Aptitude Overflow
+4 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 (21.2k points) 125 160 160 | 565 views

4 Answers

+7 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 (4.4k 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 (35.6k 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 (3.6k points)
0 votes

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

answered by (5.9k points)

Related questions

2,706 questions
981 answers
31,354 users