The area of the region given by [A=(x, y): x^2 y x+2,4-3 x] is :

Question

The area of the region given by

$$A=\left\{(x, y): x^{2} \leq y \leq \min \{x+2,4-3 x\}\right\}$$ is :

Accepted Answer

$A=\left\{(x, y): x^{2} \leq y \leq \min \{x+2,4-3 x\}\right.$

So area of required region

$$
\begin{aligned}
&A=\int_{-1}^{\frac{1}{2}}\left(x+2-x^{2}\right) d x+\int_{\frac{1}{2}}^{1}\left(4-3 x-x^{2}\right) d x \\
&=\left[\frac{x^{2}}{2}+2 x-\frac{x^{3}}{3}\right]_{-1}^{\frac{1}{2}}+\left[4 x-\frac{3 x^{2}}{2}-\frac{x^{3}}{3}\right]_{\frac{1}{2}}^{1} \\
&=\left(\frac{1}{8}+1-\frac{1}{24}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)+\left(4-\frac{3}{2}-\frac{1}{3}\right)-\left(2-\frac{3}{8}-\frac{1}{24}\right) \\
&=\frac{17}{6}
\end{aligned}
$$

Solutions · 1