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  • Let be a function satisfying <img alt="\mathrm{f}(\mathrm{x})+f(\pi

Let \mathrm{f}(\mathrm{x}) be a function satisfying \mathrm{f}(\mathrm{x})+f(\pi-\mathrm{x})=\pi^2, \forall \mathrm{x} \in \mathbb{R}. Then \int_0^\pi \mathrm{f}(\mathrm{x}) \sin \mathrm{x} \mathrm{dx} is equal to :
 

Option: 1

\frac{\pi^2}{2}
 


Option: 2

\pi^2
 


Option: 3

2 \pi^2
 


Option: 4

\frac{\pi^2}{4}


Answers (1)

best_answer

I=\int_0^\pi f(x) \sin x d x......................(1)
Apply king property
I=\int_0^\pi f(\pi-x) \sin (\pi-x) d x.......................(2)

Add

\begin{aligned} & 2 \mathrm{I}=\int_0^\pi \mathrm{f}(\mathrm{x})+\mathrm{f}(\pi-\mathrm{x}) \sin \mathrm{x} d \mathrm{x} \\ & 2 \mathrm{I}=\int_0^\pi \pi^2 \sin \mathrm{x} d \mathrm{x} \\ & \not 2{\mathrm{I}}=\pi^2(\not 2) \\ & \mathrm{I}=\pi^2 \end{aligned}

Correct Answer is option 2

 

Posted by

manish

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