Acing AP Calculus AB and BC

Chapter 6
Techniques of Integration

6.1 Basic Integration Rules

In this section, we will study several integration techniques for fitting an integrand into one of
the basic integration rules. The basic integration rules are reviewed in Table 6.1 on page 252.

Procedures for Fitting Integrands to Basic Rules

Procedure Example

1. Separating numerator 1=− 2x 1 − 2x
2. Adding and subtracting terms in numerator 1+ x2 1+ x2 1+ x2
3. Dividing improper fractions
4. Completing the square =1 1− ex =+ ex 1− ex + ex
1− ex 1− ex 1− ex 1− ex

x3 − 3x= x − 2x
x2 −1 x2 −1

1= 1

4x − x2 4 − (x − 2)2

Other integration techniques, such as the simple substitution method, were covered in section 4.8.
Using trigonometric identities, trigonometric substitution, Method of Partial Fractions and Integration
by Parts will be covered later in this chapter.

1. Separating numerator

Example 1 □ Evaluate ∫ 1− 2x dx .
1+ x2

Solution ∫ ∫ ∫□ =11−+ 2x2x dx 1 dx + −2x dx Separate the numerator.
1+ x2 1+ x2

∫ 1 dx = arctan x Basic integration rules.
1+ x2

∫ −2x dx = ∫ −du = − ln u Let u= 1 + x2 , then du = 2x dx .
1+ x2 u

∫Therefore 1− 2x =dx arctan x − ln(1+ x2 ) + C .
1+ x2

192 Chapter 6

Table 6-1 Differentiation Rules and Basic Integration Rules

Differentiation Rules Basic Integration Rules

1. d (xn ) = nxn−1 1. ∫ du= u + C
dx

2. d [ f=(x)g(x)] f (x)g′(x) + g(x) f ′(x) ∫2. un=du un+1 + C , n ≠ −1

dx n +1

d  f (x)  g(x) f ′(x) − f (x)g′(x) 3. ∫ =du ln u + C
dx  g(x)  u
3. = [g(x)] 2

∫4. eu d=u eu + C

4. d f [g(x)] = f ′[g(x)] g′(x) ∫5. =au du 1 au + C
ln a
dx

5. d (ex ) = ex 6. d (ax ) = (ln a)ax 6. ∫ sin u du =− cos u + C
dx dx

7. d ln x = 1 8. d log=a x 1 ⋅1 7. ∫ cos u=du sin u + C
dx x dx ln a x

9. d (sin x) = cos x 10. d (csc x) = − csc x cot x 8. ∫ sec2 u=du tan u + C
dx dx

11. d (cos x) = − sin x 12. d (sec x) = sec x tan x 9. ∫ csc2 u du =− cot u + C
dx dx
10. ∫ sec u tan u=du sec u + C

13. d (tan x) = sec2 x 14. d (cot x) = − csc2 x 11. ∫ csc u cot u du =− cscu + C
dx dx

15. d (sin−1 x) = 1 12. ∫ tan u du =− ln cos u + C
dx 1− x2

16. d (cos−1 x) = − 1 13. ∫ co=t u du ln sin u + C
dx 1− x2

17. d (csc−1 x) = − 1 14. ∫ sec u du= ln sec u + tan u + C
dx x x2 −1
15. ∫ csc u du =− ln csc u + cot u + C

18. d (sec−1 x) = 1 ∫16. d=u sin−1 u + C a
dx x x2 −1 a2 − u2

19. d (tan −1 x) = 1 ∫17. =du 1 tan−1 u + C
dx +x a2 + u2 aa
1 2

20. d (cot−1 x) = − 1 ∫18. =du 1 sec−1 u + C a
dx 1+ x2 u u2 − a2 a

Techniques of Integration 193

2. Adding and subtracting terms in numerator

Example 2 □ Evaluate ∫ 1 dx .
1− ex

1 dx = 1− ex + ex dx Add and subtract ex in the numerator.
∫ ∫Solution □ 1− ex Separate the numerator.
1− ex
Use the basic integration rules.
1− ex ex
1− ex 1− ex
∫ ∫=
dx + dx

= ∫ dx + ∫ ex dx
1− ex

=x − ln(1− ex ) + C

3. Dividing improper fractions

∫Example 3 □ Evaluate x3 − 3x dx .
x2 −1

Solution □ ∫ x3 − 3x =dx ∫ (x − 2x dx Divide an improper fraction.
x2 −1 ) Use the basic integration rules.

x2 −1

= ∫ x dx − ∫ 2x dx
x2 −1

= 1 x2 − ln(x2 −1) + C
2

4. Completing the square

Example 4 □ Evaluate ∫ 1 dx .
4x − x2

Solution □∫ 1 dx = ∫ 1 dx Complete the square.
4x − x2 4 − (x − 2)2 Let u= x − 2 , then du = dx .
Use the basic integration rules.
=∫ 1 du
4 −u2

= sin−1(u ) + C
2

= sin−1( x − 2) + C
2

194 Chapter 6

Exercises - Basic Integration Rules
Multiple Choice Questions

1. ∫ 1+ sin x dx =
cos2 x

(A) tan x − sec x tan x + C
(B) tan x + sec x + C
(C) tan x + sec2 x + C
(D) ln(1+ cos2 x) + C

∫2. e2x dx =

1+ ex

(A) e2x + ln(1+ ex ) + C
(B) e2x − ln(1+ ex ) + C
(C) 2e2x − ln(1+ ex ) + C
(D) ex − ln(1+ ex ) + C

3. ∫ 2 tan x ln(cos x) dx =

(A) cos x [ln(cos x)] + C
(B) sin x[ln(cos x)] + C
(C) −[ln(cos x)] 2 + C
(D) [ln(sin x)] 2 + C

Techniques of Integration 195

∫4. 3 x2 1 dx =
2 − 4x +5

π (B) 1− π (C) 1+ π (D) 1+ π
(A) 4 6 4

4

5. ∫ 2x dx =
x2 + 2x +1

(A) −arc cot x − 1 + C
x +1

(B) arctan x + 1 + C
x +1

(C) 2 ln x +1 −2 +C
(x +1)2

(D) 2 ln x +1 + 2 + C
x +1

Free Response Questions

6. The region bounded by y = sin x , x = 0 , x = π , and the x-axis is revolved around the x-axis.
cos x 4

What is the volume of the resulting solid?

