How do you evaluate the definite integral #int v^(1/3)dv# from [-3,3]?

1 Answer
Morgan
Mar 2, 2017

See below.

Explanation:

#intv^(1/3)dx# from #[-3,3]#.

To integrate #v^(1/3)#, imagine undoing a derivative. When we take the derivative of a term raised to a power, we multiply the coefficient of the term by the power, and reduce the power by one. To undo this, we start by adding one to the power of #v^(1/3)#, or #1/3+1=4/3#.

We now have #v^(4/3)#.

But we're not done! If we were to try to take the derivative of #v^(4/3)#, we would get #4/3v^(1/3)#. We want a coefficient of one, so we need a coefficient for #v^(4/3)# that will come out to one when multiplied by #4/3#. This would be the inverse of that power, #3/4#.

So, #intv^(1/3)dx=3/4v^(4/3)#.

We can check our answer by taking the derivative of #3/4v^(4/3)# to see if we get the original integrand.

#d/dx(3/4v^(4/3))= 1*v^(1/3) => v^(1/3)#

Now we can evaluate the antiderivative on the interval #[-3,3]#.

#3/4[(3)^(4/3)-(-3)^(4/3)]#

But this comes out to #0#!

Thus, #intv^(1/3)dx# from #[-3,3] = 0#.

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