Mathematics - Area Under Curves Question with Solution | TestHub

Given the equations of the parabola $y^2=4x$ and the line $y=2x-4$.

From the equation of the line, we have $x=\frac{y+4}{2}$.

Substituting this into the equation of the parabola to find the points of intersection:

$y^2=4\left(\frac{y+4}{2}\right)$

$y^2=2y+8$

$y^2-2y-8=0$

$(y-4)(y+2)=0$

$y=-2,4$

The area enclosed between the curves is given by integrating with respect to $y$ from $y=-2$ to $y=4$:

$A={\int }_{-2}^4\left(x_{\text{line}}-x_{\text{parabola}}\right)dy$

$A={\int }_{-2}^4\left(\frac{y+4}{2}-\frac{y^2}{4}\right)dy$

$A={\left[\frac{y^2}{4}+2y-\frac{y^3}{12}\right]}_{-2}^4$

$A=\left(\frac{16}{4}+8-\frac{64}{12}\right)-\left(\frac{4}{4}-4-\frac{-8}{12}\right)$

$A=\left(4+8-\frac{16}{3}\right)-\left(1-4+\frac{2}{3}\right)$

$A=\left(12-\frac{16}{3}\right)-\left(-3+\frac{2}{3}\right)$

$A=\frac{20}{3}-\left(-\frac{7}{3}\right)$

$A=\frac{27}{3}=9$ sq. units

Answer: 9 sq. units