If the area of the region (x,y):1+x^2 y minx+7,11-3x is A, then 3A is equal to :

Question

If the area of the region ${(x,y):1+{x}^{2}\le y\le min{x+7,11-3x}}$ is $A$, then $3A$ is equal to :

Accepted Answer

$$\begin{aligned}
& A=\int_{-2}^1\left(x+7-x^2-1\right) d x+\int_1^2\left(11+3 x-x^2-1\right) d x \
& =\left[\frac{x^2}{2}+6 x-\frac{x^3}{3}\right]_{-2}^1+\left[10 x-\frac{3 x^2}{2}-\frac{x^3}{3}\right]_1^2 \
& =\frac{50}{3} \Rightarrow 3 A=50 \quad \text { Option (1) }
\end{aligned}$$

Solutions · 1