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How do you integrate by substitution #int 2piy(8-y^(3/2))dy#?

Answer 1

# int 2piy(8-y^(3/2))dy = 2pi{ 4y^2 - 2/7y^(7/2)} + C#

You do not need to use a substitution as the integral can be calculated "as is":

# int 2piy(8-y^(3/2))dy = 2piint 8y-y^(5/2)dy # # :. int 2piy(8-y^(3/2))dy = 2pi{ 8y^2/2 -y^(7/2)/(7/2)} + C# # :. int 2piy(8-y^(3/2))dy = 2pi{ 4y^2 - 2/7y^(7/2)} + C#

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Answer 2

Luke Alvarez

To integrate ( \int 2\pi y(8 - y^{3/2}) , dy ) by substitution, let ( u = y^{3/2} ). Then, ( du = \frac{3}{2}y^{\frac{1}{2}} , dy ). Rearranging this gives ( dy = \frac{2}{3}u^{-\frac{1}{2}} , du ). Substituting these into the integral, we get:

[ \int 2\pi y(8 - y^{3/2}) , dy = \int 2\pi y(8 - u) \frac{2}{3}u^{-\frac{1}{2}} , du ]

This simplifies to:

[ \frac{4\pi}{3} \int (8u - u^2) , du ]

Now integrate ( 8u - u^2 ) with respect to ( u ):

[ \frac{4\pi}{3} \left(4u^2 - \frac{1}{3}u^3\right) + C ]

Finally, substitute ( u = y^{3/2} ) back in:

[ \frac{4\pi}{3} \left(4y^3 - \frac{1}{3}y^{9/2}\right) + C ]

So, the integrated expression is ( \frac{4\pi}{3} \left(4y^3 - \frac{1}{3}y^{9/2}\right) + C ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.