Question

How do you find `int_1^e 2x^2lnx`?

Answer 1

You can use integration-by-parts to get `2/9+ 4/9 e^(3) approx 9.1491275`

Explanation:

Usually, when your integrand involves a logarithm, it's a good idea to try integration by parts with `u=ln(x)` and `du = 1/x dx`. In this case, we also get `dv = 2x^2 dx` so that `v=2/3 x^3`.

The integration-by-parts formula is `int u\ dv=uv- int v\ du`, so

`int 2x^2 ln(x)\ dx=2/3 x^{3}ln(x)-2/3 int x^2 dx`

`=2/3 x^{3}ln(x) - 2/9 x^{3}+C`.

Therefore, the definite integral is

`int_{1}^{e}2x^{2}ln(x)\ dx=(2/3 e^{3} * 1 - 2/9 e^{3}) - (0 - 2/9)`

`2/9+ 4/9 e^(3) approx 9.1491275`

Answer 2

`int_1^e 2x^2 ln(x) dx =2/9( 2e^3 + 1 )`

Explanation:

`int_1^e 2x^2 ln(x) dx = 2/3 int_1^e frac{d}{dx}(x^3) ln(x) dx`

`=2/3[ [x^3 ln(x)]_1^e - int_1^e x^3 frac{d}{dx}(ln(x)) dx ]`

`=2/3[ (e^3 - 0) - int_1^e x^3 (1/x) dx ]`

`=2/3[ e^3 - int_1^e x^2 dx ]`

`=2/3[ e^3 - [x^3/3]_1^e ]`

`=2/9[ 3e^3 - [x^3]_1^e ]`

`=2/9[ 3e^3 - (e^3-1) ]`

`=2/9( 2e^3 + 1 )`