anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
\[\int\limits_{0}^{1} (9x ^{e}+e ^{x})dx\] Do I use u substitution?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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

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anonymous
  • anonymous
Wait I'm confused on the last post. \[\frac{ x ^{n+1} }{ n+1 }\] Isn't that the anti-derivative?
anonymous
  • anonymous
is that what an integral is?
anonymous
  • anonymous
Exactly! An integral is the anti-derivative.
hartnn
  • hartnn
anonymous
  • anonymous
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.
anonymous
  • anonymous
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?
anonymous
  • anonymous
so does that mean you would have: \[\frac{ 9x ^{e+1} }{ e+1 }\] and \[\frac{ e ^{x+1} }{ x+1 }\] is that right?
hartnn
  • hartnn
the first part is correct, for e^x \(\int e^xdx=e^x+c\)
anonymous
  • anonymous
But since this is a definite integral (it has bounds) you can ignore the c, which will cancel out anyway.
anonymous
  • anonymous
oh yeah that's right because the integral of e^x is itself right?
anonymous
  • anonymous
Yup
anonymous
  • anonymous
Ok so what's next? or is that all I have to do with it?
anonymous
  • anonymous
It just says evaluate the integral
anonymous
  • anonymous
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.
anonymous
  • anonymous
Oh wait I just realized I typed the problem out. There is no 9 in front of the x^e
anonymous
  • anonymous
*typed it out wrong lol
anonymous
  • anonymous
The answer you posted above was correct for the anti-derivative of x^e, just take out the 9 then.
anonymous
  • anonymous
stupid question how do you get e^-1 from e^1-e^0?
anonymous
  • anonymous
so e^1 = e e^0 = 1 So e^1 - e^0 = e - 1
anonymous
  • anonymous
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

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