Question

Can someone help me with this integral question?

Answer 1

What's the question?

Answer 2

\[\int\limits_{-1}^{1}(e ^{2x}+e ^{-x})dx\]

Answer 3

\[\int_{-1}^{1} \left(e^{2x} + \frac{1}{e^x}\right) dx\]
Let u = e^x
du = e^x dx
1/e^x du = dx
1/u du = dx
Your integral then changes to
\[\int_{e^{-1}}^{e}\left(u^2+\frac{1}{u}\right) \cdot \left(\frac{1}{u} du\right)\]
Does that help?

Answer 4

so you change \[\int\limits_{-1}^{1}\]

Answer 5

Yes. You're integrating with respect to u now, so the limits [-1,1] no longer apply.

Answer 6

Since u = e^x,
u(1) = e^1 = e
u(-1) = e^(-1) = 1/e

Answer 7

so then it would be\[1/3\int\limits_{e ^{-1}}^{e}(u ^{3}+lnu)(lnudu)\]

Answer 8

but then what would be the next step?

Answer 9

No, that's not it. Do you have any question about the substitution I made? Because I'm not sure why you're getting what you're getting.

Answer 10

yes, i have never substituted in the integral before

Answer 11

Okay, then. Let's leave that step out. I might have been over-complicating the question. Sorry about that.

You start off with \[\int_{-1}^{1} (e^{2x}+e^{-x})dx \]
You can split the integral into two, and then integrate separately.
\[\int_{-1}^{1} e^{2x}dx+\int_{-1}^{1}e^{-x}dx \]
However, you'll still have to do a substitution here. Are you sure you never encountered an integral that requires a u-sub?

Answer 12

yes, positive. professor never mentioned at least

Answer 13

Alright. You'll probably learning those soon enough.
You know that
\[\frac{d}{dx}e^{2x} = 2e^{2x},\]
right?

Answer 14

yes!

Answer 15

The integral can be considered the "inverse" operation of the derivative, or antiderivative. Using this fact, you can rewrite the last equation. Anti-differentiate both sides with respect to x:
\[\int\frac{d}{dx}[e^{2x}]dx = \int2e^{2x}dx\]
Since the antiderivative of the derivative of a function is just the function itself, you have
\[e^{2x} = \int2e^{2x}dx\]
Factoring out the constant on the right side and dividing, you get
\[e^{2x} = 2\int e^{2x}dx\\
\frac{1}{2}e^{2x} = \int e^{2x}dx\]
Similarly, you can find the second integral.

Now, evaluating over the interval [-1,1] means you have some idea of what the fundamental theorem of calculus is. Is that right?

Answer 16

i think so. is this right? \[\int\limits_{-1}^{1}1/2e ^{2x}-e ^{-x}\]

Answer 17

Your integration itself is correct, yes. But the notation is a bit off.

Since you found the general antiderivative, you no longer have an integral. Instead, it should be written like this:
\[\left[\frac{1}{2}e^{2x} - e^{-x}\right]_{-1}^{1}\]
Then you just apply the fundamental theorem.

Answer 18

great! thanks!