U Ye Min Htoo(YTU)
Grade 12
Mathematics
CHAPTER 11
Application of Integration
Rules for Definite Integration
Suppose f and g are continuous on [a, b] and k ∈ R.
Z b
Z b
1.
kf (x) dx = k
f (x) dx.
a
Z b
[f (x) ± g(x)] dx =
2.
a
Z b
f (x) dx ±
a
Z a
3.
0
Z b
g(x) dx.
a
10
1
a
f (x) dx = 0.
a
Z a
f (x) dx = −
f (x) dx.
a
b
5.
Z b
Z c
Z b
09
45
1
4.
23
Z b
f (x) dx
f (x) dx +
f (x) dx =
where c ∈ (a, b).
c
a
a
Example 1.
H
Evaluate the following definite integrals.
Z 2
Z π
2
3
(a)
x dx
(b)
sin x dx
−1
0
Z π
| cos x| dx
Z 1
(c)
M
−1
UY
π
4
Solution
3 2
Z 2
8 (−1)
23 (−1)3
x
2
x dx =
(a)
−
= −
= 3.
=
3 −1
3
3
3
3
−1
Z π
2
(b)
π
sin x dx = [− cos x]02 = − cos
0
Z 1
(c)
√
Z 1
1
(x + 2) 2 dx
x + 2 dx =
−1
π
+ cos 0 = 0 + 1 = 1.
2
−1
3 1
2
= (x + 2) 2
3
−1
1
√
x + 2 dx
(d)
U Ye Min Htoo(YTU)
=
Grade 12
Mathematics
3
2 3
(3) 2 − (1) 2
3
2 √
= (3 3 − 1).
3
Z π
(d)
π
4
| cos x| dx
0
When π4 ≤ x ≤ π2 , | cos x| = cos x.
Z π
Z π
Z π
2
| cos x| dx
=
cos x dx
+
π
4
23
π
4
10
1
When π2 ≤ x ≤ π, | cos x| = − cos x.
(− cos x) dx
π
2
π
4
2
09
45
1
π
2
= sin x π − sin x π
π π
π
− sin π − sin
= sin − sin
2
4
2
√
H
2
2
M
=2−
UY
Example 2.
Z 2
Evaluate the integral
√
x 2 − x dx.
1
Solution
Method 1
Z 2
√
x 2 − x dx
1
Let u = 2 − x. Then du = −dx.
2
U Ye Min Htoo(YTU)
Z
Grade 12
Z
√
x 2 − x dx = −
R
=−
Mathematics
√
(2 − u) u du
1
3
2u 2 − u 2 du
3
5
= − 43 u 2 + 25 u 2
3
5
2
3
5
4
2
x 2 − x dx = − (2 − x) 2 + (2 − x) 2
3
5
1
1
Z 2
√
09
45
1
14
= 0 − − 34 + 25 = 15
Method 2
Let u = 2 − x. Then du = −dx.
1→0
Z 2
M
u
UY
1→2
H
Changing the variable from x to u, we get
x
23
Next
10
1
0
= − 34 (2 − x) 2 + 25 (2 − x) 2
√
x 2 − x dx = −
Z 0
1
1
3
2u 2 − u 2 du
1
Z 1
=
1
2
2u − u
3
2
du
0
1
4 3 2 5
= u2 − u2
3
5
0
=
14
15
3
U Ye Min Htoo(YTU)
Grade 12
Mathematics
Exercise 11.1
1. Evaluate the following definite integrals.
Z 1
Z 1
√
ex
(a)
dx
(b)
x
x + 2 dx
x
0 e +1
−1
Z 4
(d)
2
Z 2
x+5
dx
2
x − 6x + 5
(e)
e
−x
Z 6
(c)
0
1
x−1
3
3
dx
Z π
cos 2x dx
(f)
tan 2x dx
π
2
0
0
Solution;
Z 1
0
ex
dx = ln |ex + 1|10
x
e +1
10
1
(a)
23
= ln |e1 + 1| − ln |e0 + 1|
Z 1
(b)
09
45
1
= ln(e + 1) − ln 2
e+1
= ln
2
√
x x + 2 dx
−1
M
When x = −1, u = 1.
H
Let u = x + 2. Then du = dx and x = u − 2.
UY
When x = 1, u = 3.
