Mathematics - Area Under Curves Question with Solution | TestHub

The given curves are $y=(x+1{)}^2$, $y=(x-1{)}^2$, and the line $y=\frac{1}{4}$.

First, finding the point of intersection of the two parabolas:

$(x+1{)}^2=(x-1{)}^2$

$x^2+2x+1=x^2-2x+1\Rightarrow 4x=0\Rightarrow x=0$

At $x=0$, $y=1$. Thus, the parabolas intersect at $(0,1)$.

Next, finding the points of intersection of the parabolas with the line $y=\frac{1}{4}$:

For $y=(x+1{)}^2$, $(x+1{)}^2=\frac{1}{4}\Rightarrow x+1=\pm \frac{1}{2}\Rightarrow x=-\frac{1}{2}$ or $x=-\frac{3}{2}$. The relevant intersection for the bounded region is $x=-\frac{1}{2}$.

For $y=(x-1{)}^2$, $(x-1{)}^2=\frac{1}{4}\Rightarrow x-1=\pm \frac{1}{2}\Rightarrow x=\frac{1}{2}$ or $x=\frac{3}{2}$. The relevant intersection is $x=\frac{1}{2}$.

The bounded region is symmetric about the y-axis, lying between $x=-\frac{1}{2}$ and $x=\frac{1}{2}$, bounded above by the parabolas and below by $y=\frac{1}{4}$.

The required area $A$ can be calculated by integrating with respect to $x$:

$A={\int }_{-1/2}^0\left((x+1{)}^2-\frac{1}{4}\right)dx+{\int }_0^{1/2}\left((x-1{)}^2-\frac{1}{4}\right)dx$

By symmetry, this simplifies to:

$A=2{\int }_0^{1/2}\left((x-1{)}^2-\frac{1}{4}\right)dx$

Evaluating the integral:

$A=2{\left[\frac{(x-1{)}^3}{3}-\frac{x}{4}\right]}_0^{1/2}$

$A=2\left(\left(\frac{(-1/2{)}^3}{3}-\frac{1/2}{4}\right)-\left(\frac{(-1{)}^3}{3}-0\right)\right)$

$A=2\left(\left(-\frac{1}{24}-\frac{1}{8}\right)-\left(-\frac{1}{3}\right)\right)$

$A=2\left(-\frac{4}{24}+\frac{1}{3}\right)=2\left(-\frac{1}{6}+\frac{1}{3}\right)$

$A=2\left(\frac{1}{6}\right)=\frac{1}{3}$ sq. units.

Answer: $\frac{1}{3}$