç elog x4 ÷ø
I = (log x) ò1 dx - ò d (log x ) ò 1 dx ò= - e log x5 dx [Q a logb = logba]
dx - e log x3

= (log x ) x - ò 1 ´ x dx çèæç x6 - x5 øö÷÷
x x4 - x3
ò= dx [Q elog x = x]
= x(logx) - ò1 dx

= x logx - x + C = ò èçæç x3 - x2 ÷øö÷ dx = ò x2(x - 1) dx
x - 1 x -1
ò21. Let I = e 2x dx

Put 2x = t 2 Þ 2dx = 2t dt Þ dx = t dt ò= x2dx = x3 + C

ò ò\ I = e t2 t dt = t× et dt 3
I II
ò25. Let I = x2 e2x dx
Using integration by parts, we get I II

= t ò etdt - ò d ò tdt Using integration by parts, we get
dt
I ( t ) e ò ò ò= x2 é d ù
e2x dx - êë dx x2 2 xdx ûú
( ) e dx
ò= tet - 1 ´ etdt
ò= x2 e2x - 2x ´ e2x dx
= tet - et + C
22
= et( t - 1) + C
ò= x2 e2x - x e2x dx
= e 2x( 2x - 1) + C [Q put t = 2x]
2 I II

22. Let I = ò 1+ x2 dx =ò 9 + x2 dx = 1 ò 32 + x2 dx Again, using integration by parts, we get
9 3 3
ò ò òIx2 é æçè d e2x dx÷öø dxùúû
1 é x 32 + x2 + 32 log|x + 32 + x2|ùú + C = 2 e2x - êë x e2xdx - dx ( x )
3 ê 2 2 û
= ë x2 é e2x e2x ù
2 ê 2 2 dxú
òé a2 + x2 + a2 log|x + ù ò= e2x - ë x - 1 ´
2 Cú û
êQ a2 + x2 dx = x x2 + a2| +
ë 2 û = x2 e2x - x e2x + e2x + C
2 24
= x 9 + x2 + 3 log|x + 9 + x2| + C
62 ò26. Let I = ex dx
ex) (2
23. Let I = ò x2 + 4x + 6 dx = ò x2 + 4x + 22 + 6 - 4 dx (1 + + ex)

ò= ( x + 2)2 + ( 2)2 dx Put ex = t Þ exdx = dt

= ( x + 2) x2 + 4x + 6 + 2 log|( x + 2) \ I = ò (1 + ex t) dt = ò (1 + 1 dt
22 t) (2 + ex t) (2 + t)

+ x2 + 4x + 6| + C Let 1 = (1 A t) + B
(1 + t) (2 + t) + (2 + t)

òé x2 + a2 dx = x x2 + a2 + a2 log|x + x2 + a2| + ù Þ 1 = A( 2 + t ) + B(1 + t ) = ( 2A + B) + t( A + B)
2 2 Cú
êëQ û On equating the coefficients of t and constant term both
sides, we get
Þ I = x + 2 x2 + 4x + 6
2 A + B = 0 and 2A + B = 1

+ log|( x + 2) + x2 + 4x + 6|+ C On solving both equations, we get

x2 +2 A = 1 and B = - 1
x +1
24. (i) Let I = ò dx \ I = ò (1 1 dt - ò(2 1 dt
+ +
t) t)

= ò æççè x -1 + 3 1 öø÷÷ dx = log|1 + t| - log|2 + t| + C
+
x = log 1 + t +C
2 + t
= ò( x - 1) dx + 3ò 1 dx
x + 1
1 + ex
= x2 - x + 3log|( x + 1)| + C = log 2 + ex +C
2

CBSE Term II Applied Mathematics XII 17

ò ò27. Let I = 1 dx = x n -1 dx Let x2 + 2x + 3 = t
xn + xn( xn +
x( 1) 1) Þ ( 2x + 2) dx = dt

Put xn = t Þ nxn-1dx = dt ò ò\ I1 = dt = t -1/ 2dt = t (-1/ 2) +1 + C1
t -1 +1
Þ xn -1 dx = 1 dt
n 2

x n -1 1 1 = 2 t + C1 = 2 x2 + 2x + 3 + C1
xn( xn + n +
ò ò\ I= dx = t( t 1) dt ...(i) dx
1) x2 + 2x + 3
òand I2 =
1 = A + B
Now, t(t + 1) t (t + 1)

Þ 1 = A(1 + t ) + Bt …(ii) ò= dx

On substituting t = 0, -1 in Eq. (ii), we get A = 1 and B = - 1 x2 + 2x + (1)2 + 3 - (1)2

\ 1 1) = 1 - (t 1 1) ò= dx
t(t + t +
(x + 1)2 + ( 2)2

\ = 1 ò ççæè 1 - 1 1) ÷øö÷ dt Let x + 1 = t Þ dx = dt
n t +
I (t [from Eq. (i)] òI2 = dt = log|t + t 2 + 2| + C2
t2 + ( 2)2
= 1 [log|t| - log|t + 1|] + C
n òé dx ù
= log|x + x2 + a2|+ Cú
= 1 [log|xn| - log|xn + 1|] + C [Q put t = xn] ëêêQ x2 + a2 ûú
n

= 1 log xn +C = log|x + 1 + ( x + 1)2 + 2| + C2
n xn+ 1
= log|x + 1 + x2 + 2x + 3| + C2

28. Let x + 2 = A d ( x2 + 2x + 3) + B On putting the values of I1 and I2 in Eq. (i), we get
dx ò x + 2 dx = 1 [ 2 x2 + 2x + 3] + log|x + 1
x2 + 2x + 3 2
Þ x + 2 = A( 2x + 2) + B

Þ x + 2 = 2Ax + ( 2A + B) + x2 + 2x + 3| + C é 1 + =C ù
ëê 2 ûú
On equating the coefficient of x and constant term both C1 C2
sides, we get
= x2 + 2x + 3 + log|x + 1 + x2 + 2x + 3| + C
2A = 1Þ A = 1
2 29. Let ( x + 3) = A d ( x2 - 2x - 5) + B
dx
and 2A + B = 2
Þ ( x + 3) = A( 2x - 2) + B Þ x + 3 = 2Ax - 2A + B
Þ 2´1 + B=2
2 On equating the coefficients of x and constant term both

Þ B=2-1 =1 sides, we get

Þ x + 2 = 1 ( 2x + 2) +1 2A = 1Þ A = 1
2 2

òx + 2 dx = 1 ( 2x + 2) + 1 and -2A + B = 3Þ B = 4
2 dx
\ò x2 + 2x + 3 Þ ( x + 3) = 1 ( 2x - 2) + 4
x2 + 2x + 3 2

= 1 ò 2x + 2 dx + ò dx \ ò x+3 dx = ò 1 ( 2x - 2) + 4 dx
2 x2 + 2x + x2 + 2x + 3 - 2x - 5 2
3 x2 x2 - 2x - 5

òLet I1 = 2x + 2 dx = 1 ò ( 2x - 2) dx + 4 ò x2 1 -5 dx
x2 + 2x + 3 2 x2 - 2x - 5 - 2x

òand I2 = dx òLet I1 = 2x - 2 dx
x2 + 2x + 3 x2 - 2x - 5

òThen, x + 2 dx = 1 I1 + I2 …(i) òand I2 = x2 1 -5 dx
x2 + 2x + 3 2 - 2x

òNow, I1 = 2x + 2 dx ò\ x2 x+3 dx = 1 I1 + 4I2 …(i)
x2 + 2x + 3 - 2x - 5 2

18 CBSE Term II Applied Mathematics XII

òNow, I1 = 2x - 2 dx = 2 log|x - 1| + 1 ( x - 1)-2 +1 - 2 log|x + 2| + C
x2 - 2x - 5 9 3 (-2 + 1) 9

Let x2 - 2x - 5 = t = 2 log x -1 - 1 ççæè 1 1 øö÷÷ + C
9 x+2 3 -
Þ ( 2x - 2) dx = dt x

ò\ I1 = dt = log|t|+ C1 = log|x2 - 2x - 5|+ C1 …(ii) é log b - log a = logèæç b ö÷ø ù
t ëêQ a ûú

òand I2 = x2 - 1 -5 dx ò31. Let I = x2 - 1 dx
2x + x2 +
x4 1
= ò 1
( x2 - 2x + 1) - 6 dx 1 - 1
x2
ò= 1 òÞ I = dx
(x - 1)2 - ( 6)2 dx 1
x2 + 1 + x2

