The intergral of (4-2x)^3 dx?
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substitute $\displaystyle u=4-2x$
Originally Posted by homerb The intergral of (4-2x)^3 dx? Use substitution: Let: $\displaystyle u=4-2x, \text{ then } du=-2\,dx$
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?
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.. $\displaystyle -\dfrac{1}{8} (4-2x)^4 + C = -\dfrac{1}{8} [-2(x-2)]^4 + C = ...$
Thanks
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