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Q.
The area of the region bounded by the parabola $y =x^{2} + 1$ and the straight line $x + y = 3$ is given by
Application of Integrals
Solution:
we have, $y=x^{2}+1 \quad\ldots\left(i\right)$
and $x + y = 3\quad\left(ii\right)$
Solving $\left(i\right) and \left(ii\right)$, we get
$x^{2} + x - 2 = 0 \Rightarrow \quad x=-2, 1$
Required area is the shaded region in the figure
$A=\int\limits_{-2}^{1}$ (line-parabola) $dx=\int\limits_{-2}^{1}\left\{3-x-\left(x^{2}+1\right)\right\}dx$
$=\left[2x-\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{-2}^{1}$
$=\left(2-\frac{1}{2}-\frac{1}{3}\right)-\left(-4-2+\frac{8}{3}\right)=\frac{9}{2}$ sq. units
