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We are told that

$$\displaystyle \int_0^x f(t) \, dt = 2\cos x + 3x + 2$$

and are asked to find $f$. If we apply the Fundamental Theorem of Calculus, then

$$\displaystyle f(x) = \frac{d}{dx}\int_0^x f(t) \, dt = -2\sin x + 3$$


$$\displaystyle \int_0^x -2\sin t + 3 \, dt = 2\cos x+3x - 2$$

It seems to me that the problem stems from using the FTC: Since I don't know anything about $f$, I cannot apply it. My question is: How exactly do I go about finding $f$ ?

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Something is not quite right... letting $x=0$ in the first equation gives $0=4$? Could be a typo, should be $-2$ instead of $+2$. – Antonio Vargas Apr 6 '12 at 1:49
It didn't occur to me to check the first equation. I knew something was fishy. I am convinced that it is a typo. Thanks. – echoone Apr 6 '12 at 2:10
up vote 4 down vote accepted

As noted, you probably need to have a $-2$ instead of $+2$, as you can see by setting $x=0$.

But the reason I am adding this answer is that, strictly speaking, there are an infinite number of such $f$!

For instance, given one such $f$ (which you found by differentiating) you can pick any point, and change the value of $f$ to be arbitrary without affecting the value of the integrals.

One can in fact show that if $f$ and $g$ are two such function, then they differ only on a set of measure $0$.

This we can show using two theorems:

1) A bounded function $f$ is Riemann integrable iff the set of discontinuities of $f$ is of measure $0$.

2) If $f$ is Lebesgue integrable on $[a,b]$ and $F(x) = \int_{0}^{x} f $ (Lebesgue integral, which is also the Riemann integral in this case), and if $f$ is continuous at $x_0$, then $F$ is differentiable at $x_0$ and $F'(x_0) = f(x)$.

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You are absolutely right. Thank you for pointing this out. – echoone Apr 6 '12 at 2:22
This is a volterra integral equation right? – Keaton Apr 6 '12 at 3:37
@Keaton: No idea. – Aryabhata Apr 6 '12 at 4:04
I only say because you can solve for eigenfunctuons and values which can be of use depending on what you want out of this. – Keaton Apr 6 '12 at 4:10
@Keaton: If you regard it as a volterra integral equation, then you will have a trivial kernel. – Jack Apr 6 '12 at 5:59

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