GATE IT 2005 | Question: 35

Question

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

• A. $-1$
• B. $0$
• C. $1$
• D. $\pi$

Comments

shortcut trick.

https://gateoverflow.in/3782/gate-it-2005-question-35?show=148523#a148523

Answer 1

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 $0.$

Comments

I am getting 12 $\prod$ -2$\prod$³ 

^{Getting same as in the link..}

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yes. That's correct. I have corrected the question now..

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Thanks for help.. :)

Answer 2

 

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

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

 

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

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

Answer 3

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

Answer 4

Answer: B

Put (x-$\pi$) = t and solve.