= 1 log x -1 - 6 + C2 …(iii) [dividing numerator and denominator by x2]
26 x -1 + 6
æçè1 - 1 ö÷ø çæè1 - 1 ÷öø dx
òé ù x2 x2
= 1 log x-a ú ò ò=
êQ dx 2a x+a + C èçæ x2 1 ÷öø dx = çèæ x2 1 2øö÷
ë x2 - a2 x2 x2
û + + 1 + + +1 -2

On substituting the values I1 and I2 from Eqs. (ii) and (iii) [adding and subtracting 2 from denominator]
in Eq. (i), we get
æçè1 1 ö÷ø
ò x+3 dx - x2 dx
- 2x - 5
x2 ò= æç x + 1 ÷ö 2 - 1

= 1 log|x2 - 2x - 5|+ 4 é 1 log x -1 - 6 ù è xø
ê x -1 + ú +C
2 êë 2 6 6 ûú
Now, put x + 1 = tÞ èæç1 - 1 ÷øö dx = dt
é 1 + =C ù x x2
ëê 2 C1 4C2 ûú
t -1
= 1 log|x2 - 2x - 5| + 2 log x -1 - 6 +C ò\ I= dt =1 log t +1 +C
2 6 x -1 + 6 t2 - 12 2

30. Let x + = A + B + C …(i) òé dx = 1 log x-a + Cùûú
(x - 1)2 (x -1 - 1)2 x+ x2 - a2 2a x+a
êëQ
2) x (x 2
x + 1 -1
Þ x = A( x - 1) ( x + 2) + B( x + 2) + C ( x - 1)2 …(ii) Þ I = 1 log x + C = 1 log x2 + 1 - x +C
2 2 x2 + 1 + x
On substituting x = 1, - 2 in Eq. (ii), we get x+1 +1

1 = B(1 + 2) and - 2 = C ( -2 - 1)2 x

Þ B = 1 and C = - 2 é t = x + 1ù
39 ëêQ x ûú

On equating the coefficient of x2 both sides in Eq. (ii), we 32. (i) We have, f ( x) = x3 + 1
x
get
ò èçæ x3 öø÷
0= A+C \ ò f ( x) dx = + 1 dx
x
Þ A=-C = 2
9 = x3 +1 + log|x| + C
3+1
ì 2 1 èçæ - 2 ø÷ö ü
ïíï 9 ïïý
\ ò x dx = ò ï 9 + 3 + ï dx = x4 + log|x| + C
( x - 1 )2 ( x + 2) x -1 (x - 1)2 x+2 4

îï ïþ ò ò(ii) Let I = e3log x dx = elog x3 dx

[from Eq. (i)]

Þ ò(x - x + 2) dx = ò x3dx
1)2( x
= x4 + C
= 2 ò 1 1 dx + 1 ò ( x 1 )2 dx - 2 ò x 1 dx 4
9 x- 3 -1 9 + 2

CBSE Term II Applied Mathematics XII 19

ò(iii) Let I = é 3x + (x 1 ù dx ò(ii) Let I = ( ex dx
ê + 2)2 ú ex )2 - 16
ë û

= ò 3x dx + ò( x + 2)- 2 dx Put ex = t

( x + 2)-2 + 1 Þ exdx = dt
-2+1
= 3x log3 + + C ò\ I= dt
t 2 - ( 4)2
1
= 3x log 3 - (x + 2) + C = 1 log t - 4 +
2´4 t + 4
C

33. (i) Let I = ò x4 x 4 dx 1 log ex - 4
- 8 ex + 4
= + C [Q t = ex]

Put x2 = t

Þ 2x dx = dt (iii) Let I = ò 3 - dx 2x = ò 3 -( x2 dx
x2 + - 2x + 1) + 1
ò\ I=1 dt
2 t 2 - 22 ò= dx

1 1 t - 2 ( 2)2 - ( x - 1)2
2 2´2 t + 2
= ´ log +C = 1 log 2 + ( x - 1) +C
´ 2 - ( x - 1)
x2 - 2 2
x2 +
= 1 log 2 + C [Q t = x2] = 1 log x+1 +C
8 2 4 3-x

Chapter Test Then, the given integral becomes standard integral

ò f (t) dt. ex
ex +
ò(i) The value of 1 dx is

Multiple Choice Questions (a) log x + C (b) log| ex + 1| + C
1 (d) None of these
1. The integral of the function f (x) = e3log x + ex + x4 is (c) - (ex + 1) 2 + C

(a) x 4 + ex + x5 + C (b) - x 4 + ex + x5 + C (ii) ò g¢( x ) dx is equal to
45 45 g( x )

(c) 3 + ex + x5 + C (d) None of these (a) log g(x ) + C (b) g2(x ) + C
x5 2

2. Evaluate ò 7 dx. (c) g(x ) + C (d) None of these
7x - 9
ò(iii) The value of log x2 dx is
(b) 7 7x - 9 + C eloge x
(a) 14 (7x - 9 )3/ 2 + C 2 (a) log x 2 + C
(c) 7 7x - 9 + C 2 (b) (log(x ))2 + C
(d) 14 7x - 9 + C

3. The simplified value of ò 1 5x - 1 dx is (c ) 1 (log(x ))2 + C (d) None of these
5x + 4 - 2

(a) 2 [(5x + 4)3/ 2 + (5x - 1)3/ 2] + C ò(iv) If 3x2 + 7 4 dx = log | x3 + 7x + k| + C, then the
15 x3 + 7x +

(b) 2 [(5x + 4)3/ 2 + (5x - 1)3/ 2] + C value of k is
75
(a) 2 (b) 3
(c) 2 [(5x + 4)3/ 2 - (5x - 1)3/ 2] + C
75 (c) 4 (d) 7

(d) None of these (v) The value of ò ( x + 1)2 dx is
x
1 dx is
ò4. The value of a2 + b2x2 (a) 3( x + 1)3 + C (b) 3 ( x + 1)3 + C
2
(c) 2 ( x + 1)3 + C
(a) blog| bx + a2 + b2x 2| + C (b) log| bx - a2 + b2x 2| + C 3 (d) None of these

(c) log| bx + a2 + b2x 2| + C (d) 1 log| bx + a2 + b2x 2| + C Short Answer Type Questions
b
7. Evaluate ò x x + 2 dx.
5. The value of ò x3 dx is
- 1) (x -
(x 2) ò8. Prove that ex dx = - 1 + C.
+ ex )2 (1 + ex )
(a) x - log| x - 1| - 4 log| x - 2| + C (1
(b) x + log| x - 1| + 4 log| x - 2| + C
(c) x - log| x - 1| + 4 log| x - 2| + C ò9. Evaluate 33x × 3x dx.

(d) None of the above 10. òFind the value of log10x dx.

Case Based MCQs Long Answer Type Question

6. The method of reducing a given integral into one of 11. Solve ò x2 + 4x + 13 dx.
the standard integrals by a proper substitution is said
to be the method of substitution. 12. Evaluate ò (x - x - dx.
-
Suppose, we have ò f (g(x))g¢(x)dx 1) (x 2) (x 3)

Substitute g(x) = t Þ g¢(x)dx = dt

Answers

1. (a) 2. (d) 3. (b) 4. (d) 5. (c) 6. (i) (b) (ii) (a) (iii) (b) (iv) (c) (v) (c)

7. 2 (x + 2)5/2 - 4 (x + 2)3/2 + C 9. 33x + C 10. 1 [x log e x - x] + C For Detailed Solutions
53 (log 3)2 log e 10 Scan the code

x+2 x 2 + 4x + 13 + 9 log|x + 2 + x 2 + 4x + 13|+ C
11.
22

12. 1 log |x - 1|- 2 log |x - 2|+ 3 log |x - 3|+ C
22

CBSE Term II Applied Mathematics XII 21

CHAPTER 02

Definite
Integrals

In this Chapter...

l Fundamental Theorem of Integral Calculus
l Properties of Definite Integral

òAn integral of the form b f ( x) dx is known as definite Properties of Definite Integrals
a
ò ò(i) b f ( x) dx = b f ( t) dt
òintegral and is given by b f ( x) dx = g( b) - g( a), where f ( x) is aa
a
ò ò(ii) b f ( x) dx = - a f ( x) dx
derivative of g( x), a and b are lower and upper limits of a ab
definite integral.
ò ò ò(iii) b f ( x) dx = c f ( x) dx + b f ( x) dx, where a < c < b.
Fundamental Theorem of Integral Calculus aa c

