The area bounded by the curve $y=\frac{x^2}{4}+\frac{x}{4}-\frac{1}{2}$ with $X-$ axis in $\left[0,2\right]$

Solution:

The given curve is $y=\frac{x^2}{4}+\frac{x}{4}-\frac{1}{2}$

To find the point of intersection of the abvoe curve and $x-$axis solve the equaiton $\frac{x^2}{4}+\frac{x}{4}-\frac{1}{2}=0$

it implies that 

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

Hence, $x=1,-2$

In the given interval $\left[0,2\right]$, the curve intersect the X- axis at $x=1$

The rough sketch of the required area is as shwon below

The required area is area of shaded region. 

The area of the shaded reegion is $A={\int }_0^{2}ydx$

$\begin{array}{rcl}A& =& {\int }_0^{2}ydx\\ & =& \left|{\int }_0^1\left(\frac{x^2}{4}+\frac{x}{4}-\frac{1}{2}\right)dx\right|+\left|{\int }_1^2\left(\frac{x^2}{4}+\frac{x}{4}-\frac{1}{2}\right)dx\right|\\ & =& {\left|\frac{x^3}{12}+\frac{x^2}{8}-\frac{x}{2}\right|}_0^1+{\left|\frac{x^3}{12}+\frac{x^2}{8}-\frac{x}{2}\right|}_1^2\\ & =& \left|\frac{1}{12}+\frac{1}{8}-\frac{1}{2}\right|+\left|\frac{8}{12}+\frac{4}{8}-\frac{2}{2}-\frac{1}{12}-\frac{1}{8}+\frac{1}{2}\right|\\ & =& \left|\frac{2+3-12}{24}\right|+\left|\frac{7}{12}+\frac{3}{8}-\frac{1}{2}\right|\\ & =& \frac{7}{24}+\left|\frac{14+9-12}{24}\right|\\ & =& \frac{7}{24}+\frac{11}{24}\\ & =& \frac{18}{24}=\frac{3}{4}\end{array}$