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roselin
Group Title
need help with :definite integral from 0 to 1 x^3/ Square root of x^4+9
 one year ago
 one year ago
roselin Group Title
need help with :definite integral from 0 to 1 x^3/ Square root of x^4+9
 one year ago
 one year ago

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ChmE Group TitleBest ResponseYou've already chosen the best response.1
\[\int\limits_{0}^{1}{x^3 \over \sqrt{x^4+9}}dx\]Is this the correct equation?
 one year ago

roselin Group TitleBest ResponseYou've already chosen the best response.0
yes, that is the one.
 one year ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
usubstitute for x^4 + 9
 one year ago

roselin Group TitleBest ResponseYou've already chosen the best response.0
let u be x^4+9?
 one year ago

roselin Group TitleBest ResponseYou've already chosen the best response.0
but how do you know which one to put as" u"?
 one year ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
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 reevaluate 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
 one year ago

roselin Group TitleBest ResponseYou've already chosen the best response.0
let me try replacing it
 one year ago

roselin Group TitleBest ResponseYou've already chosen the best response.0
dw:1355885411954:dw
 one year ago

TweT226 Group TitleBest ResponseYou've already chosen the best response.0
to reevaluate the limits plug in 0 and 1 into u=x^4 + 9
 one year ago

roselin Group TitleBest ResponseYou've already chosen the best response.0
then, what do I need to do after that?
 one year ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
I don't know why you did that.. you need to integrate the last equation I gave you that has u in it
 one year ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
\[\int\limits\limits\limits_{9}^{10}{du \over 4\sqrt{u}}\]
 one year ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
After you do usubstitution to reevaluate the limits see Twe's post
 one year ago

roselin Group TitleBest ResponseYou've already chosen the best response.0
when i plug in 0 i get 9 and whne i plug in 1 , I get 10
 one year ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
\[\int\limits\limits\limits\limits_{9}^{10}{1 \over 4}u^{1/2}du\]
 one year ago

roselin Group TitleBest ResponseYou've already chosen the best response.0
how did u get u^1/2?
 one year ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
\[{{1 \over 4}u^{1/2} \over 1/2}_9^{10}\]\[{1 \over 2}u^{1/2}_9^{10}\]\[{1 \over 2}10^{1/2}{1 \over 2}9^{1/2}\]
 one year ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
\[\sqrt x=x^{1/2}\]\[{1 \over x^n}=x^{n}\]
 one year ago

roselin Group TitleBest ResponseYou've already chosen the best response.0
oh,okay.i see
 one year ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
reevaluating 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
 one year ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
\[\int\limits\limits_{0}^{1}{x^3 \over \sqrt{x^4+9}}dx\]usubstitution: u = x^4 + 9 du = 4x^3 dx du / 4x^3 = dx Reevaluating the limits u = (0)^4 + 9 = 9 u = (1)^4 + 9 = 10\[\int\limits\limits\limits_{9}^{10}{x^3 \over \sqrt{u}}{du \over 4x^3}\]\[\int\limits\limits\limits\limits_{9}^{10}{du \over 4\sqrt{u}}\]\[\int\limits\limits\limits\limits\limits_{9}^{10}{1 \over 4}u^{1/2}du\]\[{{1 \over 4}u^{1/2} \over 1/2}_9^{10}\]\[{1 \over 2}u^{1/2}_9^{10}\]\[{1 \over 2}10^{1/2}{1 \over 2}9^{1/2}\]
 one year ago
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