Question

If $A_{1}$ = Area enclosed by the curve $y = -x^{2}$ and the straight line $x + y + 2 = 0$
$A_{2}$ = Area bounded by the curve $y=\sqrt{x}, x=2y+3$ in the first quadrant and $x$-axis.
Then which of the following is $CORRECT?$

• A. $A_{1}+A_{2}=9$
• B. $A_{1}-A_{2}=2$
• C. $A_{1}\times A_{2}=9$
• D. $\frac{A_{2}}{A_{1}}=2$

Answer 1

Answer: (D)

We have, $y = -x^{2} \ldots\left(i\right)$,
and $x + y + 2 = 0 ...\left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we get $- x - 2 = -x^{2}$
$\Rightarrow x^{2}-x-2=0$
$\Rightarrow \left(x-2\right)\left(x+1\right)=0$
$\Rightarrow x=2, -1$

Thus the area of the shaded region
$A_{1}=\left|\int\limits_{1}^{2}\left(-x-2+x^{2}\right)dx\right|$
$=\left|\int\limits_{-1}^{2}\left(x^{2}-x-2\right)dx\right|$
$=\left|\left[\frac{x^{3}}{3}-\frac{x^{2}}{2}-2x\right]_{-1}^{2}\right|$
$=\left|\frac{8}{3}-\frac{4}{2}-4+\frac{1}{3}+\frac{1}{2}-2\right|$
$=\left|\frac{16-12-24+2+3-12}{6}\right|$
$=\left|-\frac{27}{6}\right|=\frac{9}{2}$ sq. units
Also, we have, $y=\sqrt{x}\quad\ldots\left(i\right)$
and $x = 2y+3 \quad\ldots\left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we get $y=\sqrt{2y+3}$
Squaring on both sides, we get $y^{2}-2y-3=0$
$\Rightarrow \left(y+1\right)\left(y-3\right)=0$
$\Rightarrow y=-1,3$

Thus the required area of the shaded region
$A_{2}=\int\limits_{0}^{3}\left(2y+3-y^{2}\right)dy $
$= \left[y^{2}+3y-\frac{y^{3}}{3}\right]_{0}^{3}$
$=\left[9+9-9-0\right]=9$ sq. units