Question

Find the area enclosed by the curve $y=-x^2$ and the straight line $x+y+2=0$

Answer 1

Given curves:
We have $y=-x^2\dots$ $\left(1\right)$ which is downward parabola
$x+y+2=0$
Solving $\left(1\right)$ and $\left(2\right)$ we get
$\Rightarrow x^2=x+2$

$\Rightarrow x^2-x-2=0$

$\Rightarrow (x-2)(x+1)=0$

Either $(x-2)=0$ or $(x+1)=0$

$\Rightarrow x=2\text{or}x=-1$

So, the area bounded by curves is shaded in the diagram below:



$\text{Area}={\int }_{x_1}^{x_2}(y_2-y_1)dx$

$\Rightarrow \text{Area}={\int }_{-1}^2\{-x^2-(-x-2\left)\right\}dx$

$\Rightarrow \text{Area}={\int }_{-1}^2(x+2-x^2)dx$

$\Rightarrow \text{Area}={[\frac{x^2}{2}+2x-\frac{x^3}{3}]}_{-1}^2$

$[\therefore {\int }_a^b{x}^{n}dx={\left[\frac{x^{n+1}}{n+1}\right]}_a^b]$

$\Rightarrow \text{Area}=(6-\frac{8}{3})-(\frac{1}{2}-2+\frac{1}{3})$

$\Rightarrow \text{Area}=\frac{10}{3}+\frac{7}{6}=\frac{27}{6}$

$\Rightarrow \text{Area}=\frac{9}{2}Sq\text{units.}$

Hence the required area is $\frac{9}{2}Sq$ units.