Question

The area bounded by the curve $ y = 2x- x^2 $ and then straight line $ y = - x $ is given by

• A. $ \frac{9}{2} $
• B. $ \frac{43}{6} $
• C. $ \frac{35}{6} $
• D. $ \frac{23}{5} $

Answer 1

Answer: (A)

Given curves are
$y=2 x-x^{2} \,\,\,...(i)$
$y=-x\,\,\,...(ii)$
On solving Eqs. (i) and (ii) we get,
$-x=2 x-x^{2}$

$\Rightarrow x^{2}-3 x=0 \Rightarrow x(x-3)=0$
$\Rightarrow x=0$ or $x=3$
The area $(A)$ bounded by the curve is the shaded region shown in the figure.
Now, $A=\int\limits_{0}^{3}[(2 x\left.\left.-x^{2}\right)-(-x)\right] d x $
$=\int\limits_{0}^{3}\left(-x^{2}+3 x\right) d x$
$=\left|\frac{-x^{3}}{3}+\frac{3 x^{2}}{2}\right|_{0}^{3}=\frac{-1}{3}(3)^{3}+\frac{3}{2}(3)^{2}$
$=-9+\frac{27}{2}=\frac{9}{2}$ sq units