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Q.
The area bounded by $x = -4y^{2}$ and $x - 1 = 5y^{2}$ is
Application of Integrals
Solution:
We have $x = -4y^{2} \, ...\left(i\right)$, a parabola with vertex $\left(0,0\right)$
and $- 5y^{2} = x - 1 \, ...\left(ii\right)$, a parabola with vertex $\left(1/5,0\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we get $y=\pm\, 1$
Area $=\int_{-1}^{1}\left(1-5y^{2}-\left(-4y^{2}\right)\right)dy
=\int_{-1}^{1}\left(1-y^{2}\right)dy$
$=2\int_{0}^{1}\left(1-y^{2}\right)dy$
$=2\left[y-\frac{y^{3}}{3}\right]_{0}^{1}$
$=2\left(1-\frac{1}{3}\right)$
$=\frac{4}{3}$ sq. units
