The area (in sq units) of the region bounded by $x=-1, x=2, y=x^2+1$ and $y=2 x-2$ is

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Aicharya · Accepted Answer

Given curve is $y=x^2+1 \Rightarrow x^2=y-1$ and line $\quad y=2 x-2$

The intersection point of curve and line is $\begin{aligned} & \qquad x^2=2 x-2-1 \\ & \begin{array}{l}\text { Now, } \quad x^2-2 x+3=0 \\ \text { Hence, there is no point of intersection }\end{array} \\ & \therefore \text { Required area }=\int_{-1}^2\left(y_2-y_1\right) d x \\ & \qquad=\int_{-1}^2\left[\left(x^2+1\right)-(2 x-2)\right] d x \\ & \qquad=\left[\frac{x^3}{3}+x\right]_{-1}^2-\left[x^2-2 x\right]_{-1}^2 \\ & =\left[\frac{8}{3}+2-\left(-\frac{1}{3}-1\right)\right]-[4-4-(1+2)] \\ & =\frac{14}{3}+\frac{4}{3}-[-3] \\ & =6+3=9\end{aligned}$

The area (in sq units) of the region bounded by $x=-1, x=2, y=x^2+1$ and $y=2 x-2$ is

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