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anonymous
 4 years ago
need help with :definite integral from 0 to 1 x^3/ Square root of x^4+9
anonymous
 4 years ago
need help with :definite integral from 0 to 1 x^3/ Square root of x^4+9

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes, that is the one.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0usubstitute for x^4 + 9

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but how do you know which one to put as" u"?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u = 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let me try replacing it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1355885411954:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0to reevaluate the limits plug in 0 and 1 into u=x^4 + 9

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then, what do I need to do after that?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't know why you did that.. you need to integrate the last equation I gave you that has u in it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits\limits\limits_{9}^{10}{du \over 4\sqrt{u}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0After you do usubstitution to reevaluate the limits see Twe's post

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0when i plug in 0 i get 9 and whne i plug in 1 , I get 10

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits\limits\limits\limits_{9}^{10}{1 \over 4}u^{1/2}du\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how did u get u^1/2?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[{{1 \over 4}u^{1/2} \over 1/2}_9^{10}\]\[{1 \over 2}u^{1/2}_9^{10}\]\[{1 \over 2}10^{1/2}{1 \over 2}9^{1/2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt x=x^{1/2}\]\[{1 \over x^n}=x^{n}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0reevaluating 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
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits\limits_{0}^{1}{x^3 \over \sqrt{x^4+9}}dx\]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\[\int\limits\limits\limits_{9}^{10}{x^3 \over \sqrt{u}}{du \over 4x^3}\]\[\int\limits\limits\limits\limits_{9}^{10}{du \over 4\sqrt{u}}\]\[\int\limits\limits\limits\limits\limits_{9}^{10}{1 \over 4}u^{1/2}du\]\[{{1 \over 4}u^{1/2} \over 1/2}_9^{10}\]\[{1 \over 2}u^{1/2}_9^{10}\]\[{1 \over 2}10^{1/2}{1 \over 2}9^{1/2}\]
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