anonymous
  • anonymous
Why does ∫x(2-x^2)^3 dx not equal ∫du where u=5-x^2?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
...you mean u=2-x^2? Because after you find du = -2x dx, you have to solve that differential equation for dx to sub back into the integral: dx = du/(-2x). Put that expression instead of dx, put 'u' instead of (2-x^2) and the extra 'x' will cancel out.
anonymous
  • anonymous
Then, you'll end up with -1/2 * ∫ u^3 du. Evaluate, and when you have the finished expression you can plug "u" back into it.
anonymous
  • anonymous
I'm sorry, the 2 in the integral was supposed to be a 5

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
But...the same concept applies. Thank you again for all of your help Quantum! :)
anonymous
  • anonymous
First need to expand x(2-x^2)^3 which equals -x^7 + 6x^5 -12 x^3 + 8x.Then you integrate the function. Or you can ... u = (2-x^2) du = -2xdu -du/2 = dx \[- 1/2\int\limits_{?}^{?} u^3 du =(-1/2) u^4/4 =- (2-x^2)^4/8 + C\]
anonymous
  • anonymous
Thank you so much Kynosis...excellent structure. I really appreciate it.

Looking for something else?

Not the answer you are looking for? Search for more explanations.