Mathematics - Area Under Curves Question with Solution | TestHub

Given equations of the curves are $y-x=2\Rightarrow y=x+2$ and $x^2=y$.

To find the points of intersection, substitute $y=x^2$ in $y=x+2$:

$x^2=x+2$

$x^2-x-2=0$

$(x-2)(x+1)=0$

$x=-1,2$

The required area $A$ is bounded by the line and the parabola between $x=-1$ and $x=2$.

$A={\int }_{-1}^2(y_{\text{line}}-y_{\text{parabola}})dx$

$A={\int }_{-1}^2(x+2-x^2)dx$

$A={\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]}_{-1}^2$

$A=\left(\frac{4}{2}+4-\frac{8}{3}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)$

$A=\left(6-\frac{8}{3}\right)-\left(\frac{5}{6}-2\right)$

$A=\frac{10}{3}-\left(-\frac{7}{6}\right)$

$A=\frac{20}{6}+\frac{7}{6}=\frac{27}{6}=\frac{9}{2}$ sq. units.

Answer: $\frac{9}{2}$