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 $\left[0,\pi\right]$


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...


  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?


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

  • 1
    $\begingroup$ The thing with ILATE is that it works most of the time. I personally, never use ILATE. See the thing with Integration using by-parts is that the its always the second function that is integrated. Just see where you'll be benefited now...which function as the second func makes life easier? I'll give you hint for the second question you asked...Integration is the process of finding the "anti-derivative"...take it from here on your own now. Also one tip to solve the q...just see if it works...add and subtract $f'(x)$ before integrating (i.e. in the first step). $\endgroup$
    – HarshDarji
    Commented Oct 2, 2021 at 6:54
  • 1
    $\begingroup$ I was saying ILATE doesn't work all the time because I've seen questions where it has betrayed me XD...As for evaluating the answer, you should consider trying it out...I don't mind doing it for you but this isn't a homework assignment website and besides I'll take sometime to give the answer...if you have the patience then fine $\endgroup$
    – HarshDarji
    Commented Oct 2, 2021 at 7:04
  • $\begingroup$ K yeah...my bad...I thought something will cancel out and make life easier...but that isn't the case. Your solution is perfect. Do you want me to elaborate on my "hint"? I can write it as an answer if that interests you. $\endgroup$
    – HarshDarji
    Commented Oct 2, 2021 at 7:17
  • 2
    $\begingroup$ Since the statement is using $f''$, it is implicitely assumed that $f$ is at least $C^{2}$, therefore in particular continuous. $\endgroup$
    – Tobsn
    Commented Oct 2, 2021 at 10:11
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    $\begingroup$ You mentioned you do not know Latex very much for that you can click on edit of others Q&A without submitting your edit However, Not knowing latex is OK bt the important is to Never Stop Learning! $\endgroup$
    – user960916
    Commented Oct 4, 2021 at 5:50

2 Answers 2


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

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.

  • $\begingroup$ Jesus!! I just read the title of your question and with curiosity just started solving it without reading your question I'll try to answer your question by editting the above answer. $\endgroup$
    – user960916
    Commented Oct 4, 2021 at 5:09
  • 1
    $\begingroup$ nice work +1. I like how you generalized the question itself $\int(f + f'')S$ $\endgroup$ Commented Oct 4, 2021 at 5:30
  • $\begingroup$ @Vega Thanks! :) It saves giving time to LaTex. ¯_( ͡❛ ͜ʖ ͡❛)_/¯ $\endgroup$
    – user960916
    Commented Oct 4, 2021 at 5:45

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

  • 1
    $\begingroup$ Thank you, this does give a different approach to the problem, but my question is something else! $\endgroup$ Commented Oct 2, 2021 at 8:58
  • $\begingroup$ For question 1, IBP can be used without knowing the Rule. For question 2, f” exists guarantee f’ exists and hence f is continuous. $\endgroup$
    – Lai
    Commented Oct 3, 2021 at 3:55
  • $\begingroup$ "IBP can be used without knowing the Rule" is this something I have to outright remember? $\endgroup$ Commented Oct 3, 2021 at 8:35
  • 1
    $\begingroup$ I meant that I use IBP without knowing what ILATE is. We can first observe what part of integrand have primitive function and integrate it first. In your question, both $ \sin x $ and $f”(x)$ can be integrated twice. Therefore both can work. $\endgroup$
    – Lai
    Commented Oct 3, 2021 at 9:03

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