Question

Can someone give me a quick recap of integrals? I have an example and there are 5 problems that are done similarly. Example is in the post with equation converter helper.

Answer 1

\[\int\limits_{0}^{1} (9x ^{e}+e ^{x})dx\]
Do I use u substitution?

Answer 2

I can't tell what the power of x is but it looks like an e?

Either way, you can break this integral into two parts.

\[\int\limits_{0}^{1}9x^e + \int\limits_{0}^{1}e^x\]

Then just take integrate the individual integrals.

Answer 3

The rule of integration for X^n where n is a constant is

\[x^{n+1}/(n+1)\]

And the integral of e^x is just e^x

Answer 4

Wait I'm confused on the last post. \[\frac{ x ^{n+1} }{ n+1 }\]
Isn't that the anti-derivative?

Answer 5

is that what an integral is?

Answer 6

Exactly! An integral is the anti-derivative.

Answer 7

recap of integrals ?
try this....
http://openstudy.com/users/hartnn#/updates/50960518e4b0d0275a3ccfba

Answer 8

Well not exactly, but a key component of taking the integral of something is finding the anti-derivative. If there are no bounds, then an integral is essentially the anti-derivative.

Answer 9

Oh ok = ) So that's how you solve an integral. you use the rule that you can split the integrals up and then solve each one by finding the anti-derivative.

Oh yeah bc you have to incorporate the integral 0 to 1 right?

Answer 10

so does that mean you would have:
\[\frac{ 9x ^{e+1} }{ e+1 }\]
and
\[\frac{ e ^{x+1} }{ x+1 }\]
is that right?

Answer 11

the first part is correct,
for e^x
\(\int e^xdx=e^x+c\)

Answer 12

But since this is a definite integral (it has bounds) you can ignore the c, which will cancel out anyway.

Answer 13

oh yeah that's right because the integral of e^x is itself right?

Answer 14

Yup

Answer 15

Ok so what's next? or is that all I have to do with it?

Answer 16

It just says evaluate the integral

Answer 17

Once you've taken the anti-derivative of it you plug in your bounds for x

So for the integral of e^x for example:

you'd get

\[e^1 - e^0\]

Which simplifies into e-1

But you have to remember that you're adding this to the integral of the first half.

Answer 18

Oh wait I just realized I typed the problem out. There is no 9 in front of the x^e

Answer 19

*typed it out wrong lol

Answer 20

The answer you posted above was correct for the anti-derivative of x^e, just take out the 9 then.

Answer 21

stupid question how do you get e^-1 from e^1-e^0?

Answer 22

so e^1 = e

e^0 = 1

So e^1 - e^0 = e - 1

Answer 23

oh ok I thought the -1 was the exponent of e from the rule of two bases being the same the exponents divide you know what I'm trying to say? lol