4
U Ye Min Htoo(YTU)
Z 1
Grade 12
Z 3
√
x x + 2 dx =
−1
Mathematics
√
(u − 2) u du
1
Z 3
1
(u − 2)u 2 du
=
Z1 3 3
1
u 2 − 2u 2 du
1
3
2 5 4 3
= u2 − u2
5
3
1
2 5 4 3
2 5 4 3
2
2
2
2
=
(3) − (3)
−
(1) − (1)
5
3
5
3
√
√
2
4
2 4
=
·9 3− ·3 3 −
−
5
3
5 3
√
√
2 4
18 3
−4 3− +
=
5√
√ 5 3
18 3 − 20 3 2 4
− +
=
5
5 3
√
2 3 2 4
=−
− +
5√
5 3
2( 3 + 1) 4
=−
+
3
√5
−6( 3 + 1) + 20
=
√15
20 − 6 3 − 6
=
15
√
14 − 6 3
=
15
(c)
0
Let
1
x−1
3
3
dx
UY
Z 6
M
H
09
45
1
23
10
1
0
=
u = 13 x − 1
du
1
=
dx
3
3du = dx
Z 1
x−1
3
3
Z
dx =
u3 · 3du = 3
u4
4
5
U Ye Min Htoo(YTU)
3
=
4
Z 6
1
x−1
3
1
x−1
3
Mathematics
4
3
" 4 # 6
3 1
dx =
x−1
4 3
0
4
4
3 1
3 1
=
×6−1 −
×0−1
4 3
4 3
3
3
= (2 − 1)4 − (0 − 1)4
4
4
3 3
= −
4 4
10
1
0
0
Grade 12
=0
Z 6
0
09
45
1
23
Another Method
1
x−1
3
3
1
dx =
4
UY
M
H
3
=
4
Z 4
(d)
2
x+5
=
1
x−1
3
4
1
6
× 1
4 6
0
3 0
3
=
(2 − 1)4 − (0 − 1)4
4
3
= [1 − 1]
4
=0
x+5
dx
x2 − 6x + 5
x2 − 6x + 5
1
x−1
3
x+5
A
B
=
+
(x − 1)(x − 5)
x−1 x−5
x+5
A(x − 5) + B(x − 1)
=
x2 − 6x + 5
(x − 1)(x − 5)
x + 5 = A(x − 5) + B(x − 1)
6
U Ye Min Htoo(YTU)
Grade 12
Mathematics
When x = 1, 1 + 5 = A(1 − 5)
6 = −4A
A=−
3
2
When x = 5, 5 + 5 = B(5 − 1)
5
2
10
1
B=
0
10 = 4B
Z
Z
x+5
3 4 1
5 4 1
dx = −
dx +
dx
x2 − 6x + 5
2 2 x−1
2 2 x−5
4
4
5
3
= − ln |x − 1| + ln |x − 5|
2
2
2
2 3
3
5
5
= − ln |4 − 1| + ln |2 − 1| +
ln |4 − 5| − ln |2 − 5|
2
2
2
2
3
5
= − ln 3 − ln 3
2
2
1
= −4 ln 3 = ln 4
3
09
45
1
Z 4
23
5
− 32
x+5
2
=
+
∴ 2
x − 6x + 5
x−1 x−5
Z 2
e−x cos 2x dx
UY
(e)
M
H
2
0
Let u = e−x ,
dv = cos 2x dx
Z
Z
du
−x
= −e ,
dv = cos 2x dx
dx
1
du = −e−x dx, v = sin 2x
2
Z
e
−x
1
cos 2x dx = e−x sin 2x −
2
1
1
= e−x sin 2x +
2
2
Z
Z
1
sin 2x(−e−x dx)
2
e−x sin 2x dx
7
U Ye Min Htoo(YTU)
Z
Let I =
Grade 12
Mathematics
e−x cos 2x dx
1
1
Then I = e−x sin 2x +
2
2
Let
u = e−x ,
Z
Z
du
= −e−x
dx
e−x sin 2x dx
· · · (1)
dv = sin 2x dx
Z
dv =
sin 2x dx
0
1
du = −e−x dx v = − cos 2x
2
Z 1 −x
1
sin 2x dx = − e cos 2x −
− cos 2x (−e−x dx)