196 Chapter 6

6.2 Trigonometric Integrals

Trigonometric Identities

sin2 x + cos2 x =1 tan2 x +1 =sec2 x cot2 x +1 =csc2 x

sin2 x = 1− cos 2x cos2 x = 1+ cos 2x
2 2

sin 2x = 2sin x cos x

∫Guidelines for Evaluating sinm x cosn x dx .

1. If m is odd, save one sine factor and use sin2 x= 1− cos2 x to express the remaining factor
in terms of cosine.

∫Example 1 □ Evaluate sin3 x cos2 x dx .

∫Solution □ sin3x cos2 x dx One sine factor is saved.
∫= sin2 x cos2 x (sin x) dx sin2 x= 1 − cos2 x
Multiply.
∫= (1− cos2 x) cos2 x (sin x) dx u = cos x , du = −sin x dx
∫= (cos2 x − cos4 x) sin x dx

∫= (u2 − u4 ) (−du)

=− 1 cos3 x + 1 cos5 x + C
35

2. If n is odd, save one cosine factor and use cos2 x= 1− sin2 x to express the remaining factor
in terms of sine.

Example 2 □ Evaluate ∫ cos5 x dx .

Solution □ ∫ cos5 x dx

= ∫ cos4 x(cos x) dx One cosine factor is saved.
cos2 x= 1 − sin2 x
∫= (1− sin2 x)2 cos x dx u = sin x , du = cos x dx

∫= (1− u2 )2 du Multiply.

∫= (1 − 2u 2 + 4 ) du

u

=u − 2 u3 + 1 u5 + C
35

=sin x − 2 sin3 x + 1 sin5 x + C
35

Techniques of Integration 197

3. If both m and n are even, substitute sin2 x = 1− cos 2x and cos2 x = 1+ cos 2x to reduce
22

the integrand to lower powers of cos 2x .

∫Example 3 □ Evaluate sin2 x cos2 x dx .

∫Solution □ sin2 x cos2 x dx

=∫  1− cos 2x   1+ cos 2x  dx sin2 x = 1 − cos 2x , cos2 x = 1 + cos 2x
 2   2  22

( )= 1 Multiply.
4 ∫ 1− cos2 2x dx cos2 2x = 1 + cos 4x

= 1 ∫ 1 − 1+ cos 4x  dx 2
4 2  Simplify.

= 1 ∫ (1− cos 4x) dx
8

=1 x − 1 sin 4x + C
8 32

∫Guidelines for Evaluating tanm x secn x dx .

1. If m is odd, save a secant-tangent factor and convert the remaining factors to secants, and
substitute ta=n2 x sec2 x −1 .

∫Example 4 □ Evaluate tan3 2x sec2 2x dx .

∫Solution □ tan3 2x sec2 2x dx

= ∫ tan2 2x sec2x (tan2x sec2x) dx A secant-tangent factor is saved.
ta=n2 x sec2 x −1
= ∫ (sec2 2x −1) sec2x (tan2x sec2x) dx Multiply.

= ∫ (sec3 2x − sec2x) (tan2x sec2x) dx u = sec 2x , du = 2sec 2x tan 2x dx

= 1 ∫ (u3 − u) du
= 2

1  1 u4 − 1 u2  + C
2  4 2 

= 1 sec4 2x − 1 sec2 2x + C
84

198 Chapter 6

2. If there are no secant factors and m is even, convert a tangent-squared factor to a secant-squared
factor by substituting ta=n2 x sec2 x −1 .

Example 5 □ Evaluate ∫ tan4 x dx .

Solution □ ∫ tan4 x dx

∫= tan2 x (tan2 x) dx

∫= tan2 x (sec2 x −1) dx ta=n2 x sec2 x −1

= ∫ tan2 x sec2 x dx − ∫ tan2 x dx

= ∫ tan2 x sec2 x dx − ∫ (sec2 x −1) dx

= ∫ tan2 xsec2 x dx − ∫ sec2 x dx + ∫ 1 dx
=Set u ta=n x, du sec2 x dx

= 1 tan3 x − tan x + x + C
3

3. If n is even, save a secant-squared factor and convert the remaining factors to tangents,
and substitute sec2 x= 1+ tan2 x .

∫Example 6 □ Evaluate π 4 sec4 x tan2 x dx .
0

∫Solution □ sec4 x tan2 x dx A secant-squared factor is saved.
∫= sec2 x tan2 x (sec2 x) dx sec2 x= 1 + tan2 x
Multiply.
∫= (1+ tan2 x) tan2 x (sec2 x) dx u = tan x , du = sec2 x dx
∫= (tan2 x + tan4 x) (sec2 x) dx

∫= (u2 + u4 ) du

= 1 u3 + 1 u5 + C
35

= 1 tan3 x + 1 tan5 x + C
35

Therefore

∫ π 4 sec4 x tan2 x dx
0

1 1 x π 4
 3 5 0
= tan3 x + tan5

=1+1= 8
3 5 15

4. If there are no tangent factors and n is odd use integration by parts as illustrated in the section 6.5.

Techniques of Integration 199

Exercises - Trigonometric Integrals
Multiple Choice Questions

1. ∫ sin3 nx dx =

(A) 1 sin3 nx − 1 sin nx + C
3n n

(B) 1 cos3 nx − 1 cos nx + C
3n n

(C) 1 sin3 nx − 1 sin nx + C
3n n

(D) 1 cos3 nx − 1 sin nx + C
3n n

2. ∫ cos3 x sin x dx =

(A) 1 (cos x)3 − 2 (cos x)5 2 + C
35

(B) 2 (cos x)3 2 − 2 (sin x)7 2 + C
37

(C) 2 (sin x)3 2 − 2 (sin x)7 2 + C
37

(D) 2 (sin x)3 2 − 2 (cos x)5 2 + C
35

∫3. π 4sin4 θ dθ = 3π (C) 2π 5π
0 (B) (D)
(A) π
2 2

200 Chapter 6

∫4. π 4 4 tan2 θ dθ =
0

(A) 2 − π (B) 2 − π (C) 2 + π (D) 4 − π
2 2

5. ∫ sec4 x dx =

(A) 1 tan3 x + tan x + C
3

(B) 1 tan3 x + sec x + C
3

(C) 1 tan3 x + sec2 x + C
3

(D) 1 sec5 x − sec x tan x + C
5

Free Response Questions
6. Find the area bounded by the curves y = sin x and y = sin3 x between x = 0 and x = π .

