2nd PUC Maths Question Bank Chapter 8 Application of Integrals Ex 8.1

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Karnataka 2nd PUC Maths Question Bank Chapter 8 Application of Integrals Ex 8.1

2nd PUC Maths Application of Integrals NCERT Text Book Questions and Answers Ex 8.1

Question 1.
Find the area of the region bounded by the curve y² = x and the lines x = 1, x = 4 and the x-axis.
Answer:
We know that y² = x represents right hand parabola with vertex (0,0)
Required area is shaded area:-

If not mentioned above/below/any quadrant, so we should take above and below
\(\Rightarrow 2 \times \frac{14}{3}=\frac{28}{3} \text { sq. units. }\)

Question 2.
Find the area of the region bounded by y² = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
Answer:
We know y² = 9x is right handed parabola with vertex (0,0)

Question 3.
Find the area of the region bounded by x² = 4y, y = 2, y = 4 and the y – axis in the first quadrant.
Answer:
We know x² = 4y is upward parabola with vertex (0,0).
Required area is shaded area:-

Question 4.
Find the area of the region bounded by the ellipse
\(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\)
Answer:
We know curve is ellipse vertex (0,0) major axis is ‘x’ and major ‘y’
Area of given area is = 4 x (Area of shaded)

Question 5.
Find the area of the region bounded by the ellipse
\(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\)
Answer:
We know curve is ellipse vertex (0,0) major axis is ‘y’ and major ‘x’
Area of given area is = 4 x (Area of shaded)
\(=4 \times \int_{0}^{2}|y| d x\)

Question 6.
Find the area of the region in the first quadrant enclosed by x-axis, line \(x=\sqrt{3}\) y and the circle x² + y² = 4.
Answer:

Question 7.
Find the area of the smaller part of the circle x² + y² = a² cut off by the line \(x=\frac{a}{\sqrt{2}} \)
Answer:
Circle x² + y² = a² is circle with centre (0,0) and radius

Question 8.
The area between x = y² and x = 4 is divided into two equal parts by the line x = a, find the value of a.
Answer:
(1) x = y² is right hand parabola with vertex (0, 0) ⇒ y² = x
As area between y² = x, x = 4 is divided into 2 parts by x = a,

so,(area under y² = x, x = 0 to x = a) =(area under y² = x, x = a, x = 4)
⇒area (A) = area (B)

Question 9.
Find the area of the region bounded by the parabola y = x² and y = |x|
Answer:
(1) As we know x² = y is a upward parabola with vertex (0,0)
(2) line ⇒ y = |x|
So if y = 1,x = 1 (1,1)
y = -1,x = 1 (-1, 1)
y = 2, x = 2 (2, 2)
y = 2, x = 2 (-2,2)

Area required is the shaded area
= 2 x (area of one half (lets take it of 1^st quadrant))
= 2 x [(area under line y = |x|) – area under parabola y = x]

Question 10.
Find the area bounded by the curve x² = 4y and the line x = 4y – 2.
Answer:
(1) Curve x² = 4y is upward parabola vertex (0, 0)
(2) Line x = 4y – 2 (put 4y = x² in it)
⇒ x = x² – 2 ⇒ x²– x – 2 = 0
⇒ x = 2, -1

So when x = 2 , y = 1 (2,1)
\(x=-1, \quad y=\frac{1}{4} \quad\left(-1, \frac{1}{4}\right)\)
So, Area required is shaded area:-
= [Area under line x = 4y – 2 from (-1 to 2) (x axis) – [Area under x² = 4y from
-1 to 2 (x axis)]
\(\int_{-1}^{2}\left(\frac{x+2}{4}\right) d x-\int_{-1}^{2} \cfrac{x^{2}}{4} d x\)
As solving along ‘x’ axis ,so we take values in terms of x to solve

Question 11.
Find the area of the region bounded by the curve y² = 4x and the line x = 3.
Answer:
(1) Curve y² = 4x is right handed parabola with vertex (0,0) y
(2) Line x = 3, so points are (3,0) (3, 1) (3, 2)
\(\Rightarrow y^{2}=12 \Rightarrow y=\pm 2 \sqrt{3}\)

(3) But as we solve with respect to ‘x’ axis, so need only ‘x’ values:-
⇒Required area is shaded area = 2 X area of upper half of shaded region

Choose the correct answer in the following Exercise 12 and 13

Question 12.
Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is
(A) π
(B)\(\frac{\pi}{2}\)
(C)\(\frac{\pi}{3}\)
(D)\(\frac{\pi}{4}\)
Answer:
(1) Circle is x² + y² = 4 with centre (0, 0) and radius 2.
se required area is shaded area of circle from


so answer is (A) π unit ……
OR
As circle of radius 2. so area of circle = πr²
π x 2² = 4π
As we need area of only 1 quadrant
\(\Rightarrow \frac{1}{4} \times \text { area of circle }=\frac{1}{4} \times 4 \pi=\pi \mathrm{sq.} \text { units. }\)
option (A)

Question 13.
Area of the region bounded by the curve y² = 4x, y – axis and the line y = 3 is
(A) 2
(B) \(\frac{9}{4}\)
(C) \(\frac{9}{3}\)
(D) \(\frac{9}{2}\)
Answer:
(1) Curve y² = 4x is right handed parabola
(2) line y = 3 ⇒ points in line are (0,3), (1,3), (2,3),(3, 3), (-1,3) (-2, 3) etc

(3) Pt where line meets y² = 4x is solved by . solving the equation
as we know y = 3 \(\text { so } x=\frac{3^{2}}{4} \Rightarrow \frac{9}{4}\)
\(\text { so pt is }\left(\frac{9}{4}, 3\right)\)
(If we solve along ‘y’ axis)
[Here to solve along ‘x’ axis, it is tough as to get equation of line in terms of ‘x’ is tough.
(4) So area is area under parabola from y = 0 to y = 3.