Question

What is `int_(1/3)^(2/3) \ x^3sqrt(4-9x^2) \ dx`?

Answer 1

`int_(1/3)^(2/3) \ x^3sqrt(4-9x^2) \ dx = 11/405 sqrt(3)`

Explanation:

We seek:

`I = int_(1/3)^(2/3) \ x^3sqrt(4-9x^2) \ dx`

We can perform a substitution of the form:

`u = 4-9x^2 iff 9x^2=4-u`, and `(du)/dx = -18x`

This substitution will require an associated change of limits from `x` to `u`, and we have:

When `x= { (1/3), (2/3) :} => u= { (3), (0) :}`

And we manipulate the integral as follows:

`I = int_(1/3)^(2/3) \ (9x^2)/9(-18x)/(-18)sqrt(4-9x^2) \ dx`
`\ \ = -1/162 \ int_(1/3)^(2/3) \ (9x^2)(-18x)sqrt(4-9x^2) \ dx`
`\ \ = -1/162 \ int_(3)^(0) \ (4-u)sqrt(u) \ du`
`\ \ = 1/162 \ int_(0)^(3) \ (4-u)u^(1/2) \ du`
`\ \ = 1/162 \ int_(0)^(3) \ 4u^(1/2) - u^(3/2) \ du`
`\ \ = 1/162 \ [ (4u^(3/2))/(3/2) - (u^(5/2))/(5/2) ]_(0)^(3)`

`\ \ = 2/162 [ (4u^(3/2))/(3) - (u^(5/2))/(5) ]_(0)^(3)`

`\ \ = 1/81 (4/3 (3)^(3/2) - 1/5 (3)^(5/2) )`

`\ \ = 1/81 (4/3 (3sqrt(3)) - 1/5 (9sqrt(3)) )`

`\ \ = 1/81 (12/3-9/5) sqrt(3)`

`\ \ = 1/81 (11/5) sqrt(3)`

`\ \ = 11/405 sqrt(3)`

So the last answer is correct.