Question

Integrate `int_1^2x\sqrt{x-1}dx` ?

Answer 1

`= 16/15`

Explanation:

let `u = x-1`

`du = dx`

`x=2 => u = 1`

`x = 1 => u = 0`

Rearrange: `x = u+1`

`=> int_0 ^1 (u+1) sqrt(u) du`

`= int_0 ^1 (u^(3/2) + u^(1/2)) du`

`= [ 2/5 u^(5/2) + 2/3 u^(3/2) ]_0 ^1`

`= 2/5 + 2/3 = color(red)(16/15`

Answer 2

Use substitution, let `w=x-1` then `dw=dx` and the integral becomes `int_0^1(w+1)sqrt(w )dw`
answer: `(16/15)`

Explanation:

If you use this substitution the limits of integration are found by plugging `x=1` into the substitution `w=x-1=1-1=0` and `w=2-1=1`
x is found by solving `w=x-1` for x. `x=w+1`
Simplify the integrand algebraically
`int_0^1[wsqrt(w)+sqrt(w)]dw`
`int_0^1[w^(3/2)+w^(1/2)]dw`
Now integrate using the power rule.
`=(2/5)w^(5/2)+(2/3)w^(3/2)` from 0 to 1
`=(2/5)+(2/3)=(16/15)`