Differentiating equation using the Fundamental Theorem of Calculus

[Andrew Rubin]

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:



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?

 

[Steven G]

[math]\dfrac{d}{dx}\int_{g(x)}^{h(x)} f(t)dt =f(h(x))*h'(x) - f(g(x))*g'(x)[/math]
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]

 

[Andrew Rubin]

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.

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

[math]=-f(t)=-Cke^{-kt}[/math]

 

[Steven G]

[math]\int _t^\infty f(x)dx = \lim_{n->{\infty}}\int _t^n f(x)dx = \lim_{n->{\infty}}[f(n)*n' - f(t)*t'] = ????[/math]
What does the above simplify to?