Prove that

$$\lim_{x\rightarrow 1} \frac{\int_0^xg(t)dt-\int_0^1g(t)dt-\int_0^1f(t)dt(x-1)}{(x-1)^2}=\frac{f(1}{2}$$

Now I now this is a limit of the form $\frac{"0"}{"0"}$ which means I can use L'hopital along with the fundamental theorem of calculus. This is the first time that I've done something with two variable, t and x. Which one am I differentiating the terms for in this case?

  • $\begingroup$ When you say $\int_0^1 f(t) dt(x-1)$, do you mean $x-1$ times the integral? I just want to make sure. $\endgroup$ Jun 26 '16 at 14:14
  • $\begingroup$ @NobleMushtak yes. If I differentiate the first integral (g(t)dt) by x, does the term disappear, or do I use the FTC and it becomes $\frac{d}{dx} \int_0^xg(t)dt=G(x)-G(1)$ $\endgroup$
    – RonaldB
    Jun 26 '16 at 14:16
  • 1
    $\begingroup$ The derivative of $\int_0^x g(t) \ dt$ is simply $g(x)$. The second integral, $\int_0^1 g(t) \ dt$, however, disappears because it is a constant with respect to $x$. $\endgroup$ Jun 26 '16 at 14:17
  • $\begingroup$ @NobleMushtak got it, thanks $\endgroup$
    – RonaldB
    Jun 26 '16 at 14:26
  • $\begingroup$ I think there is a typo in question. what role does $g$ play in the question? Ideally it should be replaced by $f$. $\endgroup$ Jun 26 '16 at 17:47

The limit says $x \to 1$, so take the derivative with respect to $x$. Note that the $t$ variables are all variables that inside integrals, so taking the derivative with respect to $t$ doesn't make sense.


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