2

Techniques of Integration 201

6.3 Trigonometric Substitutions

Trigonometric Substitution
1. For integrals involving a2 − u2 , let u = a sinθ . Then

a2 − u2 = a2 − a2 sin2 θ = a2 (1− sin2 θ ) = a2 cos2 θ = a cosθ

2. For integrals involving a2 + u2 , let u = a tanθ . Then
a2 + u2 = a2 + a2 tan2 θ = a2 (1+ tan2 θ ) = a2 sec2 θ = a secθ

3. For integrals involving u2 − a2 , let u = a secθ . Then
=u2 − a2 a2 sec2=θ − a2 + a2 (se=c2 θ −1) a=2 tan2 θ a tanθ

a u a2 + u2 u u u2 − a2

θ θ θ
a2 − u2 a a

u = a sinθ u = a tanθ u = a secθ
arccscx = arcsin 1
Note: arc sec x = arccos 1
x x

∫Example 1 □ Evaluate 2 4 − x2 dx .
0

Solution □ ∫ 4 − x2 dx

= ∫ 4 − 4sin2 θ 2cosθ dθ x = 2sinθ , dx = 2cosθ dθ
4 − 4sin2 θ =4(1 − sin2 θ ) =2cosθ
= 4∫ cos2θ dθ
cos2 θ = 1 + cos 2θ
= 4∫ 1+ cos 2θ dθ 2
2

= 2∫ (1+ cos 2θ ) dθ 2 x x = 2sinθ

=2(θ + 1 sin 2θ ) + C =2θ + sin 2θ + C θ
2 4 − x2

=2θ + 2sinθ cosθ + C

=2sin−1( x ) + 2( x )( 4 − x2 ) + C x = 2sinθ ⇒ sinθ = x ⇒ θ = sin−1( x )
2 22 22

= 2sin−1( x ) + x 4 − x2 + C
22

202 Chapter 6

Therefore

 4 − x2  2
=2 sin−1( 
∫2 4 − x2 dx x ) + x =2sin−1(1) =2 ⋅ π =π

0  2 2  0 2

Example 2 □ Evaluate ∫ dx .
9 + x2

Solution □∫ dx
9 + x2

=∫ 3sec2 θ dθ x = 3tanθ , dx = 3sec2 θ dθ
9 + 9 tan2 θ

=∫ 3sec2 θ dθ
3secθ

= ∫ secθ dθ 9 + x2 x x = 3tanθ

= ln secθ + tanθ + C θ
3
= ln 9 + x2 + x + C
33

∫Example 3 □ Evaluate dx .
x2 x2 − 4

∫Solution □ dx
x2 x2 − 4

=∫ 2secθ tanθ dθ x = 2secθ , dx = 2secθ tanθ dθ
4sec2 x 4sec2 θ − 4 4sec2=θ − 4 4(sec2=θ −1) 2 tanθ

=∫ 2secθ tanθ dθ
4sec2 θ 2tanθ

= 1 ∫ dθ
4 secθ

= 1 ∫ cosθ dθ x x2 − 4 x = 2secθ
4
θ
= 1 sinθ + C 2
4

= 1 x2 − 4 + C
4x

Techniques of Integration 203

Exercises - Trigonometric Substitutions
Multiple Choice Questions

1. If the substitution x = 2 tanθ is made in ∫ x3 dx , where − π < θ < π , the resulting
x2 + 4 22
integral is

(A) 4∫ tan2θ secθ dθ
∫(B) 4 tan2θ sec2 θ dθ
(C) 8∫ tan3θ dθ
(D) 8∫ tan3θ secθ dθ

∫2 1 dx =
2 x x2 −1
2.

π π π π
(A) (B) (C) (D)

18 12 6 4

∫3. 1 dx =

x2 25 − x2

(A) − 25 − x2 + C
5x2

(B) − 25 − x2 + C
25

(C) − 25 − x2 + C
25x

(D) 25 − x2 + C
25x2

204 Chapter 6

4. If the substitution x = secθ is made in ∫ x2 −1 dx , where − π < θ < π , the resulting
x4 2 2
integral is

(A) ∫ sec2 θ tanθ dθ + C
(B) ∫ secθ tan2 θ dθ + C
(C) ∫ sinθ cos2 θ dθ + C
(D) ∫ sin2 θ cosθ dθ + C

5. The average value of f (x) = 4 on the interval [0, 4] is

9 + x2

(A) ln 2 (B) ln( 2 −1) (C) ln 3 (D) ln( 2 +1)

Free Response Questions

6. Let f be the function given by f (x=) (9 − x2 )3 2 on the closed interval [0,3] .
(a) Find the average value of f on the closed interval [0,3] .

(b) Substitute x = 3sinθ for f . Set up, but do not integrate, an integral expression in terms

of θ for the average value of f on the closed interval [0,3] .

Techniques of Integration 205

6.4 L’Hospital’s Rule

L’Hospital’s Rule
Suppose f and g are differentiable and g′(x) ≠ 0 near x = c (except possibly at c ).
If the limit of f (x) as x approaches c produces the indeterminate form 0 or ∞ , then

g(x) 0 ∞
lim f ( x) = lim f ′( x)
x→c g( x) x→c g′( x)

provided the limit on the right side exists.

L’Hospital’s Rule can be applied only to quotients leading to indeterminate forms such as
0 , ∞ , 0 ⋅ ∞, 1∞ , ∞0 , 00 , and ∞ − ∞.
0∞
L’Hospital’s Rule does not apply when either the numerator or denominator has a finite
nonzero limit.

Example 1 □ Find lim ex −1 .

x→0 sin x

Solution □ lim ex −1 Indeterminate form 0 .
x→0 sin x 0
= lim ex
x→0 cos x L’Hospital’s Rule: d (ex −1) =ex , d (sin x) = cos x .
=1 dx dx

e0 = 1 and cos 0 = 1 .

Example 2 □ Find lim sec x + 9 .
x→π 2 tan x

Solution □ lim sec x + 9 ∞
x→π 2 tan x Indeterminate form .
= lim sec x tan x
x→π 2 sec2 x ∞
L’Hospital’s Rule: d (sec x + 9) =sec x tan x ,
= lim tan x
x→π 2 sec x dx
d (tan x) = sec2 x .
= lim sin x dx

x→π 2 Simplify.

=1

206 Chapter 6

Example 3 □ Find lim x tan 1 .
x→∞ x

Solution □ lim x tan 1 Indeterminate form ∞ ⋅ 0 .
x→∞ x Rewrite x = 1 .

