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