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

  1. $-1$
  2. $0$
  3. $1$
  4. $\pi$
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4 Answers

Best answer
28 votes
28 votes
The 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 the product of odd and even functions is an odd function and integrating an odd function from the same negative value to positive value gives $0.$
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\[ \int\limits_0^{2a} f(x)dx = 2 \times \int\limits_0^a f(x) \space dx \space\space if \space f(2a-x)=f(x) \]

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

 

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

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

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$I1=\int_{0}^{2\pi }(x-\pi )^2(sinx)dx\: \: . I2=\int_{0}^{2\pi }(2\pi -x-\pi )^2(sin2\pi -x)dx. \; \; I2=\int_{0}^{2\pi }-(x-\pi )^2(sinx)dx. \; \; 2I=I1+I2=\int_{0}^{2\pi }0dx \; \; I=0$
Answer:

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