# Math Help - Intergrals

1. ## Intergrals

The intergral of (4-2x)^3 dx?

2. substitute $u=4-2x$

3. Originally Posted by homerb
The intergral of (4-2x)^3 dx?
Use substitution:

Let:   $u=4-2x, \text{ then } du=-2\,dx$

4. thats what I did and I got -1/8*(4-2x)^4 + c, but in the book says -2(x-4)^4, am I doing something wrong?

5. Your answer is correct.. its needs simplification..

Originally Posted by homerb
thats what I did and I got -1/8*(4-2x)^4 + c, but in the book says -2(x-4)^4, am I doing something wrong?
thats the same thing.. take -2 as a common factor in the numerator and simplify..

$-\dfrac{1}{8} (4-2x)^4 + C = -\dfrac{1}{8} [-2(x-2)]^4 + C = ...$

6. Thanks