2
2
Z
1 −x
1
= − e cos 2x −
e−x cos 2x dx · · · (2)
2
2
09
45
1
23
e
−x
10
1
Z
Substituting (2) into (1), we get
UY
M
H
Z
1 −x
1
1 −x
1
−x
I = e sin 2x +
− e cos 2x −
e cos 2x dx
2
2
2
2
Z
1 −x
1 −x
1
I = e sin 2x − e cos 2x −
e−x cos 2x dx
2
4
4
1
1
1
−x
I=e
sin 2x − cos 2x − I
2
4
4
5
1
I = e−x (2 sin 2x − cos 2x)
4
4
1 −x
I = e (2 sin 2x − cos 2x)
5
Z
∴
1
e−x cos 2xdx = e−x (2 sin 2x − cos 2x)
5
Z 2
e
0
−x
2
1 −x
cos 2x dx = e (2 sin 2x − cos 2x)
5
0
1 −2
1 0
= e (2 sin 4 − cos 4) − e (0 − 1)
5
5
1
1
= 2 (2 sin 4 − cos 4) +
5e
5
8
U Ye Min Htoo(YTU)
Z π
6
(f)
Z π
6
tan 2x dx =
π
8
π
8
1
=−
2
Z π
6
π
8
Grade 12
Mathematics
sin 2x
dx
cos 2x
−2 sin 2x
dx
cos 2x
π
1
6
= − ln | cos 2x| π
2
8
!
23
1
2
− ln
ln
2
2
09
45
1
=
√
10
1
√ !
1
2
ln − ln
2
2
1
=−
2
1
=
2
0
1
1
π
π
+ ln cos
= − ln cos
2
3
2
4
1 √
ln 2
2
√
= ln 4 2
0
M
H
2. Find the integral of
Z 2π
cos mx sin nx dx
UY
for (a) m = n and (b) m ̸= n.
9
U Ye Min Htoo(YTU)
Grade 12
Mathematics
Solution;
(a) For m = n,
Z 2π
Z 2π
cos mx sin nx dx =
0
cos mx sin mx dx
0
09
45
1
=0
(b) For m ̸= n,
Z 2π
Z
1 2π
2 cos mx sin nx dx
cos mx sin nx dx =
2 0
Z
1 2π
=
[sin(m + n)x − sin(m − n)x] dx
2 0
2π
1
cos(m + n)x cos(m − n)x
=
−
+
2
m+n
m−n
0
1
1
cos(m + n)2π cos(m − n)2π
1
=
+
+
−
− −
2
m+n
m−n
m+n m−n
UY
M
H
0
23
10
1
0
Z
1 2π
=
2 sin mx cos mx dx
2 0
Z
1 2π
=
sin 2mx dx
2 0
2π
cos 2mx
1
−
=
2
2m
0
1
cos 4mπ
1
=
−
− −
2
2m
2m
1
1
1
−
=
− −
2
2m
2m
=0
3. Evaluate the following integrals.
Z 4 √
| 2x − 2| dx
(a)
0
Z 2π
Z 1
|e − 1| dx
x
(b)
−1
| sin x| dx
(c)
0
Solution;
10
U Ye Min Htoo(YTU)
Z 4
(a)
Grade 12
Mathematics
√
| 2x − 2| dx For the intersection point of x-axis and the curve,
0
√
2x − 2 = 0
√
2x = 2
x=2
√
√
When 0 ≤ x ≤ 2, | 2x − 2| = −( 2x − 2)
√
√
When 2 ≤ x ≤ 4, | 2x − 2| = ( 2x − 2)
q
0
|ex − 1| dx
1
2
10
1
Z 2
09
45
1
23
√
H
Z 4 q
1
| 2x − 2| dx =
(−( 2x + 2)dx +
( 2x 2 − 2)dx
0
0
2
"
#2 " √ 3
#4
3
√ x2
2x 2
+ 2x +
= − 2
− 2x
3
2
2
0
2
!
