Mathematics - Area Under Curves Question with Solution | TestHub

Given equations are $x^2=8y$ and $x-8y+2=0$.

From the line equation, we have $8y=x+2$.

Substituting this into the equation of the parabola, we get:

$x^2=x+2$

$x^2-x-2=0$

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

The points of intersection are at $x=-1$ and $x=2$.

The required area $A$ bounded by the line and the parabola is given by:

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

$A={\int }_{-1}^2\left(\frac{x+2}{8}-\frac{x^2}{8}\right)dx$

$A=\frac{1}{8}{\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]}_{-1}^2$

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

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

$A=\frac{1}{8}\left(\frac{10}{3}-\left(-\frac{7}{6}\right)\right)$

$A=\frac{1}{8}\left(\frac{20}{6}+\frac{7}{6}\right)$

$A=\frac{1}{8}\left(\frac{27}{6}\right)=\frac{1}{8}\times \frac{9}{2}=\frac{9}{16}$ sq. units.

Answer: $\frac{9}{16}$ sq. units