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I have trouble doing proofs using the fundamental theorem of calculus and I think seeing an example would help.

Suppose we have a continuous function $ f $ defined on a real interval and a function $ F $ that is also defined on the same interval; and $F(x)=\int^b_x fdt$. Then for $ x $ in that interval, $ F'(x)=-f(x) $.

I know this can be done using the fundamental theorem of calculus; how is it applied?

By the way, the theorem says this:

Let $ f $ be integrable on $[a,b]$. For $x\in[a,b]$ put $F(x)=\int^x_a f(t)dt$. Then $ F$ is continuous on $[a,b]$ and furthermore if $ f $ is continuous at a point $x_0$ of $[a,b]$ then $F$ is differentiable at $x_0$ and $F'(x_0)=f(x_0)$.

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We name the function from FTC $F$ and our new function $G$ then $$G(x) = \int_x^b f(t)\ \mathrm dt = \underbrace{\int_a^b f(t)\ \mathrm dt}_{\text{constant}} - \int_a^x f(t) \ \mathrm dt = C - F(x)$$ So $G'(x) = -F'(x) = -f(x)$ by FTC.

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  • $\begingroup$ Thanks. Why is the second integral constant? $\endgroup$ – Rubi Mar 4 '15 at 19:28
  • $\begingroup$ @Rubi Because it's an integral independent of $x$. It has constant bounds ($a$ and $b$) and a function independent of $x$. $\endgroup$ – AlexR Mar 4 '15 at 19:34
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if F(x)=∫bxfdt, then -F(x)=∫xbfdt. by the FTC if -F(x)=∫xbfdt, then (-F(x))'=f(x), but (-F(x)')=-F'(x), so F'(x)=-f(x).

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  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – AlexR Mar 4 '15 at 19:17
  • $\begingroup$ Cool, thanks. I was wondering how people were getting nice integrals. I tried copying and pasting, and it copied an integral sign, but then it didn't format how I was hoping it would. $\endgroup$ – SE318 Mar 4 '15 at 19:19

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