√
"
#
8
8
16 2
=
− +4 −0 +
−8 −
−4
3
3
3
√
8
16 2
8
=− +4+
−8− +4
3√
3
3
16 2 16
=
−
3
3
16 √
= ( 2 − 1)
3
Z 4
(b)
−1
M
Z 1
UY
For the intersection point of x-axis and the curve,
ex − 1 = 0
ex = 1
x=0
When −1 ≤ x ≤ 0, |ex − 1| = −(ex − 1)
When 0 ≤ x ≤ 1, |ex − 1| = ex − 1
11
U Ye Min Htoo(YTU)
Grade 12
Z 1
Z 0
|e − 1| dx = −
x
−1
Mathematics
Z 1
(e − 1)dx +
(ex − 1)dx
x
−1
0
09
45
1
x=π
23
sin x = 0
10
1
(c) For the intersection point of x-axis and the curve,
0
= − [(ex − x)]0−1 + [(ex − x)]10
= − (e0 − 0) − (e−1 + 1) + (e1 − 1) − (e0 − 0)
1
= − 1 − − 1 + [e − 1 − 1]
e
1
= +e−2
e
When 0 ≤ x ≤ π, | sin x| = sin x
When π ≤ x ≤ 2π, | sin x| = − sin x
Z 2π
H
Z π
Z 2π
| sin x| dx =
UY
M
0
(− sin x) dx
sin x dx +
0
= (− cos x)
π
π
+ cos x
0
2π
π
= [(− cos π) − (− cos 0)] + [cos 2π − cos π]
= [1 + 1] + [1 − (−1)]
=4
12
U Ye Min Htoo(YTU)
Grade 12
Mathematics
Example 3.
Find the area enclosed between the graph of
(a) f (x) = 3 and the x-axis over [1, 4],
2
(b) f (x) = 3x + 2 and the x-axis over − , 0 ,
3
(c) f (x) = −x2 + 4 and the x-axis over [−2, 2].
10
1
0
Solution
Z 4
23
(a)
1
1
= 12 − 3
09
45
1
4
3 dx = 3x
= 9 unit2
M
H
(b)
0
3x2
+ 2x
(3x + 2) dx =
2
− 23
− 32
2
3
2
2
=0−
−
−2 −
2
3
3
2
= unit2
3
UY
Z 0
13
U Ye Min Htoo(YTU)
Grade 12
Mathematics
(c)
2
x3
(4 − x ) dx = 4x −
3 −2
−2
8
8
= 8−
− −8 +
3
3
2
= 10 unit2
3
Z 2
2
Example 4. (a) Compute the definite integral
0
10
1
Z 5
0
2
(x2 − 2x) dx. (b) Find the total
23
2
area enclosed
between the graph of f (x) = x − 2x and the x-axis
5
over 0, .
2
09
45
1
Solution
(a)
3
π2
x
2
(x − 2x) dx =
−x
3
0
0
1 π 3 π 2
=
−
3 2
2
1
= −1
24
2
UY
M
2
H
Z 5
(b) For the intersection point of x-axis and the curve, f (x) = 0
14
U Ye Min Htoo(YTU)
Grade 12
Mathematics
x2 − 2x = 0
x(x − 2) = 0
x = 0, 2
10
1
0
Thus the graph of f (x) cuts x-axis
at
0 and 2, we divide the domain in two intervals: the
5
interval [0, 2] and the interval 2, .
2
2
3
x
2
−x
(x − 2x) dx =
3
0
0
8
=
−4 −0
3
4
=−
3
Z 2
09
45
1
23
2
H
On [0, 2], the function f ≤ 0, so the value of integration f gives negative.
3
25
x
2
(x − 2x) dx =
−x
3
2
2
125 25
8
=
−
−
−4
24
4
3
7
=
24
M
Z 5
UY
2
2
5
On 2, , the function f ≥ 0, so the value of integration f gives positive.
2
4
7
5
Area = − +
= 1 unit2
3
24
8
15
U Ye Min Htoo(YTU)
Grade 12
Mathematics
Example 5.
Find the area enclosed between the graph of f (x) = cos x and the lines
(a) x = 0 and x =
(b) x =
π
2
π
3π
and x =
2
2
(c) x = 0 and x =
3π
2
cos x dx.
10
1
2
(d) Find the definite integral
0
Z 3π
UY
(a)
M
Solution;
H
09
45
1
23
0
Z π
2
cos x dx = sin x
π
2
0
0
π
= sin − sin 0
2
=1
Area = 1 unit2 .