= lim tan(1 x) 1x
x→∞ 1 x d (tan x) = sec2 x
dx
= lim (−1 x2 ) sec2 (1 x)
−1 x2 Simplify.
x→∞ sec(0) = 1

= lim sec2 (1 )
x→∞ x

=1

Example 4 □ Find lim ex .
x→0 x

Solution □ li=m ex li=m ex 1 Incorrect use of L’Hospital’s Rule
x→0 x x→0 1

This application is incorrect since e0 = 1 is not an indeterminate form.
00

ex = 1 Make direct substitution.
lim Answer
x→0 x 0

=∞

Techniques of Integration 207

Exercises - L’Hospital’s Rule
Multiple Choice Questions

1. lim ex −1− x = (B) 1 (C) 1 (D) ∞
x→0 x2 2
(A) 0

2. lim sin−1 x = (B) 0 π (D) 1
x→0 x (C)
(A) −∞
2

3. lim sinθ = (B) − 1 (C) 0 (D) 1
θ →π θ − π 2 2
(A) −1

4. lim (tan x)x = (B) 0 π (D) 1
(C)
x→0+
4
(A) −∞

5. lim (x)1 x = (B) e (C) 1 (D) ∞

x→∞

(A) −∞

208 Chapter 6

6. lim  1 x − 1  =
 tan x 
x→0

(A) −∞ (B) − 1 (C) 0 (D) ∞
2

7. lim  2 − x x =
 x2 −1 −1 
x→1−

(A) −∞ (B) − 3 (C) −1 (D) ∞
2

Free Response Questions

8. Use L’Hospital’s Rule to find the exact value of lim x[ln(x + 3) − ln x] . Show the work that leads
x→∞
to your answer.

9. Use L’Hospital’s Rule to find the exact value of lim  x − x2 + x  . Show the work that leads
 
x→∞

to your answer.

Techniques of Integration 209

6.5 Integration by Partial Fractions BC

A method for rewriting a rational function into the sum of simpler rational functions is called
the method of partial fractions. For instance, the rational function 5x +1 can be written

x2 + x − 2
as 5x +=1 2 + 3 .

x2 + x − 2 x −1 x + 2
We call the fractions 2 and 3 partial fractions because their denominators are only

x −1 x + 2
part of the original denominator. To integrate the rational function 5x +1 , we simply sum

x2 + x − 2
the integrals of the partial fractions.

∫ 5x +1 =dx ∫ x 2 dx + ∫ 3 =dx 2 ln x −1 + 3ln x + 2 + C
x2 + x − 2 −1 x+2

∫Example 1 □ Evaluate x3
dx .

x2 −1

Solution □ If the degree of the numerator is greater than or equal to the degree of
the denominator, divide numerator by the denominator to get
a polynomial plus a proper fraction.

x

x2 −1 x3
− x3 − x
x

Then we write the improper fraction as a polynomial plus a proper fraction.

x3 = x + x
x2 −1 x2 −1

Write =x x= A+B .
x2 −1 (x +1)(x −1) x +1 x −1

Multiplying both sides by (x +1)(x −1) , we have x= A(x −1) + B(x +1)

Let x = 1 . Then 1= A(1−1) + B(1+1) ⇒ B = 1 .
2

Let x = −1 . Then −1= A(−1−1) + B(−1+1) ⇒ A = 1 .
2

Thus, ∫ x x3 1 dx =∫ x dx + 1 ∫ x 1 dx + 1 ∫ 1 dx
2− 2 +1 2 x −1

=1 x2 + 1 ln x +1 + 1 ln x −1 + C .
22 2

210 Chapter 6

Example 2 □ Evaluate ∫ x +10 dx .
(x − 4)(x + 3)

Solution □ If the denominator is a product of distinct linear factors, then the partial
fraction decomposition has the form

x +10= A + B .
(x − 4)(x + 3) x − 4 x + 3

To find the values of A and B, multiply both sides of the equation by the
least common denominator and get
x +10= A(x + 3) + B(x − 4) .

To solve for A, choose x = 4 , to eliminate the term B(x − 4) .
4 +10= A(4 + 3) + B(4 − 4) ⇒ 14 = 7 A ⇒ A = 2

To solve for B, choose x = −3 , to eliminate the term A(x + 3) .
−3 +10 = A(−3 + 3) + B(−3 − 4) ⇒ 7 = −7B ⇒ B = −1

∫ x=+10 dx ∫ 2 4) dx + ∫ −1 dx
(x − 4)(x + 3) (x − (x + 3)

= 2 ln x − 4 − ln x + 3 + C

Techniques of Integration 211

Exercises - Integration by Partial Fractions BC
Multiple Choice Questions

1. ∫ dx =
x2 + x − 6

(A) 1 ln x −1 + C
5 x+6

(B) 1 ln x + 3 + C
5 x−2

(C) 1 ln x − 2 + C
5 x+3

(D) 1 ln (x − 2)(x + 3) + C
5

∫7 5 dx =

2. 4 (x − 2)(2x +1)

(A) ln 9 (B) ln 10 (C) ln 3 (D) ln 9
10 9 2 4

3. ∫ x dx =
x2 + 5x + 6

(A) −2 ln x + 2 + 3ln (x + 3) + C
(B) 2 ln x + 2 + 3ln (x + 3) + C
(C) 2 ln (x + 3) − 3ln x + 2 + C
(D) −2 ln (x + 3) − 3ln x + 2 + C

212 Chapter 6

∫4. 2e2x dx =
(ex −1)(ex +1)

(A) ln ex (e2x −1) + C

(B) ln 2ex (e2x −1) + C

(C) ln 1 +C
e2x −1

(D) ln (ex −1)(ex +1) + C

Free Response Questions

5. Let f be the function given by f (θ ) = ∫ sinθ dθ .
cosθ (cosθ −1)

(a) Substitute x = cosθ and write an integral expression for f in terms of x .
(b) Use the method of partial fractions to find f (θ ) .

Techniques of Integration 213

6.6 Integration by Parts BC

Integration by Parts Formula
If u and v are functions of x and have continuous derivatives, then

∫ u d=v uv − ∫ v du .

Guidelines for Integration by Parts
1. For integrals of the form

∫ ∫ ∫xneax dx , xn sin ax dx , or xn cos ax dx

let u = xn and let dv = eax dx, sin ax dx, or cos ax dx.

