Question

Find the integral of `(x-10)^2`?

Answer 1

`int(x-10)^2dx=(x-10)^3/3+c`

Explanation:

We have to find `int(x-10)^2dx`

let `x-10=u` then `dx=du` and

`int(x-10)^2dx`

= `intu^2du`

= `u^3/3+c`

= `(x-10)^3/3+c`

Answer 2

Please see below.

Explanation:

Method 1

`(x-10)^2 = x^2-20x+100`, so

`int (x-10)^2 dx = int (x^2-20x+100) dx`

`= x^3/3-10x^2+100x +C`

Method 2

Let `u = x-10`. The results in `du = 1 dx`. Change the integral

`int underbrace((x-10))_u^2 underbrace(dx)_(du) = int u^2 du`

`= u^3/3 +C_2`. Now undo the substitution to get

`= (x-10)^3/3 +C_2`

Note

Expanding the second answer, we get

`(x-10)^3/3 +C = (x^3-30x^2+300x-1000)/3 +C_2`

`= x^3/3-10x^2+100x-1000/3 +C_2`

So the two solution methods lead to different constants.

`C = C_2-1000/3`