(b)
16
U Ye Min Htoo(YTU)
Grade 12
Mathematics
Z 3π
2
π
2
3π
2
cos x dx = sin x π
2
= sin
3π
π
− sin
2
2
= −2
0
Area = | − 2| = 2 unit2 .
3π
is 3 unit2 .
2
10
1
(c) Area between the graph f (x) = cos x and the lines x = 0 and x =
09
45
1
23
(d)
Z 3π
2
cos x dx = sin x
3π
2
0
0
3π
= sin
− sin 0
2
M
Example 6.
H
= −1
UY
The shaded area is 16 unit2 . Find the value of k.
Solution
17
U Ye Min Htoo(YTU)
Grade 12
Mathematics
Z 4
√
2k 3 4
k x dx =
x2
3
0
0
2k 3
· 42
=
3
16k
=
3
16k
= 16, so k = 3.
3
Example 7.
M
Solution
H
09
45
1
23
10
1
Find the area between y = x2 − 2x + 2 and y = 2x − 1.
0
By the problem,
UY
To find the x-coordinate of intersection points,
x2 − 2x + 2 = 2x − 1
x2 − 4x + 3 = 0
(x − 1)(x − 3) = 0
x = 1, 3
18
U Ye Min Htoo(YTU)
Grade 12
Mathematics
Z 3
[(2x − 1) − (x2 − 2x + 2)] dx
Area =
Z1 3
(−x2 + 4x − 3) dx
1
3
3
x
2
= − + 2x − 3x
3
1
4
= 0− −
3
1
= 1 unit2
3
Example 8.
10
1
π
5π
≤x≤
.
4
4
H
09
45
1
23
Find the area between y = sin x and y = cos x for
0
=
Solution
UY
M
From the figure, we see that sin x ≥ cos x when
π
5π
<x<
.
4
4
Z 5π
4
Area =
π
4
(sin x − cos x) dx
5π
= [− cos x − sin x] π4
4
5π
π
π
5π
− sin
− − cos − sin
= − cos
4
4
4
4
√
√ !
√
√ !
2
2
2
2
=
+
− −
−
2
2
2
2
√
= 2 2 unit2
19
U Ye Min Htoo(YTU)
Grade 12
Mathematics
Exercise 11.2
UY
M
H
09
45
1
23
10
1
0
1. Find the shaded areas.
Solution:
20
U Ye Min Htoo(YTU)
Grade 12
Mathematics
(a) For the intersection point for x-axis and the curve, f (x) = 0
x2 − 2x − 3 = 0
(x + 1)(x − 3) = 0
x = −1, 3
3
3
x
2
2
(x − 2x − 3) dx =
− x − 3x
3
−1
−1
3
3
(−1)3
2
2
=
−3 −3·3 −
− (−1) − 3(−1)
3
3
1
= (9 − 9 − 9) − − − 1 + 3
3
2
= −9 − 1
3
2
= −10
3
23
10
1
0
Z 3
2
2
= 10 unit2
3
3
(b) To find the x-coordinate of intersection point
09
45
1
The shaded area = −10
cos x = sin x
M
H
1=
tan x = 1
π
x=
4
Z π
UY
Z π
4
shaded area =
sin x
cos x
(cos x − sin x) dx +
0
2
π
4
(sin x − cos x) dx
π
π
= [sin x + cos x]04 + [− cos x − sin x] π2
i h 4
h π
π
π
π π
π i
− (0 + 1) + − cos − sin
− − cos − sin
= sin + cos
4# "
2
2
4
4
"√ 4 √
√
√ !#
2
2
2
2
=
+
− 1 + (0 − 1) − −
−
2
2
2
2
√
√
= 2−1−1+ 2
√
=2 2−2
√
= 2( 2 − 1) unit2
21
U Ye Min Htoo(YTU)
Grade 12
Mathematics
(c)
4
0
x
x3
2
(x + x − 2x) dx =
+
−x
4
3
2
2
16 8
=0−
+ −4
4
3
48 − 32 − 48
=0−
12
32
=
12
8
=
3
4
1
Z 1
x
x3
3
2
2
(x + x − 2x) dx =
+
−x
4
3
0
0
1 1
=
+ −1 −0
4 3
3 + 4 − 12
=
12
5
=−
12
Z 0
0
2
09
45
1
23
10
1
3
5
8
5
32 + 5
37
1
8
+ −
= +
=
=
=3
unit2 .
