Question

Question #9ddf4

Answer 1

`int_1^5 x/sqrt(2x-1)\ dx=16/3`

Explanation:

I'm interpreting the question as:
`int_1^5 x/sqrt(2x-1)\ dx`

We begin by introducing a u-substitution, `u=sqrt(2x-1)`. We divide by the derivative and adjust the bounds to integrate with respect to `u`:
`(du)/dx=2/(2sqrt(2x-1))=1/u`

`x=5=>u=3`

`x=1=>u=1`
`int_1^5 x/sqrt(2x-1)\ dx=int_1^3x/cancelu*cancelu\ du`

We can't integrate `x` with respect to `u`, so we have to solve for `x` in terms of `u`:
`u=sqrt(2x-1)`

`u^2=2x-1`

`u^2+1=2x`

`(u^2+1)/2=x`

`int_1^3 (u^2+1)/2\ du=[1/2*u^3/3+1/2u]_1^3=3^3/6+3/2-(1/6+1/2)=`

`=27/6+9/6-1/6-3/6=32/6=16/3`

Answer 2

`int_1^5 x/sqrt(2x-1)*dx=16/3`

Explanation:

`int_1^5 x/sqrt(2x-1)*dx`

I used `u=sqrt(2x-1)`, `x=(u^2+1)/2` and `dx=u*du` transforms. According to it, `u=1` for `x=1` and `u=3` for `x=5`.

Hence,

`int_1^3 ((u^2+1)/2)/u*u*du`

=`1/2int_1^3 (u^2+1)*du`

=`[u^3/6+u/2]_1^3`

=`16/3`