# 2nd PUC Maths Question Bank Chapter 8 Application of Integrals Ex 8.2

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## Karnataka 2nd PUC Maths Question Bank Chapter 8 Application of Integrals Ex 8.2

### 2nd PUC Maths Application of Integrals NCERT Text Book Questions and Answers Ex 8.2

Question 1.
Find the area of the circle 4x2 + 4y2 = 9 which is interior to the Parobla x2 = 4y.
x2 = 4y is up-ward parabola with vertex (0, 0)
Puting x2 = 4y in 4x2 + 4y2 = 9

4 (4y) + 4y2 = 9
⇒ 4y2 + 16 y – 9 = 0

= 2x[(area under circle from (x = o to x = $$\sqrt{2}$$ – area under parabola (from x = o to x = $$\sqrt{2}$$]

Question 2.
Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y2 = 1.
(1) (x – 1)2 + y2 =1 is a circle with centre (1,0) and radius 1
(2) x2 + y2 = 1 is circle with circle with centre (0,0) and radius (1)

• As shaded area is required one and is a symmetrical convex curve, so, 2 x area of half of shaded region
• It is symmentrical because radius of both are same

2x (area of curve x2 + y2 = 1 from x = $$\frac{1}{2}$$to x = 1
+ area curve (x -1)2 + y2 = 1 from x = 0 to x = $$\frac{1}{2}$$)
As if we do along ‘x’ axis or ‘y’ axis, we get area of 1st quadrant, so to get from 1st and 4th quadrant for a curve is this question we need to multiply by 2.

Question 3.
Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 3.
Curve y = x2 + 2 is parabola with vertex (0, 2) (upward parabola)
y = x, line points are (1, 1) (0, 0) (-1, -1) (2, 2) etc
x = 0, line points are (0, 0) (0, 1) (0,-1), (0, 2) (0,-2) etc
x = 3, line points are (3, 0) (3, 1), (3, 2), (3, 3) (3-1) etc…..

so lines meet parabola at :-
line x = 0, y = 2 from y… y = x2 + 2
line x = 3, y = 11
and line x = y, y2 – y + 2 = 0 → y = not pts
so does not meet
= (Area under curve y = x2 + 2 from x = 0 to x = 3] – [Area under line x = y from x = 0 to x = 3]

Question 4.
Using integration find the area of region bounded by the triangle whose vertices are
(-1, 0), (1, 3) and (3, 2).
We need line eq: for line AB, BC and AC

so line AB,

⇒[Area under line AB from x = -1 to x = 1
+ Area under BC from x = 1 to x = 3
– Area under AC from x = -1 to x = 3]

Question 5.
Using integration find the area of the triangular region whose sides have the equations y 2x + 1, y = 3x + 1 add x = 4.
y = 2x – 1

points are (0, 1), (1, 3) (-1, -1) (4, 9) etc y = 3x + 1
points are (0, 1), (1,4), (-1, -2) (4, 13) etc x = 4
points are (4, 0), (4, 1), (4, 2) etc….

= [Area under line y = 3x + 1 from (x = 0 to x = 4)]
– [Area under line y = 2x + 1 from (x = 0 to x = 4)]

Choose the correct answer in the following exercises 6 and 7.

Question 6.
Smaller area enclosed by the circle x2 + y2 = 4 and the line x + y = 2 is
(A) 2 (π – 2)
(B) π- 2
(C) 2π – 1
(D)2(π+2)
Circle x2 + y2 = 4 his centre (0, 0) and radius 2
line x + y = 2 pts are (1,1) (2, 0) (0, 2) etc………
circle and line intersect at pts.
It can be found by solving the equation
x2 + y2 = 4 put (y = 2 – x)
⇒ x2 + (2 -x)2 = 4 ⇒ 2x2 – 4x + 0 = 0
⇒ 2x2 – 4x = 0 ⇒ x2 – 2x = 0
⇒ (x -2) x = 0  ⇒ x = 0 or x = 2
so y = 2 – x , x = 0 , y = 2    (0, 2)
x = 2 , y = 0 , (2,0)
so they meet at (0, 2) and (2, 0)

(Area under circle x2 + y2 = 4 from x = 0 to x = 2)
– (Area under line x+y = 2 from x = 0 to x = 2)

Question 7.
Area lying between the curves y2 = 4x and y = 2x is
(A)$$\frac{2}{3}$$
(B)$$\frac{1}{3}$$
(C)$$\frac{1}{4}$$
(D)$$\frac{3}{4}$$