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$$\int^{x}_{0}tf(t)dt = x\sin(x)+\cos(x)-1$$ find $f(\pi), f'(x)$

This question confuses me because usually, the way I have seen questions like these, they have been in the form:

$$H(x) = \int^{x}_{a} f(t)dt$$

This form is kind of different so I am not sure how I would solve it.

My thinking:

$f(\pi) = x\sin(\pi)+\cos(\pi)-1$

$f'(x) = x\cos(x)$

Is this correct?

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    You can write the fundamental theorem as “If $H(x)=\int_a^x g(t)\,dt$, then $H'(x)=g(x)$”. Now apply it with $g(t)=tf(t)$ and $a=0$. – 2017-02-04
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    I want to edit the question, but at the same time I don't want spoil the way the OP has posted it. – 2017-02-04
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    I know that it is standard to assume some facts such as $f $ is real valued function, but make sure that you write the assumptions explicitly (In this case that $xf(x)$ is continuous or something like that). Otherwise we will run into problems as mentioned in the answers. – 2017-02-04

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I'm afraid your answer is incorrect.

If you differentiate both sides, like normal, then using the product rule, we get that $$ xf(x)=x\cos x \iff f(x)=\cos x$$ So $f(\pi)=-1, f'(x)=-\sin x$.

EDIT

As pointed out by @celtschk, we require that $f(x)$ is continous in order for this to work. This is because for us to know for certain $$\frac {d}{dx}\int_{0}^{x} g(t) \mathrm {d}t =g(x)$$ for all $x$ we require something such as the continuity of $xf(x)$ for this solution to be true. Also, we need $f (x) $ to be continuous because of the issues in $f (0) $.

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    So in the end, would I just have $f(x) = \cos(x)$?, dividing by $x$? – 2017-02-04
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    @KSplitX Yes since the given equation holds for some $x \ne 0$. – 2017-02-04
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    @KSplitX Yes. We do. – 2017-02-04
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    Although I agree with your stragegy, why is my method wrong? – 2017-02-04
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    @KSplitX I personally cannot understand you method. What is your method? – 2017-02-04
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    @KSplitX $f(x) \neq \int^{x}_{0}tf(t)dt$. – 2017-02-04
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    Please refer [here]( http://math.stackexchange.com/q/2128882) for a correct solution. – 2017-02-04
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    @AkshayHegde You are correct. Will fix accordingly. – 2017-02-04
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    @AkshayHegde The problem has been rectified. – 2017-02-04
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    @S.C.B.: You need to require that $f(x)$ is continuous, not just that $xf(x)$ is continuous. While requiring $xf(x)$ to be continuous fixes the integral issue, the equation $x f(x) = x\cos x$ still does not fix the value of $f(x)$ at $x=0$ (that is, your $\iff$ in the first displayed equation is not an $\iff$). – 2017-02-04
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Why is my method wrong?

To do correctly, we have to use the concept of differentiation under the integral sign. We thus have, $$\frac {d}{dx} (\int_{0}^{t} tf (t) \mathrm {d}t) = \frac {d}{dx}(x\sin x +\cos x -1) $$ $$\Rightarrow xf (x) -0 = x\cos x +\sin x - \sin x$$ $$\Rightarrow \boxed {f (x) = \cos x} $$

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Well, strictly speaking the integral equation doesn't fix $f(\pi)$, nor does it guarantee even the existence of $f'(x)$. For example, for arbitrary $\alpha$ be $$f_\alpha(x) = \cases{ \alpha & for $x\in\pi\mathbb Q$\\ \cos x & otherwise}$$ Then $f_\alpha$ fulfills the integral equality, $f_\alpha(\pi)=\alpha$, and $f_\alpha$ is nowhere differentiable.

However if we add the requirement that $f$ is continuous, it is uniquely defined. By deriving both sides of the integral equation by $x$, we get that $x f(x) = x\cos x$ must hold almost everywhere. Since the complement of a measure-zero set is dense (if it weren't, the measure zero set would contain an open interval, but open intervals have positive measure), continuity then fixes $f$ to be given by $f(x)=\cos x$ (note that the continuity requirement also fixes the value $f(0)$, which wouldn't even be fixed by $x f(x) = x\cos x$ if that equation needed to hold everywhere).

Obviously, if $f(x)=\cos x$, then $f(\pi)=-1$ and $f'(x)=-\sin x$.

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    $f'(x) = -\sin(x)$ – 2017-02-04
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    @KSplitX: Oops, yes, of course. Fixed. – 2017-02-04
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    +1 Both of the above answers(including accepted answer, at the time of writing this comment) are wrong due to the _reason_ above. Maybe question should be reframed to address the issue. I have one question @celtschk, is it necessary for $f$ to be continuous to hold above conclusion? (i.e. are there any weaker conditions?) – 2017-02-04
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    @AkshayHegde: I think it might suffice to demand that for every $x$, either the left limit or the right limit exists and agrees with $f(x)$. A non-continuous function with this property is $$f(x)=\cases{0 & for $x<0$\\1 & for $x\ge 0$}$$ – 2017-02-04