(i) Let f be a continuous function defined on the closed ò ò(iv) b f ( x) dx = b f ( a + b - x) dx
interval [ a, b] and A( x) be the area of function, aa

òi.e. A( x) = x f ( x) dx. ò ò(v) a f ( x) dx = a f ( a - x) dx
a 00

Then, A¢( x) = f ( x), " x Î[ a, b] ò ò ò(vi) 2a f ( x) dx = a f ( x) dx + a f ( 2a - x) dx
0 00

(ii) Let f be a continuous function defined on the closed f ( x) dx = ïíì 2 a f ( x) dx,
interval [ a, b] and F be an anti-derivative of f. ïî
2aò ò(vii) 0 if f ( 2a - x) = f ( x)
0 if f ( 2a - x) = - f ( x)
0,
òThen, b b
a f ( x) dx = [ F( x)] a f ( x) dx = íìï 2 a f ( x) dx,
ïî
= F( b) - F( a) aò ò(viii) 0 if f ( -x) = f ( x), i. e. even
-a if f ( -x) = - f ( x), i. e. odd
0,

22 CBSE Term II Applied Mathematics XII

Solved Examples

Example 1. Evaluate ò4 x 3 dx. Sol. We have, f ( x) =|x|
2
Now, f ( -x) =|- x|=|x| = f ( x),
é ù4
òSol. Let I =4x 3 dx = ê x4 ú which is even function.
ë 4 û2
2 4ò ò\ = 2 4
-4 f ( x) dx 0 f ( x) dx

= 1 [ 44 - 24 ] = 1 [ 256 - 16] ò ò[Q a f ( x) dx =2 a f(x) dx, if f ( x) is an even function]
44 -a
0

= 1 ´ 240 = 60 ò= 2 4
4
|x| dx
0

Example 2. òEvaluate 2(8x 3 - 15x 2 + 12x + 7) dx. ò= 2 4 dx = 2 ´ é x2 ù4
1 ê 2 ú
x ë û0

0

òSol. Let I = 2( 8x3 - 15x2 + 12x + 7) dx and = 42 - 0 = 16
1

f ( x) = 8x3 - 15x2 + 12x + 7 Example 5. ò1 x 3 +|x|+ 1 dx.
-1 x 2 + 2|x|+ 1
ò òThen, 2 f ( x)dx = 2( 8x3 - 15x2 + 12x + 7)dx Evaluate
11

= é 8x4 - 15x3 + 12x2 + ù2 òSol. Let I = 1 x3 +| x| + 1 dx
ê 4 3 2 7xú -1 x2 + 2| x| + 1
ë
û1 x3
2| x| + | x| + 1
òé xndx = x n+1 + ù ò ò= 1 + 1 dx + 1 + 2| x| + 1 dx
n+1 Cú
êQ -1 x2 -1 x2
ë û
ò= 1 | x| + 1
= [ 2x4 - 5x3 + 6x2 + 7x]21 0 + 2 0 x2 + 2| x| + 1 dx

= [ 2( 24 - 1) - 5( 23 - 1) + 6( 22 - 1) + 7( 2 - 1)] ò òé 1 ìîïíï2 1 f ( x) dx, if f ( x) is even function ù
-1 0 ú
= [ 2(16 - 1) - 5( 8 - 1) + 6( 4 - 1) + 7(1)] êQ f ( x) dx = 0, if f ( x) is odd function ûú
êë
= 30 - 35 + 18 + 7
(x + 1)
= 20 ò= 2 1 x2 + 2x + 1 dx [Q| x| = x, if x ³ 0]
0
ò8 10 - x dx.
Example 3. Find the value of 2 x + 10 - x ò= 2 1 (x + 1) dx
0 (x + 1)2
8 10 - x
òSol. Let I =2 x + 10 - x dx … (i) ò= 2 1 1
0 +
x 1 dx

òÞI= 8 10 - ( 8 + 2 - x) = 2[log( x + 1 )] 1
2 ( 8 + 2 - x) + 10 - ( 8 + 2 - x) dx 0

ò òé b ù = 2[log2 - log1]
a ûú
ëêQ
b f ( x) dx = f(a + b- x) dx = 2log2

a

I=òÞ 8 x dx …(ii) Example 6. Prove that ò1 log ççèæ 2 - x ÷ö÷ø dx = 0.
2 10 - x + x -1 2 + x

On adding Eqs. (i) and (ii), we get = æçèç 2 - x ÷ö÷ø,
2 + x
8 10 - x + x 8 Sol. Let f ( x) log then
2 10 - x + x
ò ò2I = dx = 8
dx = [ x] 2
2
ççæè ÷÷öø æççè ø÷ö÷ -1 æççè ÷÷öø
2 + x 2 - x 2 - x
Þ 2I = ( 8 - 2) f( -x) = log 2 - x = log 2 + x = (- 1) log 2 + x

Þ I=6 =3 = - f(x)
2

Example 4. If absolute function f (x) =|x|is defined as Thus, f ( x) is an odd function.

ìò|x|= x, x³ 0, -44|x| dx. So, using property (viii),
í - x, x< 0
î then find ò1 log èæç 2 - x ø÷ö dx = 0 Hence proved.
-1 2 + x

CBSE Term II Applied Mathematics XII 23

Chapter
Practice

PART 1 ò7. 7 dt
Objective Questions The value of + 5t is
41

(a) 1 log 4 (b) 1 log 7
45 5 12

l Multiple Choice Questions (c) 1 log12 (d) None of these
57

ò1. Evaluate 3 3x dx. 8. If f ( x ) = - x + , then find the value of
2 1) (x 2)
(a) 9 (b) 18log1 ( x
log 3 3
ò4 f (x) dx.
(c) 18log3 (d) 18 2
log 3
(a) 1 logèçæ 4 ø÷ö (b) 1 logçèæ 27 ö÷ø
3 27 3 4
ò2. The value of 4 (x + e2x ) dx is
0 (c) logçæè 27 öø÷ (d) None of these
4
(a) 15 + e8 (b) 15 - e8
2 2 ò9. The value of 5 2x dx is

(c) e8 - 15 (d) -e8 - 15 3 x2 + 3
2 2
(a) 1 log 7 (b) log 7
ò3. The value of 2 çæ 1 ÷ö dx is 33 3

-5è x ø (c) 2log 3 (d) log 3
7 7

(a) log 2 (b) log 2 ò10. 3 x5 x 6 + 1 dx
3 5 Evaluate -1 5
(b) 5 ( 730)3/ 2
(c) log5 (d) log 4 (a) 9 ( 730)3/ 2 9
2 5 5
(d) None of these
ò4. Evaluate 2(2x 2 + 1)2 dx. (c) 5 ( 730)1/ 2
0 9

(a) 571 (b) 574 11. òThe value of 6|x|dx is
15 15 0

(c) 573 (a) 16 (b) 17
3
(d) None of these (c) 18 (d) 15

ò5. The value of 2(4x 3 - 5x 2 + 6x + 9) dx equal to 12. òEvaluate 3 1 dx.
1 1 + log
x(1 x )

(a) 64 (b) 64 (a) (1 - log3) (b) log|1 - log3|
3
(c) 2log|1 - log3| (d) log|1 + log3|
(c) 132
3 (d) 132 1

ò6. If 1(5x 2 + x + l) = 0, then the value of l is ò13. The value of the integral 1 (x - x 3 ) 3 dx is
0
1/ 3 x 4

(a) 11 (b) - 13 (c) 17 (d) 13 (a) 6 (b) 0
6 6 6 6 (c) 3 (d) 4

24 CBSE Term II Applied Mathematics XII

ò14. The value of e2 dx is a òì 2 a f (x) dx, if f is an even function
0 i.e. f (- x) = f (x)
e x log x ò f (x) dx = íïï if f is an odd function
i.e. f (- x) = - f (x)
(a) log4 (b) log1 -a ï 0,
2 îï

