## Rajasthan Board RBSE Class 12 Maths Chapter 11 Application of Integral:Quadrature Ex 11.1

Question 1.

Find the area enclosed by parabola y^{2} = 4ax and its latus rectum.

Solution:

Area bounded by parabola y^{2} = 4ax and its latus rectum is symmetric about x-axis.

Question 2.

Sketch the circle x^{2} + y^{2} = 4, find area enclosed by y – axis and x = 1.

Solution:

Circle x^{2} + y^{2} = 4, whose centre is (0,0) and radius 2.

Question 3.

Find the area enclosed by curve y = sin x and x – axis, whereas 0 < x < 2π.

Solution :

The area enclosed by curve y = sin x and x = π and x = 2π has shown in given following figure by shaded part. Table of y = sinx for various volues of x is given below:

Question 4.

Find the area enclosed by curve y = 2 and x = 0, x = 1.

Solution :

y = 2 ⇒ y^{2} = 4x, which is a parabola. The area enclosed by curve y^{2} – 4x, x = 0, x = 1 has shown by shaded part in the following figure.

If we take area between only positive coordinates, then we will solve like this

y = 2, x = 0, x = 1

Required Area = Area APOA,

Question 5.

Find the area enclosed by y = | x |, x = – 3, x = 1 and x-axis.

Solution :

y = | x |, x = – 3, x = 1

Thus y = + x and y = -x

Graph of these lines is given below

Question 6.

Find the area enclosed by curve x^{2} = 4ay, x-axis and line x = 2.

Solution:

Curve x^{2} = 4ay is a parabola. Its graph is given below.

Question 7.

Find the area enclosed by ellipse + = 1 and lie above the x – axis

Solution:

The area enclosed by ellipse + = 1 and lie above the x axis is shown in the following figure :

Question 8.

Find the total area of ellipse + = 1.

Solution:

Total area of ellipse + = 1 is shown in the following figure.

Question 9.

Find the area enclosed by line – = 2 and co-ordinate axis.

Solution:

Graph of line – = 2 is as follows:

Question 10.

Find the area enclosed by line x + 2y = 8, x = 2, x = 4 and x-axis.

Solution:

Question 11.

Find the area enclosed by the curve y – x^{2}, ordinates x = 1, x = 2 and x-axis.

Solution:

Parabola x^{2} = y is symmetric about x axis. Its vertex is origin (0, 0)

Area enclosed by lines x = 1, x = 2, x-axis and curve x^{2} = y is shown in figure by shaded part.

Question 12.

Find the area of the region in the first quadrant enclosed by y = 4x^{2}, x = 0, y = 1 and y = 4.

Solution:

∵ y = 4 x^{2}

⇒ x^{2} = y

This is a equation of parabola

Thus x = 0, y = 1, y = 4 area enclosed by and curve y = 4x^{2} is shown in figure by shaded part.

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