Question

The volume of fuel in a tank changes at a rate given by the equation V'\left(t\right)=\frac{4t}{\sqrt[3]{t^2+3}},\:t\ge0 V ′ ( t ) = 4 t t 2 + 3 3 , t ≥ 0 At t=\sqrt{[\textbf{d}]} t = [ d ] hours the tank holds 9.[\textbf{b}] 9. [ b ] gallons. What is the volume of fuel in the tank after [\textbf{c}] [ c ] hours? Round answer to one decimal place.

Answer 1

`V'(t)=\frac{4t}{\sqrt[3]{t^2+3}}`

The volume of fuel in the tank after 9 hours is given by


`\displaystyle\int_0^9V'(t)\,\mathrm dt=\int_0^9(4t)/((t^2+3)^(1/3))\,\mathrm dt`

To compute the integral, substitute
`u=t^2+3`, so that
`\mathrm du=2t\,\mathrm dt`. Then when
`t=0`,
`u=0^2+3=3`; when
`t=9`,
`u=9^2+3=84`.


`\displaystyle\int_0^9V'(t)\,\mathrm dt=2\int_(t=0)^(t=9)(2t)/((t^2+3)^(1/3))\,\mathrm dt=2\int_(u=3)^(u=84)u^(-1/3)\,\mathrm du`

`=2\cdot\frac32u^(2/3)\bigg|_(u=3)^(u=84)=3(84^(2/3)-3^(2/3))\approx51.3`

So the volume of fuel after 9 hours is about 51.3 gallons.