(c) log2 (d) log6 Based on above information, answer the following

15. If f (a + b - x) = òf (x), then b f (x) dx is equal to questions.

x
a
1
ò(a) a + b b f ( b - x) dx ò(b) a + b b f ( b + x) dx ò(i) -1 3 dx is
The value of x
2a 2a
(a) 0 (b) 1
ò(c) b - a b ò(d) a + b b
2a 2a (c) 2 (d) 3
f ( x) dx f ( x) dx

l Case Based MCQs ò(ii) The value of 2 ( x4 + x2 )dx is
-2

16. For any function f (x), we have (a) 272 (b) 274
15 15

ò ò ò òb f (x) dx = c1 f (x) dx + c2 f (x) dx + ... b f (x) dx, (c) 272 (d) None of these

aa c1 cn ò(iii) The value of -33|x4|dx is

where a < c1 < c2 < ...< cn < b. (a) 486 (b) 486
7 (d) None of these
Based on the above information, answer the
(c) 486
following questions. 5

ò(i) The value of integral 4|x - 1|dx is ò(iv) -22e|x| dx is equal to
0

(a) 5 (b) 4 (a) 2e2 - 1 (b) 2( e2 + 1)

(c) 3 (d) 7

ò(ii) The value of -21|x3 - x| dx is (c) 2( e2 - 1) (d) None of these

ò(v) 5 x5 + x2 )dx is
(a) 11 (b) 11 The value of -5 (
2 3
(a) 250
(c) 11 (d) 11 3
9 4
(b) 251
ò(iii) The value of -11e|x|dx is 3

(a) 2( e - 2) (b) 2( e - 1) (c) 248
(c) 2e (d) e - 1 3

< (d) None of the above
³
ò(iv) = ì 3, x 2, 4 f ( x) dx is
If f ( x) íî x, x 2 then
0
PART 2
(a) - 12 (b) 13 (c) 12 (d) 14 Subjective Questions

ò(v) = ì x, x ³ 0 -23|x| dx is
If |x| í -x, x < 0 , then
î

(a) 13 (b) - 13 (c) 13 (d) 15 l Short Answer Type Questions
2 2 2
ò1. Find the value of 4 (x 2 + x5 + ex )dx.
17. òAn integral of the form b f (x)dx is known as definite 0
a

òintegral and is given by b f (x)dx = g(b) - g(a), ò2. Show that 3 1 dx = 1 log æç 3 ÷ö.
a 2 x2 -1 2 è 2 ø

where f (x) is derivative of g(x) or g(x) is an

anti-derivative of f (x). Here, a and b are called ò3. Evaluate the definite integral e2 1 dx.

lower and upper limits of definite integral. If f (x) is e x log x 2
a continuous function on [-a, a], then

CBSE Term II Applied Mathematics XII 25

ò4. Evaluate 3 x dx. l Long Answer Type Questions

2 x2 +1 ò21. Simplify 1 2x + 3 dx.

ò5. Simplify 1 xex2 dx. 0 (5x 2 + 1)
0
22. òFind the value of 1 x log(1 + 2x) dx.
ò6. Solve the integral 2 6x + 3 dx. 0

0 x2 + 4 ò23. Solve 2 5x 2 dx .

ò7. 1 dx . 1 x 2 + 4x + 3
Find the solution of 0 - 2e
ex -x òEvaluate 4 {|x - 1|+|x - 2|+|x - 3|} dx.
24. 1

ò8. Evaluate 2 x x + 2 dx. ò25. Prove 3 1 dx = 2 + log 2.
0 1 x 2(x + 1) 3 3

ò9. Evaluate 2 1 dx. l Case Based Questions

0 x + 4- x2

ò10. Solve 2 çæ 1 - 1 ö÷ e2x dx. 26. Properties of definite integral is used to evaluate
the integral becomes simplest and standard form.
1 è x 2x 2 ø So that it can be solved easily.

ò ò11. Show that a f (x) g(x) dx = 2 a f (x) dx, Some of the properties are
00
ò ò òI. b f ( x)dx = c f ( x) dx + b f ( x) dx, where a < c < bc
if f and g are defined as f (x) = f (a - x) aa
and g(x) + g(a - x) = 4.
ò òII. b f ( x) dx = b f ( a + b - x) dx
12. If f (x) is a continuous function defined as aa

ìíï x + 14 On the basis of above information, solve the
ïî 3
f (x) = 5x , - 1 £ x < 1, then find the value following questions.
x ³1
ò(i) Find the value of 2|3(1 - x)|dx.
0

òof 4 f ( x )dx. ò(ii) 3 ex
-1 Find the value of 0 e3 - x + ex dx.

ò13. Solve the integral 1 x (1 - x)n dx, by using the ò(iii) Evaluate 5 (5 - x) dx.
0
x
0

property of definite integral. 27. In substitution of definite integral, we substitute
some part of the integral. Such that whose
14. òEvaluate a x dx. derivative exist and also change the limits as per the
0 x + a-x new variable.

ò15. Evaluate 5 |x + 2|dx. òSuppose b f(h(x) h¢ (x) dx
-5 a

16. If f (x) = x 2 dx, then find the value of 2 f (x) dx. Substitute h(x) = t Þ h¢ (x)dx = dt
ò1 + 5 x
-2 Then, the given integral becomes

17. If f ( x ) = ì( x+ 1), 1< x £ 3 , then find the value of òh( b) f(t) dt
í x4 , 3< x < 7 h( a)
î
On the basis of the above information, solve the
ò5 f (x) dx. following questions.
2
ò(i) Find the value of e3 ( 3 log x) 3 dx.
ò18. If 3 ( x 4 + x 3 ) dx = k(3)5 , then find the value of k.
-3 ex

ò19. Evaluate 3 |log x|dx. ò(ii) Find the value of 5 1 dx.
1/ 2
3 ( x + 1) ( x - 2)
ò20. 3 5x dx.
Evaluate +53-x ò(iii) Evaluate 2 7x4 ex5 dx.
0 5x 1

26 CBSE Term II Applied Mathematics XII

SOLUTIONS

Objective Questions = éêë16 - 40 + 12 + 18ûúù - ëêé1 - 5 + 3 + 9úùû
3 3
ò1. (d) Let I = 33x dx ççæè 3x ö÷÷ø 3
= log 3 2 = é 46 - 40 ù - ëéê13 - 5ù
2 ëê 3 ûú 3 ûú

= 1 3 ( 3x )23 = é 138 - 40 ù - é 39 - 5 ù
log ëê 3 úû ëê 3 ûú

= 1 ( 33 - 32 ) = 98 - 34 = 64
log 3 33 3

= 1 ( 27 - 9) ò6. (b) We have, 1(5x2 + x + l) = 0
log 3 0

= 18 é 5x3 x2 ù 1
log 3 ê 3 2 ú
Þ ë + + lx û0 =0

ò2. (a) Let I = 4(x + e2x ) dx = é x2 + e2x ù 4 Þ é5 + 1 + l - (0 + 0 + 0)úùû = 0
ê 2 2 ú 0 ëê 3 2
0 ë û

= ççèæ 8 + e8 ö÷÷ø - æèç 1 ö÷ø Þ 10 + 3 + l = 0
2 2 6

= 15 + e8 Þ 13 + l = 0
2 6

ò3. (b) Let I = 2 çæè 1 öø÷ dx Þ l = - 13
-5 x 6

= [log|x|] 2 ò7. (c) Let I = 7 dt
-5 41 + 5t

= [log|2| - log|- 5|] = 1 [log|1 + 5 t|] 74
5
= log2 - log5

= log 2 ëéêQ log m - log n = log m ù = 1 [log|1 + 5 ´ 7| - log|1 + 5 ´ 4|]
5 n ûú 5

ò4. (b) Let I = 2 2x 2 + 1) dx = 1 [log36 - log21]
5
(
0

ò= 2( 4x4 + 1 + 4x2 ) dx = 1 log 36 é log m - log n = log mù
0 5 21 êëQ n úû

= é 4x5 +x+ 4x3 ù 2 = 1 log12
ê 5 ú 57
ë 3 û0

= 4( 2)5 + 2 + 4( 2)3 - [ 0 + 0 + 0] 8. (b) We have, f ( x) = x +
53 (x - 1) (x
2)

= 128 + 2 + 32 By using partial fraction, we get
53
x = A + B
= 384 + 30 + 160 = 574 (x -1) ( x + 2) (x - 1) (x + 2)
15 15
Þ x = A( x + 2) + B( x - 1)
ò5. (a) Let I = 2( 4x3 - 5x2 + 6x + 9) dx
1 Þ x = ( A + B) x + 2A - B

= é 4x4 - 5x3 + 6x2 + ù2 On comparing the coefficients of x and constant terms,
ê 3 2 9xú we get
ë 4
û1 1 = A+ B

