integration

Learning Objectives:
In this chapter, you will learn about
• the concept of indefinite integral
Learning Outcomes:
•Determine integrals by reversing differentiation
Integration


if dy = f (x), then y =∫ f (x) dx dx
y = mx + c y =2x
dy = 2 dx
∫2dx= 2x+c y =2x+c
y =2x+3
y =2x−4
dy 2x+c
dx = 2
∫ 2 dx =
y =2x+c
2x+c
∫ 2 dx =
y =2x+c
dy
dx = 2


3.1.2 Integration of algebraic expressions 3.1.2 (a) Integral of Constant
∫mdx= mx+c
Integrate (a) 8 (b) 3.5 (c) −
8x+c 3.5x+c
3 3
∫− 4dx= −4x+c

3 4
∫8dx = ∫3.5 dx =


3.1.2 Integration of algebraic expressions
axn
During differentiation, we carry out two operations on each term in x: multiply the term with the index, and reduce the index by 1.
dy(axn)= naxn−1 dx
3.1.2 (b) Integral of
Differentiation
Integration
2x2 ∫2xdx= 2+c=
x2 +c
dy =2x dx
dy =2x2 dx
∫2x2 dx=
2x3 3+c
4
3 2x4 1
∫2xdx=
+c= x4+c 2
dy =2x3 dx
dy =2x4 dx
4 2x5 ∫2x dx= 5 +c=
dy = nxn dx
∫axn dx= axn+1 n+1 +c


Integrate each of the following with respect of x:
(a) x8
∫x8 dx= x8+1
8+1+c
(b)x64
6 6x−4dx
x9
= +c
∫x4 dx= ∫ 6x−4+1
9
= −4 +1 + c 6x−3
= −3 +c =−2x−3 +c
=−23 +c x


If the derivative of a function is given as dx dy = 1 ,
dx 9x3
dy = 1 x−3, dx 9
9x3
find the function y.
1 1 x−3+1
x−3 =9−3+1+c
∫
x−2
9
= −18 + c
=−1+c 18x2
dy = 1 ,


3.1.3 Determine the constant of Integration
y = mx + c y = 2x
dy = 2 dx
∫ 2 dx =
y =2x+c
y =2x−4
2x+c
dx
2x+c y =2x+c
y =2x+3 dy=2 2x+c
dy = 2 dx
∫ 2 dx =
y =2x+c
∫ 2 dx =


If dy =3x2 −6x+4 and y=5 when x=3, find the value of y when dx
x=5
y=∫ 3x2 −6x+4dx 3x3 6x2
5=27 −27+12+c 5 = 12 + c
−7 = c
y =3x3 −3x2 +4x+ c−−−−(1) therefore,
y=3 −2+4x+c
Subsitute x=3 and y=5 into (1)
5=3(3)3 −3(3)2 +4(3)+c
y=3x3 −3x2 +4x−7 when x=5
y=3(5)3 −3(5)2 +4(5)−7 y = 63


3.1.4 Equations of curve from functions of gradients
dy = 4x+5 dx
The curve passing through the point (-1, 2)
x=-1 when y=2
2=2(−1)2 +5(−1)+c 2=2−5+c
The gradient of a curve passing through the point (-1, 2) is given by Find the equation of curve.
since dy=4x+5, dx
by integration,
y = ∫4x+5dx 4x2
y= 2 +5x+c y=2x2 +5x+c
5=c
y=2x2 +5x+5
The equation of the curve is
y=2x2 +5x+5


18246–0 428160
81246–0 214860
The gradient function of a curve passing through the point (-1, 2) and (0,k) is 3x2 −10x. Find the value of k.
y 10
(0, k)
(-1, 2)
–10 –8 –6 –4 –2
8 6 4 2
–2 –4 –6 –8
– 10
2 4
6 8 10
x


dy = 3x2 −10x, dx
Therefore, the equation of the curve is
y=x3−5x2 +8 At point (0, k),
k=03−5(0)2 +8 k=8
y = ∫ 3x2 −10x dx 3x3 10x2
y=3−2+c y=x3−5x2 +c
The curve pass through (-1, 2)
Therefore, 2=(−1)3−5(−1)2 +c
8=c


Exercise
3-1-09 t0 6-1-09
1. Given that dy = 2x+7 and that y=5 when x= -1, find the value of y when x=2 dx
1. Giventhat dv =3−2t andthatv=2whent=1,findthevalueof v whent=2. dt