2. For integrals of the form

∫ xn ln x dx , ∫ xn arcsin ax dx , ∫ xn arccos ax dx , or ∫ xn arctan ax dx

let u = ln x, arcsin ax, arccos ax, or arctan ax and let dv = xn dx.

3. For integrals of the form

∫ ∫eax sin bx dx or eax cos bx dx

let u = sin bx or cos bx and let dv = eax dx.

Example 1 □ Find ∫ x sin x dx .

Solution □ Let u = x and dv = sin x dx .

Then du = dx and v = ∫ sin x dx = − cos x .

u dv u v  v  du

∫ x sin x dx =x (− cos x) − ∫ (− cos x) dx

=−x cos x + ∫ cos x dx

=−x cos x + sin x + C

Example 2 □ Evaluate ∫ arctan x dx .

Solution □ Let u = arctan x and dv = dx .

Then du = 1 dx an=d v ∫=dx x
1+ x2

∫ ∫u  dv u  v v ⋅ du 
x 1 dx
arctan x=dx arctan x ⋅ x − 1+ x2

= x arctan x − ∫ x dx
1+ x2

= x arctan x − 1 ln(1+ x2 ) + C
2

214 Chapter 6

Tabular Method

In problems involving repeated applications of integration by parts, a tabular method can help

∫ ∫to organized the work. This method works well for integrals of the form xneax dx , xn sin ax dx ,
and ∫ xn cos ax dx .

∫Example 3 □ Evaluate x2ex dx .

Solution □ Let u = x2 and dv = ex dx .

u and its v′ and its
Derivatives Antiderivatives

x2 + ex

2x − ex +x2ex
2+ ex −2xex
0 ex +2ex

Differentiate until you obtain 0 as aderivative.

∫Hence, x2ex dx = x2ex − 2xex + 2ex + C

∫Example 4 □ Evaluate e2x cos x dx .

Solution □ Let u = cos x and dv = e2x dx .

u and its v′ and its
Derivatives Antiderivatives

cos x + e2x + 1 e2x cos x
− sin x − 1 e2x 2
− cos x + 2
1 e2x − 1 e2x (− sin x)
4 4

Stop here. The variable part of the last row is same as the first row.

We stop differentiating and integrating as soon as we reach a row that is the same
as the first row except for multiplicative constants. The table is interpreted as follow.

Product of thelastrow

∫ ∫e2x cos x=dx 1 e2x cos x − 1 e2x (− sin x) + 1 e2x (− cos x) dx
24 4

∫By adding 1 e2x (cos x) dx on each side we get,

4

∫5 e2=x cos x dx 1 e2x cos x + 1 e2x (sin x) .

4 24

∫Therefore, e2x cos x dx = 2 e2x cos x + 1 e2x sin x + C .
55

Techniques of Integration 215

Exercises - Integration by Parts BC
Multiple Choice Questions

1. ∫ x sin(2x) dx =

(A) −x cos(2x) + 1 sin(2x) + C
2

(B) x cos(2x) − 1 sin(2x) + C
24

(C) − x cos(2x) + 1 sin(2x) + C
24

(D) x cos(2x) + 1 sin(2x) + C
24

∫2. 2 xex dx = (B) e2 +1 (C) e −1 (D) e +1
0
(A) e2 −1

3. If ∫ x2 cos(3x)=dx f (x) − 2 ∫ x sin(3x) dx , then f (x) =
3

(A) 2 x sin(3x)
3

(B) 1 x2 sin(3x)
3

(C) 2 x cos(3x)
3

(D) 1 x sin(3x) − 2 cos(3x)
33

216 Chapter 6

4. ∫ x2 ln x dx =

(A) x2 ln x − x2 + C
24

(B) x3 ln x − x3 + C
3

(C) x3 ln x − x3 + C
39

(D) x(ln x)2 − x3 + C
23

∫5. π 4 x sec2 x dx =
0

(A) π − ln 2 (B) π + ln 2 (C) π − ln 2 (D) π + ln 2
4 4 42 42

6. ∫ sec3 x dx =

(A) 1 sec4 x + C
4

(B) 1 sec2 x tan x + 1 ln sec x + C
22

(C) 1 sec2 x tan x + 1 ln tan x + C
22

(D) 1 sec x tan x + 1 ln sec x + tan x + C
22

Techniques of Integration 217

7. ∫ f (x) cos(nx) dx =

(A) 1 f (x) sin(nx) − 1 ∫ f ′(x) sin(nx) dx
n n

(B) 1 f (x) cos(nx) − 1 ∫ f ′(x) cos(nx) dx
n n

(C) n f (x) cos(nx) + 1 ∫ f ′(x) sin(nx) dx
n

(D) n f (x) cos(nx) − 1 ∫ f ′(x) cos(nx) dx
n

8. If ∫ arc=cos x dx x arccos x + ∫ f (x) dx , then f (x) =

(A) −x 1− x2 (B) x 1− x2 (C) − 1 (D) x
1− x2 1− x2

x f (x) g(x) f ′(x) g′(x)

1 −2 3 4 −1

3 2 −1 −3 5

9. The table above gives values of f , f ′ , g , and g′ for selected values of x .

∫ ∫If 3 f (x)g′(x) dx = 8 , then 3 f ′(x)g(x) dx =
11

(A) −4 (B) −1 (C) 5 (D) 8

218 Chapter 6

Free Response Questions
10. Find the area of the region bounded by y = arcsin x , y = 0 , and x = 1 . Show the work

that leads to your answer.

Techniques of Integration 219

6.7 Improper Integrals BC

Improper Integrals with Infinite Integration Limits

1. If f (x) is continuous on [a, ∞) , then

∫ ∫∞ b

f ( x) dx = lim f ( x) dx .
a b→∞ a

2. If f (x) is continuous on (−∞,b] , then

∫ ∫b f ( x) dx = lim b

f ( x) dx .
−∞ a→−∞ a

3. If f (x) is continuous on (−∞, ∞) , then

∫ ∫ ∫∞ c ∞

=f ( x) dx f ( x) dx + f ( x) dx ,

−∞ −∞ c

where c is any real number.

Improper Integrals with Infinite Discontinuities
1. If f (x) is continuous on [a,b) and has an infinite discontinuity at b , then

∫ ∫b c

f ( x) dx = lim f ( x) dx .

a c→b− a

2. If f (x) is continuous on (a,b] and has an infinite discontinuity at a , then

∫ ∫b b

f ( x) dx = lim f ( x) dx .

a c→a+ c

3. If f (x) is continuous on [a,b] , except for some number c in (a,b) at which f has an
infinite discontinuity, then

∫ ∫ ∫b c b

=f ( x) dx f ( x) dx + f ( x) dx ,

a ac

where c is any real number.

