JEE MAIN - Mathematics (2023 - 24th January Morning Shift - No. 8)
The area enclosed by the curves $${y^2} + 4x = 4$$ and $$y - 2x = 2$$ is :
$${{22} \over 3}$$
9
$${{23} \over 3}$$
$${{25} \over 3}$$
Explanation
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Required area $=\int_{-4}^{2}\left(\frac{4-y^{2}}{4}-\frac{y-2}{2}\right) d y$
$=\int_{-4}^{2} \frac{8-2 y-y^{2}}{4} d y$
$=\frac{1}{4}\left\{8 y-y^{2}-\frac{y^{3}}{3}\right\}_{-4}^{2}$
$=9$ square units
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