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
$y=2x-x^2...(i)$
$y=-x...(ii)$
On solving Eqs. (i) and (ii) we get,
$-x=2x-x^2$

$\Rightarrow x^2-3x=0\Rightarrow x(x-3)=0$
$\Rightarrow x=0$ or $x=3$
The area $(A)$ bounded by the curve is the shaded region shown in the figure.
Now, $A=\underset{0}{\overset{3}{\int }}[(2x-x^2)-(-x)]dx$
$=\underset{0}{\overset{3}{\int }}\left(-x^2+3x\right)dx$
$={\left|\frac{-x^3}{3}+\frac{3x^2}{2}\right|}_0^3=\frac{-1}{3}(3{)}^3+\frac{3}{2}(3{)}^2$
$=-9+\frac{27}{2}=\frac{9}{2}$ sq units