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+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)   | 421 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)  

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