If $f(\pi) = 2$ and $\int_{0}^{\pi}{\left\lbrack f(x) + f^{''}(x) \right\rbrack\sin{x\ dx = 5,}}$ then $f(0)$ is equal to, (it is given that $f(x)$ is continuous in $\lbrack 0,\ \pi\rbrack)\$

Question

Dr. Abubacker · Accepted Answer

(b)
$\int_{0}^{\pi}{f(x)\sin{x\ dx +}}\int_{0}^{\pi}{f''(x)\sin{x\ dx = 5}}$
$\Rightarrow \ \lbrack f(x)( - {\cos{x)\rbrack}}_{0}^{\pi} + \int_{0}^{\pi}{f'(x)\cos{x\ dx + \int_{0}^{\pi}{f''(x)\sin{x\ dx = 5}}}}$
$\Rightarrow \ \lbrack - f(x){\cos{x\rbrack}}_{0}^{\pi} + \lbrack f^{'}(x){\sin{x\rbrack}}_{0}^{\pi} - \int_{0}^{\pi}{f''(x)\sin{x\ dx + \int_{0}^{\pi}{f''(x)\sin{x\ dx = 5}}}}$
$\Rightarrow \ f(\pi) + f(0) = 5 \Rightarrow f(0) = 5 - f(\pi) = 3\$

Q. If $f(\pi) = 2$ and $\int_{0}^{\pi}{\left\lbrack f(x) + f^{''}(x) \right\rbrack\sin{x\ dx = 5,}}$ then $f(0)$ is equal to, (it is given that $f(x)$ is continuous in $\lbrack 0,\ \pi\rbrack)\$

Explanation

Reviewed by Dr. Abubacker

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