Question

The area bounded by the curves $y = - x^2 + 3$ and $y = 0$ is

• A. $\sqrt{3} + 1$
• B. $\sqrt{3}$
• C. $4 \sqrt{3}$
• D. $5 \sqrt{3}$

Answer 1

Answer: (C)

$4 \sqrt{3}$

Answer 2

We have,
$y =-x^{2}+3$
$\Rightarrow x^{2} =-(y-3)$
The above curve intersect $X$ -axis at the points where $y=0$
$\therefore x^{2}=3$
$\Rightarrow x=\pm \sqrt{3}$

$\therefore$ Point of intersection with $X$ -axis are $(\pm \sqrt{3}, 0)$.
$\therefore$ Required area $=2 \int\limits_{0}^{\sqrt{3}} y\, d x$
$=2 \int\limits_{0}^{\sqrt{3}}\left(-x^{2}+3\right) d x$
$=2\left[\frac{-x^{3}}{3}+3 x\right]_{0}^{\sqrt{3}}$
$=2\left[\frac{-3 \sqrt{3}}{3}+3 \sqrt{3}\right]$
$=2[-\sqrt{3}+3 \sqrt{3}]$
$=4 \sqrt{3}$ sq units