A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
need help with :definite integral from 0 to 1 x^3/ Square root of x^4+9
anonymous
 3 years ago
need help with :definite integral from 0 to 1 x^3/ Square root of x^4+9

This Question is Closed

anonymous
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0yes, that is the one.

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

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

anonymous
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0let me try replacing it

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

anonymous
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0then, what do I need to do after that?

anonymous
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits\limits\limits_{9}^{10}{du \over 4\sqrt{u}}\]

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

anonymous
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0how did u get u^1/2?

anonymous
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt x=x^{1/2}\]\[{1 \over x^n}=x^{n}\]

anonymous
 3 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
 3 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\[\int\limits\limits\limits_{9}^{10}{x^3 \over \sqrt{u}}{du \over 4x^3}\]\[\int\limits\limits\limits\limits_{9}^{10}{du \over 4\sqrt{u}}\]\[\int\limits\limits\limits\limits\limits_{9}^{10}{1 \over 4}u^{1/2}du\]\[{{1 \over 4}u^{1/2} \over 1/2}_9^{10}\]\[{1 \over 2}u^{1/2}_9^{10}\]\[{1 \over 2}10^{1/2}{1 \over 2}9^{1/2}\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.