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Q.
The area bounded by the curve $x^{2} = 4y + 4$ and line $3x + 4y = 0$ is
Application of Integrals
Solution:
We have, $x^{2} = 4y + 4\quad\ldots\left(i\right)$
and $3x + 4y = 0\quad\ldots\left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we get $x = -4,1$
Required area = area of shaded region.
$=\int_{-4}^{1}\left(-\frac{3x}{4}-\frac{x^{2}}{4}+1\right)dx$
$=-\frac{3}{8}\left(1-16\right)-\frac{1}{12}\left(1+64\right)+5$
$=\frac{45}{8}-\frac{5}{12}$
$=\frac{125}{24}$ sq. units
