Question

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

• A. $\frac{9}{2}$ sq. units
• B. $4$ sq. units
• C. $\frac{7\sqrt{17}}{2}$ sq. units
• D. $\frac{17\sqrt{17}}{6}$ sq. units

Answer 1

Answer: (D)

we have, $y=x^2-1$ $\dots (i)$
$x+y=3$ $\dots (ii)$
Solving $\left(i\right)and\left(ii\right)$, we get
$3-x=x^2-1\Rightarrow x^2+x-4=0$
$\Rightarrow x_1+x_2=-1$ and $x_1{x}_2=-4$
$\Rightarrow x_2-x_1=\sqrt{17}\dots \left(iii\right)$

Required area
$=\underset{x_1}{\overset{x_2}{\int }}\left[\left(3-x\right)-\left(x^2-1\right)\right]dx=\underset{x_1}{\overset{x_2}{\int }}\left(4-x-x^2\right)dx$
[using Equation (iii)]
$={\left[4x-\frac{x^2}{2}-\frac{x^3}{3}\right]}_{x_1}^{x_2}$
$=4\left(x_2-x_1\right)-\frac{1}{2}\left(x_2^2-x_1^2\right)-\frac{1}{3}\left(x_2^3-x_1^3\right)$
$=4\left(\sqrt{17}\right)-\frac{1}{2}\left(-1\right)\left(\sqrt{17}\right)-\frac{5\sqrt{17}}{3}=\frac{17\sqrt{17}}{6}$ sq. units