1 Marks Question

Question 11 Mark

The area bounded by the curve y = x|x|, x-axis and the ordinates x = -1 and x = 1 is given by

Answer

Required area :
$ = \left| {\int\limits_{ - 1}^1 {x\left| x \right|dx} } \right|$
$= \left| {\int\limits_{ - 1}^0 {x\left| x \right|dx} + \int\limits_0^1 {x\left| x \right|dx} } \right| $
$= \left| {\int\limits_{ - 1}^0 { - {x^2}dx} } \right| + \int\limits_0^1 {{x^2}dx} $
$= \left| {\left[ {\frac{{ - {x^3}}}{3}} \right]_{ - 1}^0} \right| - \left[ {\frac{{{x^3}}}{3}} \right]_0^1 $
$= \frac{2}{3}$sq. units

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Question 21 Mark

Area bounded by the curve $y = x^3,$ the $x-$axis and the ordinates $x = -2$ and $x = 1$ is

Answer

Required area
$ = \int\limits_{ - 2}^1 {{x^3}dx} $
$= \left[ {\frac{{{x^4}}}{4}} \right]_{ - 2}^1 $
$= \frac{{15}}{4}$

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Question 31 Mark

Find the area under the given curves and given lines: $y = x^4, x = 1, x = 5$ and $x-$axis

Answer

We can see from the figure that the area of the region bounded by the curve $y = x^4$ and the lines $x = 1, x = 5$ is shown by shaded region that is Area $ADCBA.$
Area of $ADCBA = \int_{1}^{5} y\ d y=\int_{1}^{5} x^{4} d x$
$\Rightarrow\ \left[\frac{x^{5}}{5}\right]_{1}^{5}$
$\Rightarrow\ 625-\frac{1}{5}$
$= 624.8$ sq. units.Which is the required solution.

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Question 41 Mark

Find the area under the given curves and given lines: $y = x^2, x = 1, x = 2$ and $x -$ axis.

Answer

Equation of the curve $($parabola$)$ is
$y = x^2 ...(i)$

Required area bounded by curve $(i),$
vertical line $x = 1, x = 2$ and $x -$ axis
$= \left| {\int\limits_1^2 {ydx} } \right|$
$ = \left| {\int\limits_1^2 {{x^2}dx} } \right|$
$= \left( {\frac{{{x^3}}}{3}} \right)_1^2$
$= \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$ sq units

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Question 51 Mark

Area of the region bounded by the curve $y^2 = 4x, y-$axis and the line $y = 3$ is

Answer

The area bounded by the curve $y^2 = 4x, y-$axis and the line $y = 3$ is shown by shaded region in above figure.
Thus, Area $\text{OAB} = \int_\limits{0}^{3} x \ d y=\int_\limits{0}^{3} \frac{y^{2}}{4} d y$
$=\frac{1}{4}\left[\frac{y^{3}}{3}\right]_{0}^{3}$
$=\frac{1}{12}(27)=\frac{9}{4}$
Therefore, required area is $= \frac{9}{4}$ square units

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Question 61 Mark

Area lying in the first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the lines $x = 0$ and $x = 2$ is

Answer

The area bounded by the circle and the lines, $x = 0$ and $x = 2,$ in the first quadrant is shown by shaded region in above figure.
Area of $\text{OAB} = \int_{0}^{2} y d x=\int_{0}^{2} \sqrt{4-x^{2}} d\ x$
$=\left[\frac{\mathrm{x}}{2} \sqrt{4-\mathrm{x}^{2}}+\frac{4}{2} \sin ^{-1} \frac{\mathrm{x}}{2}\right]_{0}^{-2}$
$=2\left(\frac{\pi}{2}\right)=\pi$
Therefore, required area is $= \pi$ square units.

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Question 71 Mark

Find equation of directrix for parabola $y^2=4 a x$.

Answer

Equation of directrix: $x=-a($ or $x+a=0)$

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