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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.

Respuesta :

LammettHash
[tex]V'(t)=\dfrac{4t}{\sqrt[3]{t^2+3}}[/tex]

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

[tex]\displaystyle\int_0^9V'(t)\,\mathrm dt=\int_0^9\dfrac{4t}{(t^2+3)^{1/3}}\,\mathrm dt[/tex]

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

[tex]\displaystyle\int_0^9V'(t)\,\mathrm dt=2\int_{t=0}^{t=9}\frac{2t}{(t^2+3)^{1/3}}\,\mathrm dt=2\int_{u=3}^{u=84}u^{-1/3}\,\mathrm du[/tex]
[tex]=2\cdot\dfrac32u^{2/3}\bigg|_{u=3}^{u=84}=3(84^{2/3}-3^{2/3})\approx51.3[/tex]

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