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\[\int\limits_{-1}^{1} e^-x (4-e^x) dx\]

see the thing is iruno how to break this ice

is exactly what i need help with

That's e^(-x) by the way
And I know, I'm so lost.

haha yeh this crap makes people lost in formulas man

so define the fundamental thrm of calculus; then see how that applies :)

My real question is, I don't know where to start with this problem.

start by defining the FTC and see how it applies lol.... that is the start

That does nothing for me.

expand it and you will get 4e^-x - 1

Yes, what dumbcow said. Then you can take the integral of each part.

FTC simply says it CAN be done; then you apply the techniques :)

I don't know how to apply the techniques haha, that's why I'm here!

the equation editor seems to have distorted the equation ; can you verify it?

\[\int\limits_{-1}^{1} 1/(e^x) (4-e^x) dx\]

\[\int\limits_{-1}^{1} \frac{1}{e^x} (4-e^x) dx\]

^^ that

frac{top}{bottom} in the editor makes for fancy fractions :)

Ooo, ok!

integrate {4/(e^x) - 1} dx

4 (ln(e^x)) - x

do what math93 said

4x-x = 3x
F(x) = 3x right?

no F(x) =-4e^-x - x

close lol

\[ \frac{-4}{e ^{x}}-x\] evaluated from -1 to 1

coulda thunked that 1/u integrates to ln(u)....

4e^-x is just as good i spose :)

if you sub in your values, you get
(-4e^-1 - 1)-(-4e^-1+1)

i see it..... just blind in my old age

After the values are substituted in, do I just simplify?

yes

So is the final answer (-2)?

Yeah, that's what I got

So is the answer just (-2) by itself? Or is there anything on the opposite side of the equal sign?

the integral of the original problem = -2, so "-2" is the final answer

Good luck!

Thank you!!