In each case, if the limit is finite we say that the improper integral converges and that
the limit is the value of the improper integral. If the limit fails to exists, the improper
integral diverges.

∫Example 1 □ Evaluate ∞ xe−x2 dx .
0

Solution ∫□ ∞ xe−x2 dx
0

∫= b xe−x2=dx lim − 1 e−x2  b
lim 2  0
0 b→∞
b→∞

=− 1 lim  e −b2 − e0  =− 1 (0 −1) =1
2   2 2
b→∞

220 Chapter 6

∫Example 2□ ∞ dx .
Evaluate −∞ 1 + x2

=∞ dx 0 dx + ∞ dx
−∞ 1+ x2 −∞ 1+ x2 0 1 + x2
Solution ∫ ∫ ∫□

∫ ∫= lim 0 dx + lim b dx

a→−∞ a 1 + x2 b→∞ 0 1 + x2

= lim  tan −1 x 0 + lim tan −1 x b
a 0
a→−∞ b→∞

( ) ( )= lim tan−1 0 − tan−1 a + lim tan−1 b − tan−1 0
a→−∞ b→∞

=  0 − (− π )  +  π − 0 = π
 2   2

∫5 dx .

Example 3 □ Evaluate 1 x −1

∫Solution □ 5 dx
1 x −1

∫= 5=dx 2 x − 1 5
lim b x −1 lim b

b→1+ b→1+

= lim 2 4−2 b −1

b→1+

=4

∫Example 4 □ Find 1 dx .
0 1− x

Solution ∫□ 1 dx

0 1− x

∫= b dx= − − x  b
lim 0 1− x lim ln 1 0

b→1− b→1−

= lim − ln 1−b + ln1

b→1−

=∞

Techniques of Integration 221

Exercises - Improper Integrals BC
Multiple Choice Questions

∫1. ∞ 1 dx = (B) −2 (C) 1 (D) ∞
2 x −1
(A) −∞

∫∞ 1 dx =

2. 0 (x + 3)(x + 4)

(A) − ln 4 (B) − ln 3 (C) 0 (D) ln 4
3 4

∫4 dx =
0 (x −1)2 3
3.

(A) 33 3 (B) 3(1− 3 3) (C) 3(1+ 3 3) (D) divergent

∫4. ∞ x2e−x3 = (B) 1 (C) 1 (D) divergent
0 2
(A) 1
3

222 Chapter 6
(D) divergent
∫5. 1 ln x dx = (B) −4 (C) −2

0x
(A) −6

∫6. If 1 ke− x dx = 1, what is the value of k ?

0x

(A) − 1 (B) e (C) 1 (D) There is no such value of k
2 2 2

Free Response Questions

7. Let f be the function given by f (x) = x dx .
x2 +1

∫∞

(a) Show that the improper integral f (x) dx is divergent.

1

(b) Find the average value of f on the interval [1, ∞) .

Chapter 7
Further Applications of Integration

7.1 Slope Field
A first order differential equation of the form y′ = f ( x, y) says that the slope of a solution
curve at a point (x, y) on the curve is f (x, y) . If we draw short line segments with slope
f (x, y) at several points (x, y) , the result is called a slope field.
Figure 7-1 shows a slope field for the differential equation y′ = x − y +1
Figure 7-2 shows a particular solution curve through the point (0,1) .

yy

xx

Slope field for y′ = x − y + 1 Particular solution for y′ = x − y + 1
Figure 7-1 passing through (0,1)

Figure 7-2

Example 1 □ On the axes provided, sketch a slope field y
for the differential equation y′= 1− xy . 2
1
Solution □ Make a table showing the slope at the points
shown on the graph. x

−1 O 1
−1

x −1 −1 −1 −1 0 0 0 0 1 1 1 1
y −1 0 1 2 −1 0 1 2 −1 0 1 2
y′= 1 − xy 0 1 2 3 1 1 1 1 2 1 0 −1

224 Chapter 7

Draw the line segments at the points with their respective slopes.

y
2

1

x

−1 O 1
−1

Example 2 □ On the axes provided, sketch a slope field y
for the differential equation y′= y + xy . 2
1
Solution □ Make a table showing the slope at the points
shown on the graph. x

−1 O 1
−1

x −1 −1 −1 −1 0 0 0 0 1 1 1 1
y −1 0 1 2 −1 0 1 2 −1 0 1 2
y′= y + xy 0 0 0 0 −1 0 1 2 −2 0 2 4

Draw the line segments at the points with their respective slopes.

y
2

1

x

−1 O 1
−1

Further Applications of Integration 225

Exercises - Slope Field
Multiple Choice Questions

y
6

4

2 x
24 6
−6 −4 −2
−2

−4

−6

1. Shown above is a slope field for which of the following differential equations?

(A) dy = x (B) dy = − x (C) dy = x2 (D) dy = − x2
dx y dx y dx y dx y

y
6

4

2 x
24 6
−6 −4 −2
−2

−4

−6

2. Shown above is a slope field for which of the following differential equations?

(A) dy= x + y (B) dy= x − y (C) dy =−x + y (D) d=y x2 − y
dx dx dx dx

226 Chapter 7

Free Response Questions
3. On the axis provided, sketch a slope field for the differential equation dy= y − x2 .

dx

y

x

−1 O 1

4. On the axis provided, sketch a slope field for the differential equation d=y x2 + y2 .
dx

y

x

−1 O 1

5. On the axis provided, sketch a slope field for the differential equation dy =(x +1)( y − 2) .
dx

y

x

−1 O 1

Further Applications of Integration 227

7.2 Separable Differential Equations

The equation y′ = f (x, y) is a separable equation if all x terms can be collected with dx

and all y terms with dy . The differential equation then has the form

dy = f ( x)g( y) or dy = f (x)
.
dx dx h( y)

To solve the first equation we could rewrite it in the form dy = f (x)dx , and integrate
g( y)

both sides of the equation:

∫ dy = ∫ f ( x)dx .
g( y)

To solve the second equation we could rewrite it in the form h( y)dy = f (x)dx
and integrate both sides of the equation:

∫ h( y) dy = ∫ f ( x) dx .

