Single Variable Calculus, Chapter 6, 6.1, Section 6.1, Problem 50

a.) Find the number $a$ such that the line $x = a$ bisects the area under the curve $\displaystyle y = \frac{1}{x^2}$, $1 \leq x \leq 4$.
b.) Find the number $b$ such that the line $y = b$ bisects the area in part(a)

a.) Since the area is divided into two parts, we assume that these two regions are equal, So...


$
\begin{equation}
\begin{aligned}
\int^a_1 \frac{1}{x^2} dx &= \int^4_a \frac{1}{x^2} dx\\
\\
\int^a_1 x^{-2} dx &= \int^4_a x^{-2} dx\\
\\
\left[ \frac{x^{-1}}{01} \right]^a_1 &= \left[ \frac{x^{-1}}{-1} \right]^4_a\\
\\
\frac{-1}{a} - \left[ \frac{-1}{1} \right] &= \frac{-1}{4} - \left[ \frac{-1}{a} \right]\\
\\
\frac{-1}{a} + 1 &= \frac{-1}{4} + \frac{1}{a}\\
\\
1 + \frac{1}{4} &= \frac{-1}{4} + \frac{1}{a}\\
\\
\frac{5}{4} &= \frac{2}{a}\\
\\
a &= \frac{8}{5}
\end{aligned}
\end{equation}
$


b.) The area in part(a) is $\displaystyle \int^4_1 \left(\frac{1}{x^2} \right)dx = \frac{3}{4}$ square units
Thus, the area of each sub divided region is $\displaystyle \frac{\frac{3}{4}}{2} = \frac{3}{8}$ square units.
We can use horizontal strip to evaluate the upper region to...

$
\begin{equation}
\begin{aligned}
\int^{y_2}_{y_1} \left(x_{\text{right}} - x_{\text{left}} \right) dy &= \frac{3}{8}\\
\\
\int^1_b \left( \frac{1}{\sqrt{y}} - 1 \right) dy &= \frac{3}{8}\\
\\
\left[ \frac{y^{\frac{1}{2}}}{\frac{1}{2}}\right]^1_b &= \frac{3}{8}\\
\\
2(1)^{\frac{1}{2}} - 1 - \left[ 2(b)^{\frac{1}{2}} - b\right] = \frac{3}{8}
\end{aligned}
\end{equation}
$

By factoring

$
\begin{equation}
\begin{aligned}
(\sqrt{b} -1 )^2 &= \frac{3}{8}\\
\\
\sqrt{b} &= \pm \sqrt{\frac{b}{8}} + 1\\
\\
b &= \left( \pm \sqrt{\frac{3}{8}} + 1 \right)^2
\end{aligned}
\end{equation}
$


We have, $b = 2.5997$ and $b = 0.1503$
We got two values of $b$. However, $b= 2.5997$ is outside the interval. Therefore, we choose $b = 0.1503$