3
12
3 12
12
12
12
(d) To find the x-coordinate of intersection points.
x2 = x + 2
x2 − x − 2 = 0
(x + 1)(x − 2) = 0
UY
M
H
The shaded area =
x = −1, 2
22
U Ye Min Htoo(YTU)
Grade 12
Mathematics
Z 2
−1
(x + 2 − x2 ) dx
x2 =
√
09
45
1
x4 = x
x
23
(e) To find the x-coordinate of intersection points.
10
1
2
2
x
x3
=
+ 2x −
2
3
−1
4
8
1
1
=
+4−
−
−2+
2
3
2
3
10 7
=
+
3
6
27
=
6
1
= 4 unit2
2
x4 − x = 0
x(x3 − 1) = 0
UY
M
H
x = 0, 1
Z 1
√
x − x2 dx
The shaded area =
Z0 1
√
=
x − x2 dx
0
" 3
#1
x2
x3
= 3 −
3
2
0
" 3
#1
2x 2
x3
=
−
3
3
0
2 1
−
−0
=
3 3
1
= unit2
3
23
0
The shaded area =
U Ye Min Htoo(YTU)
Grade 12
Mathematics
(f) To find the x-coordinate of the intersection points.
sin 2x = sin x
2 sin x cos x = sin x
2 sin x cos x − sin x = 0
sin x(2 cos x − 1) = 0
π
x = 0, , π
3
Z π
Z π
3
(sin x − sin 2x) dx
(sin 2x − sin x) dx +
The shaded area =
π
3
π
cos 2x
cos 2x
= −
+ cos x + − cos x +
2
2
π
0
3
1
2π
π
1
=
− cos
− − cos 0 + cos 0
+ cos
2
3
3
2
π 1
2π
1
+ − cos π + cos 2π − − cos + cos
2
3 2
3
1 1
1
1
1 1
=
+ − − +1 + 1+
− − −
4 2
2
2
2 4
1 1 1
3 1
= − +1+ + +
4 2
2 2 4
3−2+4+2+2+1
=
4
10
=
4
5
= unit2
2
10
1
π3
0
0
M
H
09
45
1
23
UY
2. Find the area under each curve between the given x-values:
(a) f (x) = ln x
from x = 1 to x = e.
(b) f (x) = 12 − 3x2
(c) f (x) = xex
2
(d) f (x) = sin x
from x = −1 to x = 2.
from x = 0 to x = 1.
from x = 0 to x =
(e) f (x) = (1 + cos x) sin x
3π
.
2
from x = 0 to x = π.
Solution.
24
U Ye Min Htoo(YTU)
(a) f (x) = ln x
Grade 12
Mathematics
from x = 1 to x = e.
Z e
The area =
ln x dx
1
= [x ln x − x]e1
= (e ln e − e) − (1 ln 1 − 1)
= (e − e) − (0 − 1)
= 1 unit2
0
from x = −1 to x = 2
Z 2
The area =
−1
(12 − 3x2 ) dx
23
= [12x − x3 ]2−1
10
1
(b) f (x) = 12 − 3x2
09
45
1
= (24 − 8) − (−12 + 1)
= 16 + 11
= 27 unit2
H
from x = 0 to x = 1
UY
Let u = x2
du
= 2x
dx
1
du = x dx
2
2
M
(c) f (x) = xex
Z
Z
x2
xe dx =
1
eu du
2
1
= eu
2
1 2
= ex
Z2
1
2
xex dx
0
1
1 x2
= e
2
0
1
= (e − 1) unit2
2
The area =
(d) f (x) = sin x
from x = 0 to x =
3π
2
25
U Ye Min Htoo(YTU)
Grade 12
Mathematics
For the intersection points of x-axis and sin x
sin x = 0
x=π
Z π
sin x dx = [− cos x]π0 = − cos π + cos 0 = 2
0
Z 3π
3π
sin x dx = [− cos x]π2 = − cos
0
π
Area = |2| + | − 1| = 3 unit2
(e) f (x) = (1 + cos x) sin x
3π
+ cos π = −1
2
from x = 0 to x = π.
