Integration, general power formula

Question - integrate the following:

(int) (4- (e^x))^3 e^x dx

attempt:

u = (4 + e^x)

n = 3

du = e^x

therefore answer is:

((4 + (e^x))^4) / 4 + c

which is incorrect, and should be:

4 / (e^(x^1/2)) + c

Stared at it for a while and can't figure it out. Any help, thanks?

Re: Integration, general power formula

Quote:

Originally Posted by

**togo** Question - integrate the following:

(int) (4- (e^x))^3 e^x dx

attempt:

u = (4 + e^x)

n = 3

du = e^x

therefore answer is:

((4 + (e^x))^4) / 4 + c

which is incorrect, and should be:

4 / (e^(x^1/2)) + c, NO indeed!

$\displaystyle \int {{{\left( {4 - {e^x}} \right)}^3}{e^x}dx = - \tfrac{1}{4}} {\left( {4 - {e^x}} \right)^4} + c$

Re: Integration, general power formula

Need to find $\displaystyle \int (4 - e^x)^3 e^x dx $.

Set $\displaystyle u = 4 - e^x $. Then $\displaystyle du = -e^x dx $. Then $\displaystyle -du = e^x dx $

$\displaystyle \int (4 - e^x)^3 e^x dx = \int -u^3 du = -\frac{u^4}{4} + C = -\frac{(4 - e^x)^4}{4} + C $

Re: Integration, general power formula

you guys are telling me the book is wrong

Re: Integration, general power formula

Try differentiating the book's answer to check that it is correct