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Sketch the region R defined by 1 ≤ x ≤ 2 and 0 ≤ y ≤ 1/x^3 .

AFind (exactly) the number a such that the line x = a divides R into two parts of equal area.
BThen find (to 3 decimal places) the number b such that the line y = b divides R into two parts of equal area.

Verified Answer :

LammettHash

For part (a), you're looking to find \(a\) such that

\(\displaystyle\int_1^a\frac{\mathrm dx}{x^3}=\int_a^2\frac{\mathrm dx}{x^3}\)

You have

\(\displaystyle\int_1^a\frac{\mathrm dx}{x^3}=-\frac1{2x^2}\bigg|_{x=1}^{x=a}=-\frac12\left(\frac1{a^2}-1\right)\)

and

\(\displaystyle\int_a^2\frac{\mathrm dx}{x^3}=-\frac1{2x^2}\bigg|_{x=a}^{x=2}=-\frac12\left(\frac14-\frac1{a^2}\right)\)

Setting these equal, you get

\(\displaystyle-\frac12\left(\frac1{a^2}-1\right)=-\frac12\left(\frac14-\frac1{a^2}\right)\implies a=2\sqrt{\dfrac25}\)

For part (b), you have

\(y=\dfrac1{x^3}\implies x=\dfrac1{\sqrt[3]y}\)

and you want to find \(b\) such that

\(\displaystyle\int_0^{1/8}\mathrm dy+\int_{1/8}^b\frac{\mathrm dy}{\sqrt[3]y}=\int_b^1\frac{\mathrm dy}{\sqrt[3]y}\)

You have

\(\displaystyle\int_0^{1/8}\mathrm dy+\int_{1/8}^b\frac{\mathrm dy}{y^{1/3}}=\frac18+\frac32y^{2/3}\bigg|_{y=1/8}^{y=b}=-frac14+\frac32b^{2/3}\)

and

\(\displaystyle\int_b^1\frac{\mathrm dy}{y^{1/3}}=\frac32y^{2/3}\bigg|_{y=b}^{y=1}=\frac32-\frac32b^{2/3}\)

Setting them equal gives

\(-\dfrac14+\dfrac32b^{2/3}=\dfrac32-\dfrac32b^{2/3}\implies b=\dfrac7{24}\sqrt{\dfrac73}\approx0.446\)

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