anonymous
  • anonymous
need help with :definite integral from 0 to 1 x^3/ Square root of x^4+9
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\int\limits_{0}^{1}{x^3 \over \sqrt{x^4+9}}dx\]Is this the correct equation?
anonymous
  • anonymous
yes, that is the one.
anonymous
  • anonymous
u-substitute for x^4 + 9

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anonymous
  • anonymous
let u be x^4+9?
anonymous
  • anonymous
but how do you know which one to put as" u"?
anonymous
  • anonymous
u = x^4 + 9 du = 4x^3 dx du / 4x^3 = dx\[\int\limits\limits_{0}^{1}{x^3 \over \sqrt{u}}{du \over 4x^3}\]\[\int\limits\limits_{0}^{1}{du \over 4\sqrt{u}}\]Can you finish it? I think you are actually supposed to re-evaluate the limits but I don't remember how to do that. It is not necessary if you replace u with what you substituted it for before you apply the limits
anonymous
  • anonymous
let me try replacing it
anonymous
  • anonymous
|dw:1355885411954:dw|
anonymous
  • anonymous
to re-evaluate the limits plug in 0 and 1 into u=x^4 + 9
anonymous
  • anonymous
then, what do I need to do after that?
anonymous
  • anonymous
I don't know why you did that.. you need to integrate the last equation I gave you that has u in it
anonymous
  • anonymous
\[\int\limits\limits\limits_{9}^{10}{du \over 4\sqrt{u}}\]
anonymous
  • anonymous
@TweT226 thx
anonymous
  • anonymous
After you do u-substitution to re-evaluate the limits see Twe's post
anonymous
  • anonymous
when i plug in 0 i get 9 and whne i plug in 1 , I get 10
anonymous
  • anonymous
\[\int\limits\limits\limits\limits_{9}^{10}{1 \over 4}u^{-1/2}du\]
anonymous
  • anonymous
how did u get u^-1/2?
anonymous
  • anonymous
\[{{1 \over 4}u^{1/2} \over 1/2}|_9^{10}\]\[{1 \over 2}u^{1/2}|_9^{10}\]\[{1 \over 2}10^{1/2}-{1 \over 2}9^{1/2}\]
anonymous
  • anonymous
\[\sqrt x=x^{1/2}\]\[{1 \over x^n}=x^{-n}\]
anonymous
  • anonymous
oh,okay.i see
anonymous
  • anonymous
re-evaluating the limits is the short method. If you don't do that you can plug what you substituted u for then use the old limits after integration. You should get the same answer
anonymous
  • anonymous
okay.
anonymous
  • anonymous
\[\int\limits\limits_{0}^{1}{x^3 \over \sqrt{x^4+9}}dx\]u-substitution: u = x^4 + 9 du = 4x^3 dx du / 4x^3 = dx Re-evaluating the limits u = (0)^4 + 9 = 9 u = (1)^4 + 9 = 10\[\int\limits\limits\limits_{9}^{10}{x^3 \over \sqrt{u}}{du \over 4x^3}\]\[\int\limits\limits\limits\limits_{9}^{10}{du \over 4\sqrt{u}}\]\[\int\limits\limits\limits\limits\limits_{9}^{10}{1 \over 4}u^{-1/2}du\]\[{{1 \over 4}u^{1/2} \over 1/2}|_9^{10}\]\[{1 \over 2}u^{1/2}|_9^{10}\]\[{1 \over 2}10^{1/2}-{1 \over 2}9^{1/2}\]
anonymous
  • anonymous
thanks

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