Mathematics - Area Under Curves Question with Solution | TestHub

To find the points of intersection, we solve the equations $y^2=2x$ and $y=4x-1$ simultaneously.

From the equation of the line, we have $x=\frac{y+1}{4}$.

Substituting this into the equation of the parabola gives:

$y^2=2\left(\frac{y+1}{4}\right)$

$2y^2=y+1$

$2y^2-y-1=0$

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

This gives the $y$-coordinates of the intersection points as $y=-\frac{1}{2}$ and $y=1$.

The region is bounded by the curves $x=\frac{y^2}{2}$ and $x=\frac{y+1}{4}$. For $y\in \left[-\frac{1}{2},1\right]$, the line lies to the right of the parabola, so $\frac{y+1}{4}\ge \frac{y^2}{2}$.

The area $A$ of the region is given by integrating with respect to $y$:

$A={\int }_{-1/2}^1\left(\frac{y+1}{4}-\frac{y^2}{2}\right)dy$

$A={\left[\frac{y^2}{8}+\frac{y}{4}-\frac{y^3}{6}\right]}_{-1/2}^1$

Evaluating at the upper limit $y=1$:

$\frac{1}{8}+\frac{1}{4}-\frac{1}{6}=\frac{3+6-4}{24}=\frac{5}{24}$

Evaluating at the lower limit $y=-\frac{1}{2}$:

$\frac{1}{8}{\left(-\frac{1}{2}\right)}^2+\frac{1}{4}\left(-\frac{1}{2}\right)-\frac{1}{6}{\left(-\frac{1}{2}\right)}^3=\frac{1}{32}-\frac{1}{8}+\frac{1}{48}$

The least common multiple of $32,8,$ and $48$ is $96$.

$\frac{3}{96}-\frac{12}{96}+\frac{2}{96}=-\frac{7}{96}$

Subtracting the lower limit value from the upper limit value:

$A=\frac{5}{24}-\left(-\frac{7}{96}\right)=\frac{20}{96}+\frac{7}{96}=\frac{27}{96}$

Simplifying the fraction gives:

$A=\frac{9}{32}$ sq. units.

Answer: $\frac{9}{32}$