# evaluate the integral using FTC

• Apr 13th 2009, 10:45 AM
williamb
evaluate the integral using FTC
∫10 (e^(-9x)) / (e^(-9x) + 6)) dx
8

Evaluate the integral using the fundamental theorem of calculus.

I don't know how to do rational numbers, i know that it is complex
because of the quotient rule for derivatives

anyway, here's mywork so far

∫from 10 to 8 (e^(-9x))/(e^(-9x)) + ∫from 10 to 8 (e^(-9x))/6

∫from 10-8 (1) + ∫from 10 to 8 (e^(-9x))/6

2 + 1/6 ∫from 10 to 8 (e^(-9x))

2 + 1/6 [((e^(-90+1))/(-90+1)) - ((e^(-72+1))/(-72+1))]

2 + 1/6 [(e^(-89)/(-89)) - (e^(-71)/(-71))]

this is wrong,
thanks,
brittany
• Apr 13th 2009, 12:40 PM
Reckoner
Just let $\displaystyle u=e^{-9x}\Rightarrow du=-9e^{-9x}\,dx.$

Quote:

Originally Posted by williamb
∫from 10 to 8 (e^(-9x))/(e^(-9x)) + ∫from 10 to 8 (e^(-9x))/6

You cannot split fractions in this way. $\displaystyle \frac a{b+c}$ is not, in general, equal to $\displaystyle \frac ab+\frac ac.$

Also, you are integrating from 8 to 10, not 10 to 8 (unless you wrote it wrong at the top of your post).

Quote:

...
2 + 1/6 [((e^(-90+1))/(-90+1)) - ((e^(-72+1))/(-72+1))]
Okay, even if we assume that your previous work was correct, this line does not make sense. You can only apply the power rule when the variable is not in the exponent. For an exponential function like $\displaystyle e^{-9x},$ you need to use the following rules.

$\displaystyle \frac d{dx}\left[e^x\right]=e^x$

$\displaystyle \int e^x\,dx=e^x+C$

If the base is something other than $\displaystyle e,$ you can use

$\displaystyle \frac d{dx}\left[a^x\right]=(\ln a)a^x$

$\displaystyle \int a^x\,dx=\frac{a^x}{\ln a}+C$