Let the area of the region enclosed by the curves y=3 x, 2 y=27-3 x and [y=3 x-x x] be A.

Question

Let the area of the region enclosed by the curves $y=3 x, 2 y=27-3 x$ and $$y=3 x-x \sqrt{x}$$ be $A$. Then $10 A$ is equal to

Accepted Answer

$$\begin{aligned}
& y^{\prime}=3-\frac{3}{2} \sqrt{x} \
& y^{\prime}=3\left(1-\frac{\sqrt{x}}{2}\right)
\end{aligned}$$

$$\begin{aligned}
& \text { Area }=\int_\limits0^3\left(3 x-\left(3 x-x^{3 / 2}\right)\right) d x+\int_\limits3^9 \frac{27-3 x}{2}-\left(3 x-x^{3 / 2}\right) d x \
& =\int_\limits0^3 x^{3 / 2} d x+\frac{1}{2} \int_\limits3^9\left(27-9 x+2 x^{3 / 2}\right) d x \
& =\left[\frac{2}{5} x^{5 / 2}\right]_0^3+\frac{1}{2}\left[27 x-\frac{9 x^2}{2}+2 \times \frac{2}{5} x^{5 / 2}\right]_3^9 \
& =\frac{2}{5} \times 9 \sqrt{3}+\frac{1}{2}\left[243-\frac{729}{2}+\frac{4}{5} \times 81 \times 3-81+\frac{81}{2}-\frac{4}{5} \times 9 \sqrt{3}\right] \
& =\frac{1}{2}\left[\frac{486-729-81}{2}+\frac{972}{5}\right]=\frac{81}{5} \
& A=\frac{81}{5} \
& \therefore 10 A=10 \times \frac{81}{5}=162
\end{aligned}$$

Solutions · 1