B2

1. AREAS
AREAS

2. AREAS

3. AREAS
1) Find the area bounded by the
parabola y=x2, the x-axis and the
lines x=-1, x=2
x
y
0
-1 1 2
x'
Y'
Solution
Given parabola is y=f(x)=x2, and
given lines are x=-1, x=2
=
x𝟑
3 −1
2
Required area =
−𝟏
𝟐
x𝟐
dx
The area bounded by the
curve y=f(x), the x−axis and
the lines x=a, x=b is 𝒃
𝒂
f(x) dx

4. AREAS
=
8
3
- −
1
3
= 3 sq.units
=
8
3
+
1
3
=
9
3

5. AREAS
1(ii) Find the area enclosed between
the curve y=x3+3, x=-1, x=2
Required area
x
x'
y
y'
-1 2
0
Given curve is y =f(x) = x3+3
Solution
2
3
11
=
−𝟏
𝟐
f(x) dx =
−𝟏
𝟐
(x𝟑+3) dx
=
x4
4
+ 3x
−𝟏
𝟐
=
16
4
+ 6 −
1
4
− 3
= 4+6 −
1
4
+ 3 =
51
4
The area bounded by
the curve y=f(x), the
x−axis and the lines
x=a, x=b is 𝒃
𝒂
f(x) dx

6. AREAS
2) Find the area cut off between y=0,
y=x2 -4x+3.
Solution x
y
0 1 3
y=x2-4x+3
Given y = x2-4x+3 = f(x), (say)
y = 0(2)
Solving (1) & (2)
x2-4x+3=0
Required area
(or) (x-2)2 =y-(-1) (1)
(x-1)(x-3)=0  x=1,3
= −
𝟏
𝟑
(x𝟐−4x+3) dx
The area bounded by
the curve y=f(x), the x-
axis and the lines x=a,
x=b is − b
a
f(x) dx if
f(x)0, x[a,b]

7. AREAS
= −
x3
3
− 4
x2
2
+3x
𝟏
𝟑
= − (9−18+9) −
1
3
−2+3
=
4
3
sq. units

8. AREAS
3(i) Find the area cut off between
x=0 and x = 4-y2
Given x = 4-y2(1)
(y-0)2 = -(x-4)
x = 0 (2)
Solving (1) & (2)
0= 4 – y2  y2=4
 y= ±2
Solution x
y
0
+2
(4, 0)
-2
Eq(1) is a left
handed parabola
with vertex (4,0)

9. AREAS
Required area =
−𝟐
𝟐
f(y) dy =
−𝟐
𝟐
(4−y𝟐) dy
= 16−
16
3
=
32
3
sq.units.
= 2
𝟎
𝟐
(4−y𝟐) dy
= 2 4y−
y𝟑
3 𝟎
𝟐
= 2 8−
8
3
−(0−0)
4-y2 is even function

10. AREAS
(or) Alternative method
Area with respect to x-axis is
= 2 ×
4−x
𝟑
𝟐
3
2
−1
𝟎
𝟒
A = 2
𝟎
𝟒
f(x) dx
A = 2
𝟎
𝟒
4−x dx
=
32
3
sq.units.
Given eq x = 4-y2
 y2 = 4-x
 y = 4−x = f(x)

11. AREAS
4(i) Find the area cut off between
x=0 and 2x = y2- 1.
Given 2x = y2 – 1(1)
x = 0 (2)
Solving (1) & (2)
0= y2 – 1
0
-
1
2
+1
-1
x
y
x'
y'
Solution
(or) (y-0)2 = 2 x− −
1
2
 y2=1  y= ±1
Eq(1) is a right
handed
parabola with
vertex (-1/2, 0)

12. AREAS
= −
−1
1
y𝟐
−1
2
dy
= y−
y𝟑
3 𝟎
𝟏
=
2
3
sq.units
= −2
0
1
y𝟐−1
2
dy ∵
y2−1
2
is even function
= −
0
1
y𝟐−1 dy =
0
1
1−y𝟐 dy
= 1−
1
3
−(0−0)
From eq(1)
2x = y2 – 1
 x=
y2−1
2
= f(y)
Required Area = –
−𝟏
𝟏
f(y) dy

13. AREAS
Area with respect to x-axis is
=
2
3
sq.units
(or) Alternative method
= 2
−𝟏
𝟐
𝟏
2x+1 dx
= 2
(2x+1)𝟑/𝟐
3
2
×2
−𝟏/𝟐
𝟎
A = 2
−𝟏
𝟐
𝟏
f(x) dx
From eq(1)
2x = y2 – 1
 y2 = 2x+1
 y =  2x+1 = f(x)

14. AREAS
Thank you…