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Q. The area bounded by the curve $x^2=y, x^2=-y$ and $y^2=4 x-3$ is

Integrals

Solution:

Area $A=2 \int\limits_0^1\left(\frac{y^2+3}{4}-\sqrt{y}\right) d y$
$A=\frac{1}{3}$ sq. units
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