# Thread: Expanding a tricky integrand involving exponents

1. ## Expanding a tricky function involving exponents

Why does the expansion of -

[(e^x)/12 - 3(e^-x)]^2

contain a (-1/2)?

Note: when expanded --> [(e^2x)/144 - 9(e^-2x) -1/2]

Shouldn't it just be the first two terms?

2. ## Re: Expanding a tricky function involving exponents

Originally Posted by bhaktir
Why does the expansion of -

[(e^x)/12 - 3(e^-x)]^2

contain a (-1/2)?

Note: when expanded --> [(e^2x)/144 - 9(e^-2x) -1/2]

Shouldn't it just be the first two terms?

Recall that $(a-b)^2=a^2-2ab+b^2$. So in this case $-2ab =-2\left(\tfrac{1}{12}e^x\right)\left(3e^{-x}\right) = -2\left(\tfrac{1}{4}\right)=-\tfrac{1}{2}$.

Does this make sense?

3. ## Re: Expanding a tricky function involving exponents

hi

the formula for the square of a difference is:

$(a-b)^2=a^2-2ab+b^2$

If $a=\frac{e^x}{12}$ and $b=\frac{3}{e^x}$

then:

$(a+b)^2=a^2+2ab+b^2=(\frac{e^x}{12})^2-2\cdot\frac{e^x}{12}\cdot\frac{3}{e^x}+(\frac{3}{e ^x}$

try it now.

4. ## Re: Expanding a tricky function involving exponents

OH! (a-b)^2

I can't believe I forgot that! Calc III is messing with my head. :/

Thanks again for the swift responses!! This site has been extremely helpful!