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, $y=x^{2}-1$ $\quad$ $\ldots (i) $
$x+y=3$ $\quad$ $\ldots (ii)$
Solving $\left(i\right) and \left(ii\right)$, we get
$3 - x = x^{2} - 1 \quad\Rightarrow \quad x^{2} + x- 4 = 0$
$\Rightarrow \quad x_{1}+x_{2}=-1$ and $x_{1}x_{2}=-4$
$\Rightarrow \quad x_{2}-x_{1}=\sqrt{17}\quad\quad\ldots\left(iii\right)$

Required area
$=\int\limits_{x_1}^{x_{2}}\left[\left(3-x\right)-\left(x^{2}-1\right)\right]dx=\int\limits_{x_1}^{x_{2}}\left(4-x-x^{2}\right)dx$
[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