• May 30th 2011, 04:15 AM
Hydralisk
I'm having difficulty with this question, I've been able to successfully get it as far as this (which is correct according to the solutions page) but I cannot get the actual stage of finding the indefinite integral part correct.
$\displaystyle \int(1-2e^{-x/2}+e^{x/2})dx$
Finding the indefinite integrals for each term at a time gives me:
$\displaystyle \int(1)dx=x+c$
$\displaystyle \int(-2e^{-x/2})dx=\int(-2)(e^{(-1)x/2})dx=(-2)\frac{1}{-1}e^{(-1)x/2}+c=2e^{-x/2}+c$
$\displaystyle \int(e^{x/2})dx=\int(e^{(1)x/2})dx=\frac{1}{1}e^{(1)x/2}+c=e^{x/2}+c$
Putting this all together gives me:
$\displaystyle \int(1-2e^{-x/2}+e^{x/2})dx=x+2e^{-x/2}+e^{x/2}+c$
But the correct solution is:
$\displaystyle \int(1-2e^{-x/2}+e^{x/2})dx=x+4e^{-x/2}+2e^{x/2}+c$
Can anybody see where I am making my mistakes?
Many thanks.
• May 30th 2011, 04:21 AM
Matt Westwood
Third line, e^{x/2} is e^{x \times 1/2} and not what you got, so you need to times its integral by 2. Similarly for the line above except it's minus it.
• May 30th 2011, 04:31 AM
Hydralisk
Matt that's clear now, thank you very much.
Shayne.