Question

Find the number a such that the line x = a bisects the area under the curve y = 1/x2 for 1 ≤ x ≤ 4. 8 5​ (b) find the number b such that the line y = b bisects the area in part (a).

Answer 1

a. We're looking for $a$ such that

$\displaystyle\int_1^a\frac{\mathrm dx}{x^2}=\int_a^4\frac{\mathrm dx}{x^2}$

$-\dfrac1x\bigg|_{x=1}^{x=a}=-\dfrac1x\bigg|_{x=a}^{x=4}$

$1-\dfrac1a=\dfrac1a-\dfrac14$

$\dfrac54=\dfrac2a\implies\boxed{a=\dfrac85}$

b. Integrating with respect to $y$ will make things easier.

$y=\dfrac1{x^2}\implies x=\dfrac1{\sqrt y}$

(where we take the positive square root because we know $x>0$)

Now we want to find $b$ such that

$\displaystyle\int_0^b\frac{\mathrm dy}{\sqrt y}=\int_b^1\frac{\mathrm dy}{\sqrt y}$

$2\sqrt y\bigg|_{y=0}^{y=b}=2\sqrt y\bigg|_{y=b}^{b=1}$

$2\sqrt b=2-2\sqrt b$

$4\sqrt b=2\implies\boxed{b=\dfrac14}$