First time here? Checkout the FAQ!
+2 votes

What is the value of $\int_{0}^{2\pi}(x-\pi)^2 (\sin x) dx$

  1. -1
  2. 0
  3. 1
  4. $\pi$
asked in Calculus by Veteran (19k points)   | 409 views

3 Answers

+4 votes
Best answer
answer is (b)

Put $x-\pi=t$ then limit $0$ changes to $-\pi$ and upper limit $2\pi$ changes to $\pi$.

$\frac{d}{dx}(x-\pi)=dt \implies dx =dt$

Integration of $t^2\sin t dt$ for limit $-\pi$ to $\pi$. One is an odd function and one is even and product of odd and even functions is odd function and integrating an odd function from the same negative value to positive value gives 0.
answered by Loyal (3.3k points)  
selected by

So, I suppose it might be a typo in question- very hard to read for me

I am getting 12 $\prod$ -2$\prod$

Getting same as in the link..

yes. That's correct. I have corrected the question now..
Thanks for help.. :)
+1 vote
Answer: B

Put (x-$\pi$) = t and solve.
answered by Veteran (34k points)  
0 votes
apply property $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$  .The equation will be $\int_{0}^{2\Pi }-\Pi ^2sinxdx=\Pi ^2[1-1]=0$
answered by Loyal (3.1k points)  
Members at the site
Top Users Feb 2017
  1. Arjun

    4672 Points

  2. Bikram

    4004 Points

  3. Habibkhan

    3738 Points

  4. Aboveallplayer

    2966 Points

  5. sriv_shubham

    2278 Points

  6. Smriti012

    2212 Points

  7. Arnabi

    1814 Points

  8. Debashish Deka

    1788 Points

  9. sh!va

    1444 Points

  10. mcjoshi

    1444 Points

Monthly Topper: Rs. 500 gift card

20,788 questions
25,938 answers
21,923 users