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.

How do I find the integral from -4 to 4 of (3*sqrt(16-x^2)^2 dx ? I think you have to use u-substitution but I don't entirely understand how that works in this case.

Calculus1
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
use nothing
well the sqrt gets eliminated with the power 2, so...Its a simple integral
\[y=\sqrt{16-x^2}\] is the upper half of a circle centered at the origin with radius 4 so \[6\int_{-4}^4\sqrt{16-x^2}dx\] is 6 times the area of the upper half of a circle with radius 4

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

oh sorry i didn't see the square at the end if it is really the square root squared, then it is easy enough, although a rather strange way to write it
it is really \[3\int_{-4}^4\sqrt{16-x^2}^2dx\]?
Assuming it's what sattelite said then this becomes very simple.
|dw:1354850130324:dw| Then this become easy to integrate.
i think the typo from asker
I think so too.

Not the answer you are looking for?

Search for more explanations.

Ask your own question