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

Question

$$\int^{x}_{0}tf(t)dt = x\sin(x)+\cos(x)-1$$ 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?

Answer 1

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) $.

Answer 2

Why is my method wrong?

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

Answer 3

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 $f'(x)=-\sin x$.