= é x4 - 5x3 + 3x2 + ù2 and 0 = 2A - B
ê 3 9xú
ë Þ A = 1 and B = 2
û1 33

= é 24 - 5( 2)3 + 3( 2)2 + ù - é 1 ) 4 - 5(1)3 + 3( 1 ) 2 + ù \ x = 1 + 2
ê 3 9( 2)ú ê( 3 9( 1 ) ú - 1)( x 3( x - 1) 3( x +
ë ë (x + 2) 2)
û û

CBSE Term II Applied Mathematics XII 27

On integrating from limit 2 to 4, we get ò\ I= 6 = é x2 ù6
ê 2 ú
òLet I = 4 x dx x dx ë û0
1) (x
2(x + + 0

2) = 1 [ 62 - 02 ] = 36 = 18
22
41 42
2 3( x - 2 3( x +
ò ò= dx + dx
1) 2)
ò12. 3 dx
= 1 [log|x -1|] 4 + 2 [log|x + 2|] 4 (d) Let I= + logx)
3 2 3 2 1 x(1

= 1 [log|4 - 1| - log|2 - 1|] + 2 [log|4 + 2| - log|2 + 2|] Now, putting logx = t Þ 1 dx = dt
33 x

= 1 [log3 - log1] + 2[log6 - log4] Lower limit When x = 1, then t = log1 Þ t = 0
33
Upper limit When x = 3, then t = log3

= 1 [log3 - 0] + 2 logçæè 6 ø÷ö òNow, I= log 3 dt = [log| 1 + t |] log 3
3 3 4 0 (1 + 0
t)

= 1 log3 + 2 logèæç 3 öø÷ = log|1 + log 3| - log|1 + 0|
3 3 2
= log|1 + log 3| - log1

1 é logæèç 3 ø÷ö 2 ù = log|1 + log 3| - 0 [Q log 1 = 0]
ê 2 ú
= 3 ëê log 3 + ûú = log|1 + log 3|

é çèæ 3÷ö 2 ù 1
ê log3 2ø ú
= 1 ëê ´ úû ò13. (a) Let I =1 (x - x3)3 dx
3 1/ 3 x4

= 1 é 27 ù æçèç x x3 ÷öø÷1/ 3
3 ëê 4 ûú x3 x3
log ( x 3 )1/ 3 -

ò= 1 dx

ò9. (b) Let I = 5 2x 3 dx 1/ 3 x4
3 x2 +
æçè 1 - 1ö÷1/ 3
Substitute x2 + 3 = t Þ 2x dx = dt ò= 1 x2 ø
x3 dx
Upper limit When x = 5, then t = (5)2 + 3 = 28 1/ 3

Lower limit When x = 3, then t =( 3)2 + 3 = 12 Put 1 -1 =t Þ -2 dx = dt
x2 x3
ò\ I= 28 1 dt = [log t ]1228
12 t 1 -1
Þ x3 dx = 2 dt
= log28 - log12

= log 28 = logæèç 7 ö÷ø Upper limit When x = 1, then t = 0
12 3
Lower limit When x = 1, then t = 8
ò10. (b) Let I = 3 x5 x6 + 1 dx 3

-15 ò\ I = - 1 0 t1/ 3 dt

Substitute x6 + 1 = t 28

Þ 6x5 dx = dt = -1 ´ èæç 34÷øö [ t 4 / 3 ] 0
2 8
Upper limit When x = 3, then t = 36 + 1 = 730
= - 3 [ 0 - ( 23)4 / 3]
Lower limit When x = - 1, then t = ( -1)6 - 1 = 0 8

ò\ I = 730 t ´ 5 dt = - 3 ´ ( -16) = 6
8
06

= 5 é t 3/ 2 ù 730 ò14. (c) Let I = e2 dx
6 ê 3/ ú 0
ë 2 û e x logx

= 5[ t 3/ 2 ] 730 Now, put log x = t Þ 1 dx = dt
9 0 x

= 5 [( 730)3/ 2 ] = 5 ( 730)3/ 2 Lower limit When x = e, then t = loge = 1
99
Upper limit When x = e2, then t = loge2 = 2loge = 2
6
ò11. (c) Let I = òNow, I = 2 dt = [logt ]12
|x| dx 1t
0

Q |x| = ì x, x³0 = log2 - log1 = log2 [Q log 1 = 0]
îí- x, x<0

28 CBSE Term II Applied Mathematics XII

ò15. (d) Let I = b …(i) ò(iii) (b) Let I = 1 e|x| dx
-1
x f ( x) dx
a

Then, by a property of definite integrals |x| = ì-x, x<0
Q í x³0
òI = b( a + b - x) f ( a + b - x) dx î x,
a
0 -xdx 1 ex dx
ò= b( a + b - x) f ( x) dx …(ii) ò ò\ I= -1 e +
a 0

[Q given f ( a + b - x) = f ( x)] = é e-x ù0 + [ ex ]01
ê -1 ú
On adding Eqs. (i) and (ii), we get ë û -1

ò2I = b( a + b) f ( x) dx = æèçç e0 - e1 öø÷÷ + ( e1 - e0 )
a -1 -1

I= a+ b b
òÞ =-1 + e + e -1
2a f ( x) dx

ò16. (i) (a) Let I = 4 |x - 1| dx. = 2e - 2 = 2( e - 1)
0
(iv) (c) We have, f ( x) = íîì3x,, x<2
Now, let us first define the given absolute function. x³2

ì ( x - 1), if x ³1 4 f ( x) dx = 2 f ( x) dx + 4
Clearly,|x - 1| = îí - ( x - 1), if x <1 ò ò ò\
f ( x) dx
00 2

2 4

3 dx x dx
Now, divide the given limit at x = 1, i.e. write I as ò ò= +
02

ò òI = 1 |x - 1| dx + 4 |x - 1|dx = 3[ x] 2 + é x2 ù4
01 0 ê 2 ú
ë û2
ò ò òé ù
ûú
êëQ
b f ( x)dx = c f ( x)dx + b f ( x)dx, if a < c < b
c
a a = 3( 2 - 0) + 1 ( 42 - 22 )
2
ò ò= 1 - ( x - 1)dx + 4 ( x - 1)dx
01 = 6 + 1 (16 - 4)
2
é x2 ù 1 é x2 ù 4
ê 2 xú ê 2 xú
= - ë - û0 + ë - û1 = 6 + 1 (12)
2
é èççæ 12 - 1÷ø÷ö ù é æçèç 42 - 4÷÷öø æèçç 12 1 ÷öø÷ ù
= - ê 2 - ( 0 - 0)ú + ê 2 - 2 - ú = 6 + 6 = 12
ëê êë ûú
úû (v) (a) We have,|x| = ì x, x ³0
îí- x, x<0
= 1 + é 4 + 1ù =5
2 êë 2 úû
-23|x| dx = -03|x| dx + 2|x| dx
ò(ii) (d) Let I = -21|x3 - x| dx ò ò ò\
0
Let f ( x) = x3 - x = x( x - 1) ( x + 1)
0 -x dx + 2
ò ò=
-3 0 x dx

f ( x) = 0Þ x = 0, 1, - 1 = - é x2 ù0 + é x2 ù2
ê 2 ú ê 2 ú
ì x3 - x, -1 <x <0 ë û -3 ë û0
\ f ( x) = ïí- ( x3 - x), 0<x <1
ï 1<x<2 = - é 0 - èçæç -32 ÷ø÷ö ù + é 22 - ù
î x3 - x, ê 2 ú ê 0ú
ëê ûú ë 2 û

0 1 - ( x3 - x) dx + 2( x3 - x) dx = 9 + 4 = 13
22 2
-1( 0 1
ò ò ò\ I = x3 - x) dx +

= é x4 - x2 ù 0 + é -x4 + x2 ù 1 + é x4 - x2 ù 2 ò17. (i) (a) Let I =1 x3dx
ê 4 2 ú -1 ê 4 2 ú 0 ê 4 2 ú 1 -1
ë û ë û ë û
Let f ( x) = x3
=0-0-1 + 1 -1 + 1+ 0-0+ 4-2-1 + 1
4 24 2 42 \ f( -x) = ( -x)3

=-1 +1+ 2+ 1 = - x3 = - f(x)
24
\ f ( x) is an odd function.

