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Q.
The area enclosed by the curves $3x^{2} + 5y = 32$ and $y=|x-2|$ is
Application of Integrals
Solution:
Given that $3x^{2} + 5y = 32$, a parabola with vertex $\left(0,6.4\right)$.
$A=\int_{-2}^{3} \frac{32-3x^{2}}{5} dx -\frac{4\times4}{2}-\frac{1\times1}{2}$
$=32-\frac{1}{5}\left(27+8\right)-\frac{17}{2}$
$=25-\frac{17}{2}$
$=\frac{33}{2}$ sq. units
