1 Marks

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 = x³, 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 = x⁴, 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 = x⁴ 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 = x², x = 1, x = 2 and x - axis.

Answer

Equation of the curve (parabola) is
y = x² ...(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 y² = 4x, y-axis and the line y = 3 is

Answer

The area bounded by the curve y² = 4x, y-axis and the line y = 3 is shown by shaded region in above figure.
Thus, Area 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 x² + y² = 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 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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