Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

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

Calculus1
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[\int\limits_{0}^{1}{x^3 \over \sqrt{x^4+9}}dx\]Is this the correct equation?
yes, that is the one.
u-substitute for x^4 + 9

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

let u be x^4+9?
but how do you know which one to put as" u"?
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
let me try replacing it
|dw:1355885411954:dw|
to re-evaluate the limits plug in 0 and 1 into u=x^4 + 9
then, what do I need to do after that?
I don't know why you did that.. you need to integrate the last equation I gave you that has u in it
\[\int\limits\limits\limits_{9}^{10}{du \over 4\sqrt{u}}\]
After you do u-substitution to re-evaluate the limits see Twe's post
when i plug in 0 i get 9 and whne i plug in 1 , I get 10
\[\int\limits\limits\limits\limits_{9}^{10}{1 \over 4}u^{-1/2}du\]
how did u get u^-1/2?
\[{{1 \over 4}u^{1/2} \over 1/2}|_9^{10}\]\[{1 \over 2}u^{1/2}|_9^{10}\]\[{1 \over 2}10^{1/2}-{1 \over 2}9^{1/2}\]
\[\sqrt x=x^{1/2}\]\[{1 \over x^n}=x^{-n}\]
oh,okay.i see
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
okay.
\[\int\limits\limits_{0}^{1}{x^3 \over \sqrt{x^4+9}}dx\]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\[\int\limits\limits\limits_{9}^{10}{x^3 \over \sqrt{u}}{du \over 4x^3}\]\[\int\limits\limits\limits\limits_{9}^{10}{du \over 4\sqrt{u}}\]\[\int\limits\limits\limits\limits\limits_{9}^{10}{1 \over 4}u^{-1/2}du\]\[{{1 \over 4}u^{1/2} \over 1/2}|_9^{10}\]\[{1 \over 2}u^{1/2}|_9^{10}\]\[{1 \over 2}10^{1/2}-{1 \over 2}9^{1/2}\]
thanks

Not the answer you are looking for?

Search for more explanations.

Ask your own question