JEE MAIN - Mathematics (2007 - No. 8)
The area enclosed between the curves $${y^2} = x$$ and $$y = \left| x \right|$$ is :
$$1/6$$
$$1/3$$
$$2/3$$
$$1$$
Explanation
The area enclosed between the curves
$${y^2} = x$$ and $$y = \left| x \right|$$
From the figure, area lies between
$${y^2} = x$$ and $$y = x$$
$$\therefore$$ Required area $$ = \int_0^1 {\left( {{y_2} - {y_1}} \right)} dx$$
$$ = \int_0^1 {\left( {\sqrt x - x} \right)dx = \left[ {{{{x^{3/2}}} \over {3/2}} - {{{x^2}} \over 2}} \right]} _0^1$$
$$\therefore$$ Required area $$ = {2 \over 3}\left[ {{x^{3/2}}} \right]_0^1 - {1 \over 2}\left[ {{x^2}} \right]_0^1$$
$$ = {2 \over 3} - {1 \over 2} = {1 \over 6}$$
$${y^2} = x$$ and $$y = \left| x \right|$$
From the figure, area lies between
$${y^2} = x$$ and $$y = x$$
$$\therefore$$ Required area $$ = \int_0^1 {\left( {{y_2} - {y_1}} \right)} dx$$
$$ = \int_0^1 {\left( {\sqrt x - x} \right)dx = \left[ {{{{x^{3/2}}} \over {3/2}} - {{{x^2}} \over 2}} \right]} _0^1$$
$$\therefore$$ Required area $$ = {2 \over 3}\left[ {{x^{3/2}}} \right]_0^1 - {1 \over 2}\left[ {{x^2}} \right]_0^1$$
$$ = {2 \over 3} - {1 \over 2} = {1 \over 6}$$
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