Example 1 □ Find the general solution of (x + 3) y′ =2 y .

Solution □ (x + 3) y′ =2 y

(x + 3) dy =2 y Rewrite y′ as dy .
dx dx

dy = 2 dx Separate the variables.
y x+3 Integrate.

∫ dy = ∫ 2 dx Let C1 = ln C .
y x+3 General solution

ln =y 2 ln x + 3 + C1
= ln(x + 3)2 + ln C

=y C(x + 3)2

Example 2 □ Find the general solution of dy = − 2x .
dx y

Solution □ dy = − 2x
dx y
y dy = −2x dx Separate the variables.
Integrate.
∫ y d=y ∫ − 2x dx
General solution, C = 2C1
1 y2 =− x 2 + C1
2

2x2 + y2 =C

228 Chapter 7

Exercises - Separable Differential Equations
Multiple Choice Questions

1. The solution to the differential equation dy = 3x2 , where y(3) = 4 , is
dx 2 y

(A=) y x3 +1 (B) y= 7 − x3 (C)=y x3 − 9 (D)=y x3 −11
3 3

2. If dy = x + sec2 x and y(0) = 2 , then y =
dx y

(A) x2 + 2sec x + 2
(B) x2 + 2 tan x + 4
(C) x2 + sec2 x + 2
(D) x2 + tan2 x + 4

3. At each point (x, y) on a certain curve, the slope of the curve is xy . If the curve contains the
point (0, −1) , which of the following is the equation for the curve?

(A) =y x2 − 2 (B)=y 3x2 − 4 x2 (D) y = −e(x2 −1)

(C) y = −e 2

4. If dy= ( y − 4) sec2 x and y(0) = 5 , then y =
dx

(A) etan x + 4 (B) 6etan x −1 (C) 2etan x + 2 (D) 4sec x +1

Further Applications of Integration 229

5. What is the value of m + b , if =y mx + b is a solution to the differential equation dy = 1 x − y +1?
dx 4

(A) 1 (B) 3 (C) 1 (D) 5
2 4 4

Free Response Questions

6. Consider the differential equation dy = x +1 .
dx y

(a) On the axis provided sketch a slope field for the given differential equation at the nine points
indicated.

y x
1

O 12
−1

(b) Let y = f (x) be the particular solution to the differential equation with the initial condition y(1) = 3 .
Write an equation for the line tangent to the graph of f at (1, 3) and use it to approximate f (1.2) .

(c) Find the particular solution y = f (x) to the differential equation with the initial condition y(1) = 3 .
(d) Use your solution from part (c) to find f (1.2) .

230 Chapter 7

7. Consider the differential equation dy = 2x + 3 .
dx ey

(a) Let y = f (x) be the particular solution to the differential equation with the initial condition y(0) = 2 .
Write an equation for the line tangent to the graph of f at (0, 2) .

(b) Find f ′′(0) with the initial condition y(0) = 2 .
(c) Find the particular solution y = f (x) to the differential equation dy = 2x + 3 with the initial

dx ey
condition y(0) = 2 .

8. Consider the differential equation dy = y2 (1− 2x) .
dx 3

(a) On the axis provided sketch a slope field for the given differential equation at the nine points
indicated.

y

x

−1 O 1

(b) Find d2y in terms of x and y.
dx2

(c) Let y = f (x) be the particular solution to the differential equation with the initial condition y(1) = 4 .
2

Does f have a relative minimum, a relative maximum, or neither at x = 1 ? Justify your answer.
2

(d) Find the particular solution y = f (x) to the differential equation with the initial condition y(1) = 4 .
2

Further Applications of Integration 231

9. Consider the differential equation dy =−2x + y +1.
dx

(a) On the axis provided sketch a slope field for the given differential equation at the nine points
indicated.

y

x

−1 O 1

(b) Find d2y in terms of x and y . Describe the region in the xy- plane in which all the solution
dx2

curves to the differential equation are concave down.

(c) Let y = f (x) be the particular solution to the differential equation with the initial condition f (0) = −1 .
Does f have a relative minimum, a relative maximum, or neither at x = 0 ? Justify your answer.

(d) Find the value of the constants m and b , for which =y mx + b is a solution to the differential
equation.

232 Chapter 7

7.3 Exponential Growth and Decay
In modeling many real-world situations, a quantity y increases or decreases at a rate proportional to
its size at a given time t . If y is a function of time t , the proportion can be written as follows.
Rate of change of y is proportional to y.

dy = ky
dt

The Law of Exponential Change
If y is a differentiable function of t such that y > 0 and y′ = ky , for some constant k ,
then

y = y0ekt , where y0 is the initial value of y .
Exponential growth occurs when k > 0 , and exponential decay occurs when k < 0 .
The number k is the rate constant of the equation.

Example 1 □ The number of bacteria in a culture increases at a rate proportional to the number
present. If the number of bacteria was 600 after 3 hours and 19,200 after 8 hours,

when will the population reach 120,000?

Solution □ Since the growth rate is proportional to population size, we use the equation y = y0ekt .

600 = y0ek⋅3 y = 600 and t = 3

y0 = 600 Solve for y0 .
e3k y = 19, 200 and t = 8

19, 200 = y0ek⋅8

19, 200 = 600 e8k Substitution
e3k

32 = e5k Simplify.

=k 1 ln 32 ≈ 0.693
5

Therefore, the exponential growth model is y = y0e0.693t .
To solve for y0 , substitute y = 600 when t = 3 and obtain
600 = y0e0.693(3) .

=y0 600 ≈ 75
e0.693(3)

So the model is y = 75e0.693t .

120, 000 = 75e0.693t ⇒ 120, 000 = e0.693t
75

⇒ e0.693t = 1600 ⇒ 0.693t = ln1600

⇒ t ≈ 10.646

Further Applications of Integration 233

Exercises - Exponential Growth and Decay
Multiple Choice Questions

1. Bacteria in a certain culture increase at a rate proportional to the number present. If the number of
bacteria doubles every four hours, in how many hours will the number of bacteria triple?

(A) ln( 27) (B) ln(81) (C) 4 ln 2 (D) 4 ln 3
2 2 ln 3 ln 2

2. Population y grows according to the equation dy = ky , where k is a constant and t is measured
dt

in years. If the population doubles every 15 years what is the value of k ?

(A) 0.035 (B) 0.046 (C) 0 .069 (D) 0.078

3. A baby weighs 6 pounds at birth and 9 pounds three months later. If the weight of baby increasing
at a rate proportional to its weight, then how much will the baby weigh when she is 6 months old?