23
Z π
10
1
2
(1 + cos x) sin x dx
The area =
Z π
09
45
1
0
(sin x + cos x sin x) dx
Z π
1
=
sin x + (2 sin x cos x) dx
2
0
Z π
1
=
sin x + sin 2x dx
2
0
π
1
= − cos x − cos 2x
4
0 1
1
= − cos π − cos 2π − − cos 0 − cos 0
4
4
1
1
= 1−
− −1 −
4
4
3 5
= +
4 4
=
UY
M
H
0
= 2 unit2
3.
Z 1 √
(a) Compute
x 1 − x2 dx.
−1
√
(b) Find the area enclosed between the graph of f (x) = x 1 − x2 and the x-axis
over [−1, 1].
26
Grade 12
Mathematics
0
U Ye Min Htoo(YTU)
10
1
Solution
Z 1 √
(a)
x 1 − x2 dx
23
−1
du
= −2x
dx
1
du = x dx
2
09
45
1
Let u = 1 − x2 .
1
x 1 − x2 dx =
u −
du
2
Z
1
=−
u1/2 du
2
Z 1 √
1
1
x 1 − x2 dx = − (1 − x2 )3/2 −1
3
−1
1
= − [0 − 0]
3
√
Z
UY
M
H
Z
=0
Z 1 √
(b)
x 1 − x2 dx
−1
27
√
U Ye Min Htoo(YTU)
Grade 12
Mathematics
Z 1 √
1
1
x 1 − x2 dx = − (1 − x2 )3/2 −1
3
−1
1
= − [1 − 0]
3
1
=−
3
The enclosed area = − 13 + 13 = 23 unit2 .
10
1
0
4. Find the area enclosed between the graphs of y = x3 − 3x and y = x for
0 ≤ x ≤ 2.
Z 2
23
Solution.
(x − (x3 − 3x)) dx
The enclosed area =
09
45
1
Z0 2
(x − x3 + 3x) dx
=
Z0 2
(4x − x3 ) dx
0
2
x4
2
= 2x −
4 0
= (8 − 4) − 0
= 4 unit2
UY
M
H
=
5. Find the area enclosed of the graphs of ln(3x − 2) and y = x2 − 2x + 1 for
1 ≤ x ≤ 2.
Solution.
Z
ln(3x − 2) dx
Let u = ln(3x − 2), dv = dx
du
1
=
× 3,
dx
3x − 2
du =
3
dx,
3x − 2
R
dv =
R
dx
v=x
28
U Ye Min Htoo(YTU)
Grade 12
Mathematics
Z
Z
3
x·
dx
3x − 2
Z 2
dx
= x ln(3x − 2) −
1+
3x − 2
2
= x ln(3x − 2) − x + ln |3x − 2|
3
Z
3
x
(x2 − 2x + 1) dx =
− x2 + x
3
10
1
0
ln(3x − 2) dx = x ln(3x − 2) −
Z 2
[ln(3x − 2) − (x2 − 2x + 1)] dx
Z 2
Z1 2
(x2 − 2x + 1) dx
ln(3x − 2) dx −
=
1
1
2 3
2
2
x
2
= x ln(3x − 2) − x + ln |3x − 2|
−
−x +x
3
3
1
1
2
8
1
= 2 ln 4 − 2 −
− (0 − 1 − 0) −
−4+2 −
−1+1
3
3
3
2 1
4
= ln 4 − 2 + 1 − +
3
3 3
4
4
= ln 4 −
3
3
4
= (ln 4 − 1)unit2
3
M
H
09
45
1
23
The enclosed area =
UY
6. The shaded area is 8a unit2 . Find the value of a.
Solution.
29
U Ye Min Htoo(YTU)
Grade 12
Mathematics
3
a
x
(x + 1) dx =
+x
3
−a
3
−a a
(−a)3
=
+a −
+ (−a)
3
3
3
a3
a
+a− − −a
=
3
3
3
3
a
a
=
+a+
+a
3
3
2a3
+ 2a
=
3
Z a
10
1
0
2
23
By the problem:
09
45
1
2a3
+ 2a = 8a
3
2a3 + 6a = 24a
2a3 − 18a = 0
a3 − 9a = 0
H
a(a2 − 9) = 0
M
a(a + 3)(a − 3) = 0
UY
a = 0 (impossible)
or
a = −3 (impossible)
∴a=3
30
or
a=3