In this page, you can find the complete solutions of the seventh exercise of Antiderivatives and its Application chapter from Basic Mathematics Grade XI book published and distributed by Sukunda Pustak Bhawan.

In the above-mentioned book, Complex Numbers is the 19th chapter and has seven exercises only. Out of which, this is the solution of the final exercise.


Check: Basic Mathematics Grade 11 (Sukunda Publication) Guide:

Grade 11 Basic Mathematics by Sukunda Pustak Vawan Notes and Solutions | Nepal


Disclaimer:

Answers mentioned here are not solved by teachers. These are the solutions written by a student of Grade 11. Answers are all correct. However, the language or process of solving the questions might be informal and in examinations, you might have to add little more language and show more calculations than what has been done here. So, we highly encourage you to view these solutions as guide rather than just copying everything mentioned here.

1) Find the area bounded by the x-axis and the following curves and ordinates.

a) $\rm x^2 = 4by; x = a, x =b$

Solution:

The lower limit is x = a and the upper limit is x = b; $\rm b > a$.

Integrating given function $\rm x^2 = 4by \Rightarrow y = \frac{x^2}{4b}$ with respect to x from a to b.

$\rm = \int^b_a y dx$

$\rm = \int_a^b \frac{x^2}{4b} dx$

$\rm = \frac{1}{4b} \left [ \frac{x^3}{3} \right ]^b_a$

$\rm = \frac{1}{4b} \left [ \frac{b^3}{3} - \frac{a^3}{3} \right ]$

$\rm = \frac{1}{4b} \times \frac{b^3 - a^3}{3}$

$\rm = \frac{b^3 - a^3}{12b}$


2) Find the areas bounded by the x-axis and the following given curves and ordinates:

i) $\rm y^2 = 8ax$ and the ordinate at the point (4a, 0)

Solution:

On the x-axis, y = 0

$\rm y^2 = 8ax \Rightarrow 0^2 = 8ax \Rightarrow x = 0$ is the lower limit. x = 4a is the upper limit of integration.

$\rm y^2 = 8ax \Rightarrow y = \sqrt{8ax} = \sqrt{8a} \cdot \sqrt{x}$

Integrating y with respect to x from x = 0 to x = 4a, we get,

$\rm = \int _0 ^{4a} y \ dx$

$\rm = \int _0^{4a} \sqrt{8a} \cdot \sqrt{x}$

$\rm =  \sqrt{8a} \int _0^{4a} x^{\frac{1}{2}}$

$\rm = \sqrt{8a} \left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ] ^{4a}_0$

$\rm = \sqrt{8a} \times \frac{2}{3} \left [ (4a)^{\frac{3}{2}} - 0^{\frac{3}{2}} \right ]$

$\rm = 2 \times \sqrt{2a} \times \frac{2}{3} \times 2^3 \times a^{\frac{3}{2}}$

$\rm = 2 \times 2^3 \times \frac{2}{3} \times \sqrt{2} \times \sqrt{a} \times a^{\frac{3}{2}}$

$\rm = \frac{32 \sqrt{2}}{3} a^2$


3) Find the area enclosed by the axis of x and the following curves:

i) y = 3x - 5x$^2$

Solution:

We solve the given equation to find the upper and lower points of integration when y = 0.

$\rm y = 3x - 5x^2$

$\rm 0 = 3x - 5x^2$

$\rm 0 = x(3 - 5x)$

Either $\rm x = 0$

Or $\rm 3 - 5x = 0 \Rightarrow x = \frac{3}{5}$

Integrating the given function y with respect to x from x = 0 to x = $\frac{3}{5}$,

$\rm \int ^{\frac{3}{5}} _0 y \ dx$

$\rm = \int ^{\frac{3}{5}} _0 (3x - 5x^2) \ dx$

$\rm = \int ^{\frac{3}{5}}_0 3x \ dx - \int ^{\frac{3}{5}} _0 5x^2 \ dx$

$\rm = 3 \left [ \frac{x^2}{2} \right ]^{\frac{3}{5}}_0 - 5 \left [ \frac{x^3}{3} \right ]^{\frac{3}{5}}_0$

$\rm = 3 \left [ \frac{ \frac{3}{5}^2}{2}  - 0 \right ] - 5 \left [ \frac{ \frac{3}{5}^3}{3}  - 0 \right ]$

$\rm = \frac{3}{2} \left [ \frac{9}{25} \right ] - \frac{5}{3} \left [ \frac{27}{125} \right $

