Question

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

Answer 1

Let's find the point of intersection of curve $y=x^2$ & $y=x+2$ i.e., finding $A$ & $B$ points

$x^2=y\Rightarrow x^2=x+2\Rightarrow x^2-x-2=0\Rightarrow (x-2)(x+1)=0\therefore x=-1,2\Rightarrow y=1,4$

So $A(-1,1)$ and $B(2,4)$

Required Area $=$ Area of trapezium $ABCD-$Area of $AOBCD$

Now, Area of trapezium$=\frac{1}{2}\times$(sum of parallel sides)$\times$height

$=\frac{1}{2}\times (1+4)\times 3$

$=\frac{15}{2}$units

Area under $AOBCD$ is

${\int }_{-1}^2{x}^{2}dx={\frac{x^3}{3}}_1^2=3$

$\therefore$ Area required $=\frac{15}{3}-3=\frac{15-6}{2}=\frac{9}{2}$units.