Question

Using integration, find the area of the region bounded by the line. $x-y+2=0,$ the curve $x=\sqrt{y}$ and $y$-axis.

Answer 1

${\int }_0^a\left[\left(x+2\right)-x^2\right]dx$

First we need to get ${}^{\prime }{a}^{\prime }$

This represent the point of intersection of the two curves, so we equate the two functions:

$\begin{array}{c}x+2=x^2\\ \Rightarrow x^2-x-2=0\\ \Rightarrow x^2-2x+x-2=0\\ \Rightarrow \left(x-2\right)\left(x+1\right)=0\end{array}$

Since, $x>0$, we reject the solution $x=-1$

$\begin{array}{c}Hence,a=2\\ So:A={\int }_0^2\left[\left(x+2\right)-x^2\right]dx\\ ={\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]}_0^2\\ =2+4-\frac{8}{3}\\ =\frac{10}{3}\end{array}$