(A) 11.9 (B) 12.8 (C) 13.5 (D) 14.6

4. Temperature F changes according to the differential equation dF = kF , where k is a constant and t
dt

is measured in minutes. If at time t = 0 , F = 180 and at time t = 16 , F = 120 , what is the value of k ?

(A) −0.025 (B) −0.032 (C) −0.045 (D) −0.058

234 Chapter 7

Free Response Questions

5. The rate at which the amount of coffee in a coffeepot changes with time is given by the differential
equation dV = kV , where V is the amount of coffee left in the coffeepot at any time t seconds. At
dt
time t = 0 there were 16 ounces of coffee in the coffeepot and at time t = 80 there were 8 ounces of
coffee remaining in the pot.
(a) Write an equation for V , the amount of coffee remaining in the pot at any time t .
(b) At what rate is the amount of coffee in the pot decreasing when there are 4 ounces of coffee remaining?
(c) At what time t will the pot have 2 ounces of coffee remaining?

Further Applications of Integration 235

7.4 Logistic Equations BC

The differential equation

=dP kP 1 − P 
dt A 

is called a logistic equation.

In this equation P(t) is the size of the population at time t , A is the carrying capacity
(the maximum population that the environment is capable of sustaining in the long run), and
k is a constant.

If 0 < P < A , then (1− P A) is positive so, dP dt > 0 and the population increases.
If P > A , then (1− P A) is negative, so dP dt < 0 the population decreases.

In logistic equations
1. lim dP = 0

t→∞ dt

2. lim P(t) = A .

t →∞

3. The population is growing the fastest when P = A .
2

(When P is half the carrying capacity.)
4. The graph of P(t) has a point of inflection at the point where P = A .

2

Figure 7-3 displays typical logistic curves.

P

P=A

population P= A
2

t

O time
Figure 7-3

Solution curves for the logistic equations
with different initial conditions

236 Chapter 7

Example 1 □ A population is modeled by a function P that satisfies the logistic differential

equation =dP P  3 − P  , where the initial population P(0) = 100 and t is the
dt 2  20 

time in years.

(a) What is lim P(t) ?

t 

(b) For what values of P is the population growing the fastest?

(c) Find the slope of the graph of P at the point of inflection.

Solution □ (a) Write the differential equation in the standard form.

dP = P  3 − P  = 3P 1 − P 
dt 2  20  2 60 

lim P(t=) A= 60

t→∞

(b) The population is growing the fastest when P = A .
2

P= A= 60= 30
22

(c) The graph of P has a point of inflection at P = A .
2

So, when P = 30 ,

dP = 30  3 − 30  = 22.5
dt P=30 2  20 

Further Applications of Integration 237

Exercises - Logistic Equations BC
Multiple Choice Questions

1. The population P(t) of a species satisfies the logistic differential equation d=P 3P − 0.0006P2 ,
dt

where the initial population is P(0) = 1000 and t is the time in years. What is lim P(t) ?

t→∞

(A) 1000 (B) 2000 (C) 3000 (D) 5000

2. A healthy population P(t) of animals satisfies the logistic differential equation =dP 5P(1− P ) ,
dt 240

where the initial population is P(0) = 150 and t is the time in years. For what value of P is the
population growing the fastest?

(A) 48 (B) 60 (C) 120 (D) 240

3. A population is modeled by a function P that satisfies the logistic differential equation

=dP P 1 − P  , where the initial population is P(0) = 800 and t is the time in years.
dt 5 150 

What is the slope of the graph of P at the point of inflection?

(A) 5 (B) 7.5 (C) 10 (D) 12.5

4. A certain rumor spreads in a small town at the rate d=y y (1− 3y) , where y is the fraction of the

dt
population that has heard the rumor at any time t . What fraction of the population has heard the
rumor when it is spreading the fastest?

(A) 1 (B) 1 (C) 1 (D) 1
6 5 4 3

238 Chapter 7

P(t)
120

60

t

O

5. Which of the following differential equations for population P could model the logistic growth shown
in the figure above
(A=) dP 0.03P2 − 0.0005P
dt
(B=) dP 0.03P2 − 0.000125P
dt
(C)=dP 0.03P − 0.001P2
dt
(D)=dP 0.03P − 0.00025P2
dt

Further Applications of Integration 239

Free Response Questions

6. Let f be a function with f (2) = 1, such that all points (t, y) on the graph of f satisfy the

differential equation=dy 2 y 1 − t  .
dt 4 

Let g be a function with g(2) = 2 , such that all points (t, y) on the graph of g satisfy the

logistic differential equation =dy y 1 − y  .
dt 5 

(a) Find y = f (t) .

(b) For the function found in part (a), what is lim f (t) ?

t→∞

(c) Given that g(2) = 2 , find lim g(t) and lim g′(t) .
t→∞ t→∞

(d) For what value of y does the graph of g have a point of inflection? Find the slope of
the graph of g at the point of inflection.

240 Chapter 7

7.5 Euler’s Method BC

Euler’s Method is a numerical approach to approximate the particular solution of the

differential equation y′ = f (x, y) with an initial condition y(x0 ) = y0 .

Using a small step h and (x0 , y0 ) as a starting point, move along the tangent line until you

arrive at the point (x1, y1) , where

x=1 x0 + h and y=1 y0 + h  dy  ,
 dx 
( x0 , y0 )

as shown in Figure 7-4.

Repeat the process with the same step size h at a new starting point (x1, y1) . The values

of xi and yi are as follows.

x=1 x0 + h y=1 y0 + h  dy 
 dx 
( x0 , y0 )

x=2 x1 + h y=2 y1 + h  dy 
   dx 
( x1 , y1 )

=xn xn−1 + h =yn yn−1 + h  dy 
 dx 
( xn−1 , yn−1 )

y Euler h  dy 
approximation lines  dx  (x2 , y2 )
O
Exact
solution

h

h h  dy 
 dx  (x1, y1 )
h

h  dy 
 dx  (x0 , y0 )

x

x0 x1 x2 x3

Figure 7-4

Example 1 □ Let f be the function whose graph goes through the point (1, −1) and
whose derivative is given y′= 2 − y . Use Euler’s method starting at x = 1
x
with a step size of 0.5 to approximate f (3) .

Solution □ Given dy= 2− y , x0 = 1 , y0 = −1 , and h = 0.5 .
dx x