Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

shinigami1mBest ResponseYou've already chosen the best response.0
use integration by parts
 2 years ago

ZarkonBest ResponseYou've already chosen the best response.0
the integrand is positive on 0 to pi/2 so the answer has to be a positive number
 2 years ago

DenebelBest ResponseYou've already chosen the best response.0
The answer is (pi^2/8)  (1/2)
 2 years ago

DenebelBest ResponseYou've already chosen the best response.0
according to the back of the book
 2 years ago

DenebelBest ResponseYou've already chosen the best response.0
Ok, how do I get there?
 2 years ago

ash2326Best ResponseYou've already chosen the best response.2
sorry it'd be x^2 cos x + \(\int\limits_{0}^{\pi/2} 2x \cos xdx\) x^2 cos x+2x sin x\(\int\ limits_{0}^{pi/2} 2 sin x dx so we get x^2 cos x+ 2x sin x+2 cos x 2x sin x cos x(x^2 2) now insert the limits [ 2* pi/2 * 10][01*(02)] pi 2
 2 years ago

ZarkonBest ResponseYou've already chosen the best response.0
that is correct if it was \[x^2\sin(x)\] but it is \[x^2\sin(2x)\]
 2 years ago

ash2326Best ResponseYou've already chosen the best response.2
sorry i thought it's x^2 sin x let me put the correct one x^2 cos 2x /2+ ∫ 0 π/2 xcos2xdx x^2 cos 2x /2 + x sin 2x/2 ∫ 0 π/2 sin2x/2dx x^2 cos 2x/2 + x sin 2x /2 +cos 2x /4 now inserting the limits [pi^2/8+01/4][1/4] pi^2/81/2 π^2 /81/2
 2 years ago
See more questions >>>
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.