565 views

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 | 565 views

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.
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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

Put (x-$\pi$) = t and solve.
apply property $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$  .The equation will be $\int_{0}^{2\Pi }-\Pi ^2sinxdx=\Pi ^2[1-1]=0$

02a   f(x)dx = 2 0a   f(x)dx  if     f(2a-x)  = f(x)

=0 if f(2a-x)  =  -  f(x)

bcz   02π  (x−π)(sinx) dx  is a odd fxn so it becomes 0 so correct option is B

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