Question

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

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

Answer 1

Answer: (D)

$\frac{9}{2}$

The two meet at $3-x=x^2+1\Rightarrow x^2+x-2=0\Rightarrow (x+2)(x-1)=0\Rightarrow x=-2,1$

$\therefore$ Required area $={\int }_{-2}^1[(3-x)-(x^2+1)]dx$

$={\int }_{-2}^1(2-x-x^2)dx={[2x-\frac{x^2}{2}-\frac{x^3}{3}]}_{-2}^1$

$=(2-\frac{1}{2}-\frac{1}{3})-(-4-\frac{4}{2}+\frac{8}{3})=\frac{9}{2}$