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

Solution:

Given, y = 1/x² for 1 ≤ x ≤ 4.

1) we have to find a such that the line x = a bisects the area under the given curve for 1 ≤ x ≤ 4.

So, \(\int_{1}^{a}\frac{dx}{x^{2}}=\int_{a}^{4}\frac{dx}{x^{2}}\)

\(\left [ \frac{-1}{x} \right ]_{1}^{a}=\left [ \frac{-1}{x} \right ]_{a}^{4}\)

\(\\(\frac{-1}{a})-(\frac{-1}{1})=(\frac{-1}{4})-(\frac{-1}{a})\\(\frac{-1}{a})+1=(\frac{-1}{4})+(\frac{1}{a})\\(\frac{1}{a})+(\frac{1}{a})=1+\frac{1}{4}\\\frac{2}{a}=\frac{5}{4}\)

5a = 8

a = 8/5

2) Integrating with respect to y will make things easier.

y = 1/x²

x² = 1/y

x = ±1/√y

We know, x > 0 so take positive square root

x = 1/√y

Now we want to find b,

\(\int_{0}^{b}\frac{dy}{\sqrt{y}}=\int_{b}^{1}\frac{dy}{\sqrt{y}}\)

\(\\\left [ 2\sqrt{y}\right ]_{0}^{b}=\left [ 2\sqrt{y}\right ]_{b}^{1}\\2(\sqrt{b}-0)=2(1-\sqrt{b})\\2\sqrt{b}=2-2\sqrt{b}\\\sqrt{b}=1-\sqrt{b}\\2\sqrt{\sqrt{b}}=1\\\sqrt{b}=\frac{1}{2}\)

Squaring on both sides,

b = 1/4

Therefore, the values of a and b are 8/5 and 1/4.

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

Summary:

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