= 3 - 1 = 11 ò\ I= 1 x 3dx = 0
44 -1

CBSE Term II Applied Mathematics XII 29

ò(ii) (a) Let I = 2 ( x4 + x2 )dx Subjective Questions
-2
Let g( x) = x4 + x2 ò1. Let I = 4 x 2 + x5 + ex) dx

(
0

Þ g( - x) = ( - x)4 + ( -x)2 = x4 + x2 = g( x) é x3 x6 ù 4
ê 3 6 ú
So, g( x) is an even function. = ë + + ex û0

ò\ I = 2 2( x4 + x2 )dx = 1 ( 43 - 0) + 1 ( 46 - 0) + e4 - e0
0 36

= é x5 + x3 ù2 = 64 + 4096 + e4 - 1
2ê 5 ú 36
3 û0
ë = 128 + 4096 - 6 + e4
6
= 2 é 32 + 8 -(0 + 0)úûù
êë 5 3
= 4218 + e4 = 703 + e4
= 2 é 96 + 40 ù 6
ëê 15 úû
é x -1 ù3
= 2 ´ 136 ò2. Let I = 3 1 dx = ê 1 log x+1 ú
15 2 x2 - 1 ë 2 û2

= 272 òé dx = 1 log x-a + ù
15 x2 - a2 2a x+a Cú
êQ û
(iii) (c) Let g( x) =|x4|, then ë

g( - x) =|( - x)4| =|x4| = g( x) = 1 log 3-1 - 1 log 2-1
2 3+1 2 2+1
\ g( x) is an even function.

3ò ò\ 3|x4| dx
-3 |x4| dx = 2 = 1 log 2 - 1 log 1
0 2 42 3

ò= 2 3 4 dx

x = 1 [log (1) - log ( 2) - log(1) + log( 3)]
0 2

= 2 é x5 ù3 = 2 [ 35 - 0] éëêQlog çæ a ö÷ = - log b úùû
ê 5 ú 5 è b ø
ë û0 log a

= 2 ( 243) = 486 = 1 log èçæ 3 ÷øö Hence proved.
55 2 2

(iv) (c) Let f ( x) = e|x|, then e2 1
e x logx2
f ( -x ) = e|-x| = e|x| = f ( x), it is an even function. ò3. Let I = dx

2ò ò\ 2ex 2
-2 e|x|dx = 2 dx = 2 [ e x ] 0 Put logx2 = t
0

= 2 [ e2 - e0 ] = 2 ( e2 -1) Þ 1 ´ 2x dx = dt
x2
ò(v) 5 x5 + x2 )dx
(a) Let I = -5 ( Þ 2 dx = dt
x
f ( x) = x5 and g( x) = x2

Now, f ( - x) = ( - x)5 and g( - x) = ( -x)2 Lower limit When x = e, then t = loge2 = 2loge = 2
Þ f ( - x) = - x5 and g( - x) = x2
Upper limit When x = e2, then t = loge4 = 4loge = 4

Þ f ( - x) = - f ( x) and g( - x) = g( x) ò\ I= 41 = 1 [log t ] 24
2 2t 2
\ f ( x) is an odd function and g( x) is an even function.
= 1 [log4 - log2]
ò ò ò\ 5 x5 5 x2 5x2 2
I= -5 dx + -5 dx = 0 + 2 dx
0

ò òé a = a f is an even functionùûú = 1 log 4 = 1 log2
-a 0 2 22
êëQ
f ( x) dx 2 f ( x) dx, if

= 0 + 2 é x3 ù 5 ò4. Let I = 3 x 1 dx
ê 3 ú 0 2 x2 +
ë û
Put x2 + 1 = t
= 2 [53 - 0]
3 Þ 2x = dt
dx
= 2(125) = 250
33 Þ dx = dt
2x

30 CBSE Term II Applied Mathematics XII

ò ò\ I = 3 x dt = 1 3 1 dt 1 dx 1 exdx
0 - 2e-x 0 e2x - 2
2 t 2x 2 2 t ò ò7. Let I = =
ex
1
= 2 [log|t |]32 Put ex = t Þ exdx = dt

= 1 [log|x2 + 1 | ] 3 [Qt = x2 + 1] Lower limit When x = 0, then t = e0 = 1
2 2
Upper limit When x = 1, then t = e1 = e

= 1 [log|32 + 1| - log|22 + 1| ] ò\I= e1 2)2 dt
2 1 t2 - (

= 1 [log|10| - log|5|] é 1 t - 2 ù e
2 ê ´ t + ú
= êë log
= 1 log 10 = 1 log2 é aù 2 2 2 ûú 1
2 52 ëêQ log a - log b = log b ûú

ò5. Let I = 1 xex2 dx = 1 é log e - 2 - log 1 - 2 ù
0 22 ê + 2 1 + ú
ëê e 2 ûú

Put x2 = t Þ 2x = dt = 1 é logæçèç e - 2 ö÷ø÷ - logèæçç 1 2 - 1 ø÷ö÷ ù
dx 22 ëêê e + 2 + 2 ûúú

Þ dx = dt ò ò8. Let I =2 2 {( x + 2) - 2} x + 2 dx
2x x + 2 dx =
x
Lower limit When x = 0, then t = 0 00

Upper limit When x = 1, then t = 1 [Q add and subtract 2]

ò ò\ I = 1 xet dt = 1 1 et dt ò= 2{( x + 2)3/ 2 - 2( x + 2)1/ 2}dx
0
0 2x 2 0
= é ( x+ 2)(3/ 2 + 1) - 2( x + 2)1/ 2+ 1 ù 2
ê ( 3/ 2) + 1 (1/ 2) + ú 0
= 1 [ et ]10 = 1 [ e1 - e0 ] ë 1 û
2 2
2[( x 4[( x
= 1 [e - 1] = 5 + 2)5 / 2 ] 2 - 3 + 2) 3/ 2 ] 2
2 0 0

ò6. Let I = 2 6x + 3 dx = 2[ 45 / 2 - 2 5 / 2 ] - 4[ 43/ 2 - 23/ 2 ]
0 x2 + 4 53

ò ò= 2 6x 4 dx + 2 3 4 dx = 2[ 25 - 25 / 2 ] - 4[ 23 - 23/ 2 ]
0 x2 + + 53
0 x2
= 2[ 32 - 4 2] - 4[ 8 - 2 2]
Put x2 + 4 = t 53

Þ 2x = dt = 8[8 - 2] - 8[4 - 2]
dx 53

Þ dx = dt = çèæ 64 - 32 øö÷ + èæçç 8 2 - 8 2 ÷÷öø
2x 5 3 3 5

Lower limit When x = 0, then t = 0 + 4 = 4 = 32 + 16 2
15 15
Upper limit When x = 2, then t = 4 + 4 = 8

8 6x dt + 2 3 ò9. Let I = 2 1
4 t 2x 0 + 0 4-
I=ò ò\ x2 4 dx x + x2 dx

8 1 dt + 3 2 1 2 1
4t + 0 - çæ 1 ö÷ 2
ò ò= 3 0 x2 22 dx ò= æç 1 ö÷ 2 dx

xù2 x + 4 - x2 +
é 2úû 0 è 2ø è 2ø
= 3 [log t ] 8 + 3 êë tan-1
4 2 ò= 2 1
0 dx
1 é çèæ 14 ÷øö ù
òé dx = 1 tan-1 x + C ù 4 + 4 - êë x 2 - x + úû
+ x2 a a ûú
ëêQ a2

= 3 [log ( 8) - log ( 4)] + 3 é tan -1 2ù ò= 2 1 2 dx
2 êë 2 ûú - æèç x
0 + 1 - 1 øö÷
çæè ø÷ö ´p é ù 4 2
4 ëêQ ûú 4

= 3 log 8 + 3 log b - log a = log b ò= 2 1
4 2 a - èæç x
0 çèçæ 2 dx
= 3 log 2 + 3p ÷ö÷ø 1 ö÷ 2
8 17 - 2ø
2

CBSE Term II Applied Mathematics XII 31

é 17 + - 1 ù2 We have, f ( x) = ìïíx + 14, -1 £ x < 1
ê x ú 12. ïî 3 x ³1
1 ê log 5 x,
= 2 2ú
17 ê 17 1 ú
2× 2 ëê 2 - x + 2 úû 0 ò ò ò\4 f( x) dx = 1 f ( x) dx + 4 f ( x) dx
-1 -1
1

