Question

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

• A. $\frac{59}{6}$
• B. $\frac{9}{2}$
• C. $\frac{10}{3}$
• D. $\frac{7}{6}$

Answer 1

The correct option is B $\frac{9}{2}$

Given, $y=x^2+1$ and $x+y=3$



On solving
$3-x=x^2+1$
$x^2+x-2=0$
$(x+2)(x-1)=0$
$\Rightarrow x=-2$ and $1$
So, Area $=\int \int dydx={\int }_{x=-2}^1{\int }_{x^2+1}^{3-x}dydx$
$={\int }_{-2}^1(3-x-x^2-1)dx$
$={\int }_{-2}^1(2dx-xdx-x^{2}dx)$
$=\frac{9}{2}$