Question

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

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

Answer 1

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


`\displaystyle\int_1^a(\mathrm dx)/(x^3)=\int_a^2(\mathrm dx)/(x^3)`

You have


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

and


`\displaystyle\int_a^2(\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√(\frac25)`

For part (b), you have


`y=\frac1{x^3}\implies x=\frac1{\sqrt[3]y}`

and you want to find
`b` such that


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

You have


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

and


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

Setting them equal gives


`-\frac14+\frac32b^(2/3)=\frac32-\frac32b^(2/3)\implies b=\frac7{24}√(\frac73)\approx0.446`