The area of the region bounded by the curve $y=4 x-x^{2}$ and the $x$ -axis is
The area of the region bounded by the curve $y=4 x-x^{2}$ and the $x$ -axis is
$\frac{16}{3}$ sq. units
$\frac{32}{3}$ sq. units
32 sq. units
16 sq. units
Solution
We have $y=4 x-x^{2}$
When $y=0$, we get $x(4-x)=0 \Rightarrow x=0,4$
Required area is shaded.
$\begin{aligned}
\therefore A &=\int_{0}^{4}\left(4 x-x^{2}\right) d x \\
&=\left[\frac{4 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{4}=\left[2 x^{2}-\frac{x^{3}}{3}\right]_{0}^{4} \\
&=\left|2(16-0)-\frac{64-0}{3}\right|=\left|32-\frac{64}{3}\right| \\
&=\frac{32}{3}
\end{aligned}$