òé dx = 1 log a + x + C ù ò ò= 1x + 14 dx + 4 5x dx
a2 - x2 2a a - x ú 3 1
êQ û -1
ë
é x2 ù 1 é x2 ù 4
ù2 1 ê 2 14xú ê 2 ú 1
= 1 é 17 + 2x - 1 ú = 3 ë + û -1 + 5 ë û
ê log 17 - 2x + 1 ûú 0
17 ëê
1 é1 æèç 1 -14øö÷ 5 [ 42 ù
1 é 17 + 3 17 - 1 ù = 3 ëê 2 + 14 - 2 + 2 - 1 2 ] úû
ê log 17 - 3 17 + 1 ú
= 17 ëê - log úû

1 log 17 + 3 ¸ 17 - 1 = 1 é1 + 14 - 1 + 14úùû + 5 [16 - 1]
17 17 - 3 17 + 1 3 ëê 2 2 2
=
= 1 [ 28] + 5 ´ 15
êéëQ log m - log n = log m ù 32
n ûú
= 28 + 75
= 1 log ( 17 + 3) ( 17 + 1) 32
17 ( 17 - 3) ( 17 - 1)
= 56 + 225 = 281
66

= 1 log 20 + 4 17 ò13. Let I = 1 x(1 - x)ndx
17 20 - 4 17 0

ò10. Let I = 2 èæç 1 - 1 ÷øö e 2 xdx òÞ I = 1 (1 - x) {1 - (1 - x)}ndx
1 x 2x2 0

ò òé f ( a - x)dxúûù

ëêQ
2 1 e2xdx - 2 1 a f ( x)dx = a
1 x 1 2x2 0
ò ò= e 2 xdx 0

Evaluate the first integral by rule of integration by parts ò ò= 1 (1 - x)xn dx = 1 ( xn - xn+1)dx
00

taking 1 as the first function, we get = é x n+1 - x n+ 2 ù 1
x ê n n ú 0
ë +1 + 2 û

ò òI= é 1 × e2x ù 2 + 2 1 × e2x dx - 21 e 2 xdx = é 1 - 1 ù - 0
ê x 2 ú 1 1 x2 2 1 2x2 ëê n + + 2 úû
ë û 1 n

= é 1 × e2x ù 2 = ( n + 2) - (n + 1)
ê x 2 ú 1 (n + 1 )( n + 2)
ë û
1
= e4 - e2 = (n + 1) (n + 2)
42
a x
= e2 æçèç e2 - 1ö÷÷ø ò14. Let I = 0 x+ a - x dx …(i)
2 2
a - x
ò11. Let I = a òI = a a - dx
…(i) 0 a - x + (a - x)
f ( x) g( x) dx
0

òÞ I = a f ( a - x) g( a - x) dx ò òé dxúûù
0
ëêQ
a f ( x) dx = a f ( a - x)
0
0

ò òé a f ( a - x)dxúùû
0
ëêQ
a f ( x)dx = ò= a a-x dx …(ii)
0 a-x + x
0

òÞ I = a f ( x) {4 - g( x)}dx …(ii) On adding Eqs. (i) and (ii), we get
0

[Q f ( x) = f ( a - x) and g ( x) = g ( a - x) = 4 (given)] a x + a - x 2I = a1 dx
0 a -x + x
0
ò ò2I =
On adding Eqs. (i) and (ii), we get dx ,

ò2I = a 4 f ( x) dx = [ x ] a = a - 0 = a
0 0

I = 2 a f ( x) dx Þ I=a
òÞ 0 Hence proved. 2

32 CBSE Term II Applied Mathematics XII

ò15. Let I = 5 |x + 2|dx ò18. Let I = 3 ( x4 + x3) dx
-5 -3

It can be seen that ( x + 2) £ 0 on [ -5, - 2] and ( x + 2) ³ 0 ò ò= 3 x 4 dx + 3 3dx
on [ - 2, 5]. -3
-3x

ò ò\ I = -2 -( x + 2) dx + 5 ( x + 2) dx By the property of definite integral,
-5 -2
ò òa dx = îìïíï2 a f ( x) dx, if, f is an even function
-a 0 0, if, f is an odd function
ò ò òé b ù f ( x)
c ûú
ëêQ
b f ( x)dx = c f ( x)dx + f ( x)dx, a < c < b
a
a

é x2 ù -2 é x2 ù5 Here, x4 is an even function and x3 is an odd function.
ê 2 2xú 2xú
Þ I = - ë + + ê + -33( x4 + x3) dx = 2 3x4dx + 0
û -5 ë û -2
0
2 ò ò\

= - é ( -2)2 + 2( -2) - ( -5)2 - 2( -5)úù = é x5 ù 3
ê 2 2 û 2ê 5 ú 0
ë ë û

+ é ( 5) 2 + - ( -2)2 - -2) ù = é ( 3)5 ù = 2( 3)5
ê 2 2 ú 2ê 5 ú 5
ë 2( 5 ) 2( û ë û

= - é 2 - 4 - 25 + 10ûúù + é 25 + 10 - 2 + 4úùû ò19. Let I = 3 |logx| dx
ëê 2 êë 2 1/ 2

= - 2 + 4 + 25 - 10 + 25 + 10 - 2 + 4 = 29 Q |logx| = ìíî-lologgx,x, 0<x <1
22 1<x£3

2 x2 I= 1 3
- 21 + 5x ò ò\
ò16. Let I = dx …(i) - logx dx + (log x ) dx
1/ 2 1

2 (2 - 2 - x)2 òSuppose, I1 = 1 logx dx [using integration by parts]
-2 1 + 5 2 - 2 - x
[Q b f ( x) dx = b f ( a + b - x) dx] II I

aa
ò ò ò= dx = - ò 1 ´ ´
x
(log x ) x x dx

ò= 2 x2 dx = x logx - ò1 dx = x logx - x
+ 5 -x
-21 \ I = [ -( x logx - x)]11/ 2 + [ x logx - x]13

òÞ I= 2 5x 1 x2 dx ... (ii) = - éêë1 -1 - èçæ 1 log 1 - 12 öø÷ ù
- 25x + 2 2 ûú
log1

On adding Eqs. (i) and (ii), we get + [ 3log3 - 3 - (1 log1 - 1)]

ò ò2I = 2 2 çæçè 1 + 5x ÷÷öø x2 dx = 2 x2 dx = - é 0 - 1 - 1 log 1 + 1ù + [ 3log3 - 3 - 0 + 1]
- 5 x +1 ëê 2 2 2 ûú
-2

ò ò òÞ 2I = 2 2x2 dx
0
[Q x2 is even, so 2 x2 dx = 2 2x2 dx] = 1 + 1 log1 - 1 + 3log3 - 2
2 22
-2 0
= -3 + 1 log1 + 3log3
Þ I = é x3 ù 2 1 ( 23 - 0) = 8 222
ê ú 3 3
ë 3 û =

0 5x
+ 53- x
17. We have, f ( x) = ì( x + 1 ), 1<x£3 ò20. Let I =3 dx …(i)
í x4 , 3<x<7
î 0 5x

ò ò ò\ 5 f ( x) dx = 3 f ( x) dx + 5 f ( x) dx I=ò\ 3 53- x dx ò ò[Q b f ( x) = b f ( a + b - x)] …(ii)
22 3 053- x + 5x aa

ò ò= 3( x + 1) dx + 5 x4 dx On adding Eqs. (i) and (ii), we get
23
5x + 53- x
= é x2 + ù3 + é x5 ù5 ò2I = 3 5x + 53- x dx
ê 2 xú ê 5 ú 0
ë û2 ë û3
ò= 31 dx = [ x ] 3
é 9 çæè 4 2ø÷ö ù 1 [55 0
= êë 2 + - 2 + úû + 5 - 35 ] 0
3
=3-0

= é 15 - 8ù + 1 [ 3125 - 243] Þ I=3
ëê 2 2 úû 5 2

= 7 + 1 ´ 2882 ò21. Let I = 1 2x + 3 dx
25 0 (5x2 + 1)

= 35 + 5764 = 5799 1 2x 1 1
10 10 0 (5x2 + 0 (5x2 + 1)
ò ò=
1) dx + 3 dx

CBSE Term II Applied Mathematics XII 33

Put 5x2 + 1 = t ò23. Let I = 2 5x2 dx
1 + 4x
Þ 10x = dt Þ dx = dt x2 + 3
dx 10x
Here, the degree of numerator and denominator is same, so
1 2x dt + 3 1
0 t 10x 5 1 + çæ 1 on dividing numerator by denominator, we get
ò ò\I= 0 ö÷ 2 dx

x2 è 5ø x2 5x2 + =5 - 20x + 15
+ 4x x2 + 4x + 3
3

1 1 dt + 3 1
0t 5 x2 + çæ 1
ò ò= 1 1 dx ò\ I= 2 é - 20x + 15 ù dx
0 ö÷ 2 1 ê5 x2 + 4x + ú
5 ë 3 û

