Question

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

Answer 1

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`.