Find the area of the region enclosed by the parabola $x^2=y$ and the line $y=x+2$.

Toolbox:
• Let $f(x)$ be a continuous function defined on $[a,b]$.Then the area bounded by the curve $y=f(x)$,the $x$-axis and the ordinate $x=a$ and $x=b$ is given by
• $A=\int_a^bf(x) dx$ or $\int_a^b ydx.$
Step 1:
Given $x^2=y$ and the line $y=x+2$
Consider the curve $x^2=y$
This is a parabola which is open upwards.
The focus of the parabola is $(4a=1\Rightarrow a=\large\frac{1}{4})$$\Rightarrow (0,\large\frac{1}{4}) y=x+2 is a straight line. Step 2: The points of intersection can be obtained by solving the two equations. x^2=x+2\Rightarrow x^2-x-2=0 On factorising we get, (x-2)(x+1)=0 Hence x=2 or -1 The points of intersection are (2,4)(-1,1) Step 3: The required area is the shaded portion shown in the fig. Area of the shaded portion can be evaluated A=\int_{-1}^2(y_2-y_1)dx. Where y_2=x+2 and y_1=x^2 Hence A=\int_{-1}^2[(x+2)-x^2]dx. Step 4: On integrating we get, A=\begin{bmatrix}(\large\frac{x^2}{2}\normalsize+2x)-\large\frac{x^3}{3}\end{bmatrix}_{-1}^2 On applying limits we get, A=[(2+4)-\large\frac{1}{2}$$-2)]-[\large\frac{8}{3}+\frac{1}{3}]$$=6+\large\frac{3}{2}$$-3$
$\;\;=3+\large\frac{3}{2}=\frac{9}{2}$sq.units.
edited Dec 22, 2013