Differentiating equation using the Fundamental Theorem of Calculus

Andrew Rubin

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I am trying to understand an application of integration in probability. Specifically, example 5 in Adams & Essex, Calculus: A Complete Course (9th ed.), p. 441-442.

The example is an application of integration to the exponential distribution:

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I get stuck at:

Differentiating this equation with respect to t (using the Fundamental Theorem of Calculus), we obtain ...

I don't understand the steps involved in the authors use of the Fundamental Theorem of Calculus to arrive at their results here, and they don't explain them either. I have tried reviewing the theorem without any luck. Could someone help me out?
 
ddxg(x)h(x)f(t)dt=f(h(x))h(x)f(g(x))g(x)\dfrac{d}{dx}\int_{g(x)}^{h(x)} f(t)dt =f(h(x))*h'(x) - f(g(x))*g'(x)
Note that if g(x) (and/or h(x)) is a constant then that term at the end will be 0, since g'(x) = 0.

If this is still unclear, then let us know.[/math]
 
Thanks for your quick reply, Jomo! I was thinking more of a reply that could help me understand the missing midsection which I try to illustrate below. Your answer might be exactly it (?) but I am unsure. Recently picked up calculus again.

tf(x)dx=Pr(Tt)=Cekt\int_{t}^{\infty} f(x) \,dx=Pr(T \geq t) = Ce^{-kt}
= [differentiating equation above w.r.t. t using the Fundamental Theorem of Calculus]

=f(t)=Ckekt=-f(t)=-Cke^{-kt}
 
tf(x)dx=limn>tnf(x)dx=limn>[f(n)nf(t)t]=????\int _t^\infty f(x)dx = \lim_{n->{\infty}}\int _t^n f(x)dx = \lim_{n->{\infty}}[f(n)*n' - f(t)*t'] = ????
What does the above simplify to?
 
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