Question

The area of the region bounded by the parabola $y=x^2+1$ and the straight line $x+y=3$ is given by

• A. $\frac{45}{7}$ sq. units
• B. $\frac{25}{4}$ sq. units
• C. $\frac{\pi }{18}$ sq. units
• D. $\frac{9}{2}$ sq. units

Answer 1

Answer: (D)

we have, $y=x^2+1\dots \left(i\right)$
and $x+y=3\left(ii\right)$
Solving $\left(i\right)and\left(ii\right)$, we get
$x^2+x-2=0\Rightarrow x=-2,1$

Required area is the shaded region in the figure
$A=\underset{-2}{\overset{1}{\int }}$ (line-parabola) $dx=\underset{-2}{\overset{1}{\int }}\left\{3-x-\left(x^2+1\right)\right\}dx$
$={\left[2x-\frac{x^2}{2}-\frac{x^3}{3}\right]}_{-2}^1$
$=\left(2-\frac{1}{2}-\frac{1}{3}\right)-\left(-4-2+\frac{8}{3}\right)=\frac{9}{2}$ sq. units