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 $f\left(0\right)$

Question

$\color{orange}{\mathrm{Question:}}$

If $f\left(\pi\right)=\pi$ and $\int_{0}^{\pi}\left(f (x)+f''(x)\right)\sin x\ dx\ =\ 7\pi$ then find $f(0)$ given that $f(x)$ is continuous in $\left[0,\pi\right]$

$\color{green}{\mathrm{Solution:}}$

Given: $$\int_{0}^{\pi}\left(f(x)+f''(x)\right)\sin x\ dx\ =\ \int_{0}^{\pi}f(x)\sin x\ dx\ +\int_{0}^{\pi}f''(x)\sin x\ dx$$ By ILATE ( Integration by parts), keeping $f''(x)$ as the first function and $\sin x$ as the second function: $$7\pi\ =\int_{0}^{\pi}f(x)\sin x\ dx\ +\ \left[\sin x\cdot f'(x)-\int_{0}^{\pi}\cos x\cdot f'(x)dx\right]$$ $$7\pi\ =\int_{0}^{\pi}f(x)\sin x\ dx\ +\ \left[\sin x\cdot f'(x)-\left[\cos x\cdot f(x)-\int_{0}^{\pi}\left(-\sin x\right)\left(f(x)dx\right)\right]\right]$$ $$7\pi=\sin x\cdot f'(x)-\cos x\cdot f(x)$$ The limits being of integration being from $0$ to $\pi$, (sorry i don't know Latex much :( ) $$7\pi=\left[\sin\pi\cdot f'(\pi)-\sin0\cdot f'(0)\right]-\left[\cos\pi\cdot f(\pi)-\cos0\cdot f(0)\right]$$ Solving this I got $f(0)=6\pi$, which is the correct answer, no issues with that but...

$\color{pink}{\mathrm{Doubt}}$

1. When using the ILATE rule, we don't know what kind of function $f(x)$ is, so how can we decide whether to take it as the first function or the second function, I just did that for my convenience because I thought that will give me the solution.
2. Secondly, what is the importance of the statement of the question: $f(x)$ is continuous?

$\color{red}{\mathrm{Edit}}$

Basically it looks like ILATE is not a very good rule and Integration by parts is OP!

Answer 1

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

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.

Answer 2

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 $$f(0)=6 \pi$$