è 5ø ò ò= 2 2 20x + 15
1 5 dx - + 1) (x + dx ...(i)
é ù1 1 (x 3)

= 1 [log t ]10 + 3 ´ 1 ê tan-1 x ú Let (x 20x + 15 3) = (x A + (x B 3)
5 5 1 ê 1 ú + 1) (x + + 1) +
ê ú
5 êë 5 úû 0 Þ 20x + 15 = A( x + 3) + B( x + 1)
Þ 20x + 15 = Ax + 3A + Bx + B
òé dx =1 tan-1 x + ù
a2 + x2 a a Cú On equating the coefficients of x and constant term on both
êQ
ë û sides, we get

= 1 [log|5 x 2 + 1 |]10 + 3 5 [tan-1 5 x ]10 20 = A + B …(ii)
5 5
and 15 = 3A + B …(iii)

= 1 [log|5 + 1| - log|0 + 1|] On solving Eqs. (ii) and (iii), we get
5
+ 3 [tan-1 5 (1) - tan-1 5 ( 0)] A = - 5 and B = 45
5 22

2 1 dx + 5 2 1 45 2 1
12 + 1) 2 +
= 1 log 6 + 3 tan-1 5 ò ò ò\ I= 5 1 (x dx - 1 (x 3) dx
55
[Qlog1 = 0 and tan-1( 0) = 0] [from Eq. (i)]

ò22. Let I = 1 x log(1 + 2x) dx = êëé5 x + 5 log|x + 1| - 45 log|x + 3|ùúû 2
0 2 2 1

é x2 ù1 1 x2 = ëéê10 + 5 log|3| - 45 log|5| - 5 -5 log|2| + 45 log|4|úûù
ê log(1 2 ú + 2 2 2 2 2
ë û0
ò= + 2x) - 1 2x × 2× dx

ò= 1 [x2 x2 Þ I = 5 + 5 log 3 - 45 log 5
2 log(1 + 2x)]10 - + 2x dx 2 22 4

1 ëêéQm log a - m log b = m log æçè a øö÷ ù
b ûú
é æç x ÷ö ù
ê ç ÷ ú
1 [1 log3 ê 1 x 2 ÷ø÷ dxú ò24. 4 {|x - 1| + |x - 2| + |x - 3|} dx
2 + ú 1
ò= - 0] - 0 çèç 2 - 1 2x û

ê ò= 2(|x - 1| + |x - 2| + |x - 3|) dx
ë 1

ò ò= 1 log3 - 1

22
1 x dx + 1 1 1 x dx ò+ 3 {|x - 1| + |x - 2| + |x - 3|} dx
02 0 + 2x 2

1 log3 1 é x2 ù1 1 1 1 ( 2x + 1 - 1) ò+ 4 {|x - 1| + |x - 2| + |x - 3|} dx
2 2 0 2 3
ò= - ê ú + ( 2x + 1) dx
2 ë 2 û0 ò= 2
- 1 - (x - 2) - (x - 3)} dx
1 {x

1 log3 1 é1 0ûùú +1 1 dx - 1 1 1 ò+ 3 - + - - -
2 2 êë 2 4 04 + 2x
ò ò= - - 01 dx {x 1 x 2 (x 3)} dx
2

= 1 log3 - 1 + 1 [ x ]10 - 1 [log|(1 + 2x)|] 1 ò+ 4 ( x - 1 + x - 2+ x - 3) dx
2 4 4 8 0 3

= 1 log3 - 1 + 1 - 1 [log3 - log1] ò ò ò= 2( -x + 4) dx + 3 x dx + 4( 3x - 6) dx
2 448 1 23

é x2 ù2 é x2 ù 3 é 3x2 ù 4
ê 2 4xú ê 2 ú 2 ê 2 6xú
= 1 log3 - 1 log3 = 3 log3 = ë - + + ë û + ë - û3
2 88 û1

34 CBSE Term II Applied Mathematics XII

ççæè - 22 8÷ö÷ø èçæ -1 4÷öø 1 ò(ii) 3 ex
= 2 + - 2 + + 2 ( 32 - 22 ) Let I = 0 e3 - x + ex dx …(i)

+ çèæ 3 ´ 42 - 6 ´ 4÷öø - æçè 3 ´ 32 - 6 ´ 3ø÷ö òI = 3 e3 - x e3 - x dx
2 2 0 e3 - (3 - x) +

7 5 èæç - 92 öø÷ ò òé ù
2 2 ûú
ëêQ
= 6 - + + ( 24 - 24) - a f ( x) dx = a f ( a - x) dx
0
0

= 12 - 7 + 5 + 0 + 9 = 19 I=òÞ 3 e3 - x dx …(ii)
2 22 0 ex + e3 -x

ò25. To prove 31 1) dx = 2 + log 2 On adding Eqs. (i) and (ii), we get
1 x2(x+ 3 3
ò2I = 3ex + e3 - x
By using partial fraction method 0 ex + e3 - x dx

Let 1 =A+ B + C …(i) òÞ 3 = 3
x2(x+ 1) x x2 x+ 1 2I = dx [ x ] 0
1

0

Þ 1 = Ax ( x + 1) + B( x + 1) + Cx2 …(ii) Þ 2I = 3 - 0

Substituting x = 0, -1 in Eq. (ii), we get Þ I=3
1 = B( 0 + 1) and 1 = C( -1)2 Þ B = 1 and C = 1 2

ò(iii) 5 (5 - x) dx
On equating the coefficients of x2 on both sides of Eq. (ii), Let I =
x
0
we get
ò= 5(5 - x) 5 - (5 - x) dx
0= A + C Þ A = - C = - 1 0

1 æççè 1 1+ 1 1 ø÷ö÷ ò òé a ù
x+ x x2 x+ 0 úû
êëQ
ò ò\ 3 dx = 3 - + dx a f ( x) dx = f ( a - x) dx
1 x2( 1
1) 0

= é - log|x| - 1 + log|x + 1|ùûú 3 ò= 5 - x) x dx
êë x 1
(5
0

é x+ 1 1 ù 3 ò= 5(5 x - x x )dx
ëê x x úû 1 0

= log - ò= 5 x1/ 2 - x3/ 2 ) dx

(5
0

= çæ log 4 - 1 ö÷ - æç log 2 - 1 ö÷ = é 5 x3/ 2 - x5 / 2 ù5
è 3 3ø è 1 1ø ê 3/ 2 5/ ú
ë 2 û 0

= çèæ log 4 - log 2÷øö + èçæ1 - 1 öø÷ = é 10 x 3/ 2 - 2 x5 / 2 ù5
3 3 ëê 3 5 úû 0

= log æèç 4 ´ 1 øö÷ + 2 é log èçæ m ø÷ö = log m - log ù = é 10 5 3/ 2 - 255 / 2 -(0 - 0)ûúù
3 2 3 ëêQ n núû ëê 3 5

= log 2 + 2 Hence proved. = é 50 5 - 2 ´ 25 ´ 5ù
33 ê 3 5 ú
ë û
ò26. (i) Let I = 2|3(1 - x)| dx
0 = é 50 5 - 10 ù
ëê 3 ûú
ò= 3 2|1 - x| dx 5
0

ò ò= 3éëê 1 - + 2 (1 - x) dxúùû = é 50 5 - 30 5 ù
x 1 ) dx 1 ê 3 ú
(
0
ë û
é 1 2 ù
ê é x2 ù é x2 ù 1 ú = 20 5
= 3 êë ê 2 - xú + êx - 2 ú úû 3
ë û0 ë û
ò ò27. (i) Let I = e3 ( 3logx)3 dx = 27 e3 (logx)3 dx
= 3 é çæ 1 -1 - 0÷ö + çæ 2 - 4 - æç1 - 1 ÷ö ö÷ ù ex ex
ëê è 2 ø è 2 è 2 ø ø ûú
Put logx = t Þ 1 dx = dt
= é - 1 + æç 0 - èçæ 1 ÷øö ÷ö ù x
3êë 2 è 2 ø ûú
Upper limit When x = e3, then t = loge3 = 3

= 3 é - 1 - 1ù = - 3 Lower limit When x = e, then t = loge = 1
ëê 2 2 ûú