$\rm = \frac{3}{2} \times \frac{9}{25} - \frac{5}{3} \times \frac{27}{125}$

$\rm = \frac{27}{50} - \frac{9}{25}$

$\rm = \frac{27}{50} - \frac{18}{50}$

$\rm = \frac{27-18}{50}$

$\rm = \frac{9}{50}$


4. Find the area of the region between:

i) the curve $\rm y^2 = 16x$ and line $y = 2x$

Solution:

$\rm y^2 = 16x \Rightarrow y = 4 \sqrt{x}$

Solving the two equations to get the upper and lower limits

$\rm y^2 = 16x$

$\rm (2x)^2 = 16x$

$\rm 4x^2 - 16x = 0$

$\rm 4x(x - 4) = 0$

Either $\rm 4 x= 0 \Rightarrow x  = 0$

Or $\rm x - 4 = 0 \Rightarrow x = 4$

Integrating the difference of the two equations with respect to x from 0 to 4, we get,

$\rm \int_0^4 \left ( 4\sqrt{x} - 2x \right ) \ dx$

$\rm = 4 \int_0^4 \left ( \sqrt{x} \ dx \right ) - 2 \int_0^4 \left ( x \ dx \right )$

$\rm = 4 \left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]_0^4 - 2 \left [ \frac{x^2}{2} \right ]_0^4$

$\rm = 4 \times \frac{2}{3} \left [ 4^{\frac{3}{2}} - 0 \right ] - \frac{2}{2}  \left [ 4^2 - 0 \right ]$

$\rm = \frac{8}{3} \times 2^3 - 16$

$\rm = \frac{64}{3} - \frac{48}{3}$

$\rm = \frac{64-48}{3}$

$\rm = \frac{16}{3}$

$\rm = 5 \frac{1}{3}$


5. Find the areas of the circle:

i) $\rm x^2 + y^2 = 16$

Solution: We have,

$\rm x^2 + y^2 = 16$

$$\rm or, y^2 = 16 - x^2$$

$$\rm or, y = \sqrt{16 - x^2}$$

Integrating with respective limits to get the area of the circle that lies in the first quadrant, we get,

$$\rm or, \int y \ dx = \int^4_0 \sqrt{16 - x^2} \ dx$$

Put $\rm x = 4 \sin \theta$ and $\rm \ dx = 4 \cos \theta \ d \theta$

Then, the limit changes as $\rm 0 to \frac{\pi}{2}$

$$\rm or, A_1 = \int^\frac{\pi}{2}_0 \sqrt{16 - (4 \sin \theta)^2} \ 4 \cos \theta \ d \theta$$

$$\rm or, A_1 = \int^\frac{\pi}{2}_0 \ sqrt{16 - 16 sin^2 \theta} \ 4 \cos \theta \ d \theta$$

$$\rm or, A_1 = \int^\frac{\pi}{2}_0 16 \cos^2 \theta \ d \theta$$

$$\rm or, A_1 = 16 \int^\frac{\pi}{2}_0 \frac{\cos 2 \theta - 1}{2} \ d \theta$$

$$\rm or, A_1 = 16 \int^\frac{\pi}{2}_0 \frac{\cos 2 \theta}{2} \ d \theta - 4 \int^\frac{\pi}{2}_0 \frac{1}{2} \ d \theta$$

$$\rm or, A_1 = 16 \left [ \frac{1}{2} \cdot \frac{\sin 2 \theta}{2} \right ]^{\frac{\pi}{2}}_0 - 16 \left [ \frac{\theta}{2} \right ]^{\frac{\pi}{2}}_0$$

$$\rm or, A_1 = 16 \cdot 0 -  16 \cdot \left [ \frac{\frac{\pi}{2}}{2} - 0 \right ]$$

$$\rm \therefore A_1 = 4 \pi sq. units.$$

Since a circle spans over the four quadrants equally, the area of the circle given by the equation is

$$\rm A = 4 \times A_1$$

$$\rm or, A = 4 \times 4 \pi sq.units.$$

$$\rm \therefore A = 16 \pi sq.units$$

ii) $\rm x^2 + y^2 = a^2$

Solution: We have,

$\rm x^2 + y^2 = a^2$

$\rm or, y^2 = a^2 - x^2$

$\rm or, y = \sqrt{a^2 - x^2}$


About the Textbook:

Name: Basic Mathematics Grade XI
Author(s): D.R. Bajracharya | R.M. Shresththa | M.B. Singh | Y.R. Sthapit | B.C. Bajracharya
Publisher: Sukunda Pustak Bhawan (Bhotahity, Kathmandu)
Telephone: 5320379, 5353537
Price: 695 /- (2078 BS)