If $f\left(\pi\right)=\pi$ and $\int_{0}^{\pi}\left(f\left(x\right)+f''\left(x\right)\right)\sin x\ dx\ =\ 7\pi$ then find $f\left(0\right)$

$$\begin{align*} \int(f + f'')S &=\int fS + \int Sf''\\ &=[f (-C) + \int (f'C)] + [Sf' - \int(Cf')]\\ &=Sf' - Cf \end{align*}$$

at $x = 0, \pi$

$Sf' = 0, 0$ $\sin(0) = \sin(\pi) = 0$

$$\begin{align*} -[\cos(x)f(x)]_0^{\pi} &= 7\pi\\ & = -(\cos(\pi)f(\pi))+f(0)\\ & \implies f(0) = 6\pi \end{align*}$$

• $1$ First of all, ILATE is not a ground-rule which must be followed you might have come across few integrals where ILATE doesn't really fit.

• $2$ The given function $f$ is an implicit function don't worry about the continuity/differentiability, Understand the demand of the question

They just want someone who really has sound knowledge of product rules or integration by parts e.i The questioner is looking for the one with a good mathematical approach which by seeing your approach it's clear!

Of course! you should never stop asking questions.

Using integration by parts twice, we express the integral in terms of $f(\pi)$ and $f(0)$.

$\begin{aligned} \int_{0}^{\pi} f^{\prime \prime}(x) \sin x d x &=\int_{0}^{\pi} \sin x d\left(f^{\prime}(x)\right) \\ &=\left[\sin x f^{\prime}(x)\right]_{0}^{\pi}-\int_{0}^{\pi} \cos x f^{\prime}(x) d x \\ &=-\int_{0}^{\pi} \cos x d(f(x)) \\ &=-[\cos x f(x)]_{0}^{\pi}-\int_{0}^{\pi} \sin x f(x) d x \end{aligned}$

$\therefore \displaystyle \int_{0}^{\pi}\left(f(x)+f^{\prime \prime}(x)\right) \sin x$ $=f(\pi)+f(0)=\pi+f(0)$

By the given information, $ \displaystyle \int_{0}^{\pi}\left(f(x)+f^{\prime \prime}(x)\right) \sin x=7 \pi.$

We can now conclude that $$f(0)=6 \pi$$