1.4k views

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

1. $-1$
2. $0$
3. $1$
4. $\pi$
in Calculus
edited | 1.4k 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.
by Active (5k points)
selected by
0

product of odd functions is also odd function ??

+1
nopes. Product should be even.
0

But selected answer says - product of odd functions is also odd function

But I m getting even Function ,, as product of two odd functions is even.

And I m not getting option match..

0
yes, even is correct. It was a mistake..
0
But now m not getting correct answer.. Can u help..
0
+2
+1

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

+1

I am getting 12 $\prod$ -2$\prod$

Getting same as in the link..

0
yes. That's correct. I have corrected the question now..
+1
Thanks for help.. :)

$\int\limits_0^{2a} f(x)dx = 2 \times \int\limits_0^a f(x) \space dx \space\space if \space f(2a-x)=f(x)$

$\int\limits_0^{2a} f(x)\space dx = 0 \space\space if \space f(2a-x)=-f(x)$

$\int\limits_0^{2\pi} f(x-\pi)^2(sin\space x)\space dx = 0$

bcz   it is a odd function so it becomes 0 so correct option is B

by Boss (11.2k points)
edited
+1 vote

Put (x-$\pi$) = t and solve.
by Boss (33.8k points)