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

$\begin{array}{1 1} \frac{9}{2}\;sq.units. \\ 18\;sq.units. \\ \frac{7}{2}\;sq.units. \\ \frac{17}{2}\;sq.units \end{array}$

Toolbox:
• The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
• $\int x^n dx=\large\frac{x^{n+1}}{n+1}$$+c. Step 1: Given y=-x^2 and the line x+y+2=0 (i.e) y=-(x+2) Consider the curve y=-x^2.Clearly this curve is a parabola open downwards,with vertex (0,0). To obtain the points of intersection,let us solve the equation,by equating them. -x^2=-(x+2) \Rightarrow x^2-x-2=0 On factorising we get, (x-2)(x-1)=0. (i.e) x=2 or -1 and y=-4 or -1 Hence the points of intersection are (2,-4) and (-1,-1). Step 2: The required area is the shaded region shown in the fig below. The area of the shaded portion A=\int_{-1}^2(y_2-y_1)dx. Where y_2=-x^2 and y_1=-(x+2) A=\int_{-1}^2(-x^2)dx-\int_{-1}^2-(x+2)dx. On integrating we get, \;\;=-\begin{bmatrix}\large\frac{x^3}{3}\end{bmatrix}_{-1}^2$$+\begin{bmatrix}\big(\large\frac{x^2}{2}\big)\normalsize+2x\end{bmatrix}_{-1}^2$
Step 3:
On applying limits we get,
$\;\;=-\begin{bmatrix}\large\frac{2^3}{3}-\large\frac{(-1)^3}{3}\end{bmatrix}+\begin{bmatrix}\large\frac{2^2}{2}-\frac{(-1)^2}{2}\normalsize+2(2)-2(-1)\end{bmatrix}$
$\;\;=-\begin{bmatrix}\large\frac{8}{3}+\frac{1}{3}\end{bmatrix}+\begin{bmatrix}\large\frac{4}{2}-\frac{1}{2}+\normalsize 4+2\end{bmatrix}$
$\;\;=[-3]+[8-\large\frac{1}{2}]$
$\;\;=\large\frac{17}{2}$$-3=\large\frac{9}{2}$
Hence the required area is $\large\frac{9}{2}$sq.units.