3.1.5 Integrate by substitution Find the integration by substitution
∫(2x+5)4 dx let u=2x+5
du Hence, dx = 2
1 dx= 2du
∫u4 .1du 2
∫u4 .1du=1u5 +c 2 25
 
= 1 u5 + c
10
= 1(2x+5)5+c 10


3.1.5 Integrate by substitution Find the integration by substitution
∫ 1 dx (3x−2)2
∫ 1 dx= ∫ 3x−2 −2dx (3x−2)2 ( )
let u=3x−2
=∫u−2.1du du 3
Hence,
= 3 dx=13du
1 u −1 =3−1 +c
dx
=−1 +c 3u
=−1+c 3(3x−2)


∫
∫
= (2x+5)4+1 (4+1)(2)
3.1.5 (a) Integral of (ax + b)n
(ax+b)n+1 (n+1)(a)
(ax+b)n dx=
(2x+5)4 dx
(2x+5)5
= 10 +c
= 1 (2x+5)5 +c 10


The slope of a curve at any point P(x, y) is given by (3 − x)5. Find the equation of the curve given that its passes through the point ( 4, -3)
dy = (3− x)5 dx
The gradient of the curve, Integrate with respect to x, we have
y = ∫ 3− x 5dx ()
(3− x)6
y= 6(−1) +c
y=−1(3−x)6 +c 6
Since the curve passes through (4, -3)
−3=−1(3−4)6 +c 17 6
−6=+c Therefore,
The equation of the curve is
y=−1(3−x)6 −17 or 66
6y=−(3−x)6 −17


Area under a curve


3.2.2(b) Area under a curve bounded by x= a and x=b Integration as Summation of Area
A
A= ∫b y dx a
The area A under a curve by y = f(x) bounded by the x-axis from x=a to x=b is given by


3.2.2(b) Area under a curve bounded by x= a and x=b
A
x3 2  1
322
A=2 +2(2) −1 +2(1)
y=x2 +2
A=  3 +2x
12 33  1  
A=∫2ydx A=8+4 −1+2 3  3 
1 
A=∫2x2 +2dx A=13unit2 13


Step (1) Find x-intercept On the x-axis, y =0
y=x2 −2x 0 = x2 −2x
x2 −2x=0
 0  x3 2 2
2
x(x − 2) = 0
x=0 andx=2
A= ∫2 x2 −2xdx 0
ab2
0
x3 2x2  A=3−2
A=3−x  0
23 2 2 02 2  A=3−2−3−0
 0   A = − 43 u n i t 2
A = 43 u n i t 2


Area under a curve bounded by
y=x2 +4x+4


Area under a curve bounded by
y = ( x − 1) 2


The area under a curve which is enclosed byy=aandy=bis
A= ∫b x dy a


2 1
A =  3 − 3  −  3 −1  
x=y2 −2y A
33
3 2 1 2
Theareaunderacurveis
A=[9−9]−1 −1 3
 A= 0 −−2 
2 []3 2
A=∫y −2ydy 1
y3 2y2 2 A=3−2
 1

2
2 3
A= unit


Area under a curve bounded by curve


Area under a curve bounded


Exercise 19-2-09 23-2-09
2
1


Exercise Text Book Page 71 23-2-09
10- ( a) ( b ) ( c )
11- ( a) ( b ) ( c )
12- ( a) ( b )


Exercise Text Book Page 72
13- ( a) ( b ) ( c )
14
17 ( a ) (b ) ( c )


Volume of Revolutions


The resulting solid is a cone
To find this volume, we could take slices (the yellow disk shown above), each dx wide and radius y:


The volume of a cylinder is given by V = πr2h
Because radius = r = y and each disk is dx high, we notice that the volume of each slice is:
V = πy2dx
Adding the volumes of the disks (with infinitely small dx), we obtain the
formula:
y = f(x) is the equation of the curve whose area is being rotated a and b are the limits of the area being rotated
dx show that the area is being rotated abount the x-axis.




Example 2
Find the volume if the area bounded by the curve y = x3 + 1, the x-axis and the limits of x = 0 and x = 3 is rotated around the x-axis..


When the shaded area is rotated 360° about the x-axis, we again observe that a volume is generated:


y = 2x
V =π4x3 2 3
 1
4(2)3  4(1)3 
V=π 3 − 3     
∫2
V =π 1 y2dx
3 3    
V =π 28 3
∫2
V=π (2x)2dx
1
V =π∫24x2dx 1

 V =π32−4
V =91π unit3 3