I have added a pic of the notes that I am looking at!

I’m a little confused by them.In integration calcs, I’m used to seeing the integral of $f(x)$ to get the difference between two points. Here it’s integrating $f^\prime(x)$ to get the difference.

I’m trying to picture the problem graphically - I’m assuming that $M$ is along the $y$-axis and $t$ is on the $x$-axis. In which case wouldn’t we just integrate $M(t)$? Unless $M(t)$ is supposed to be the CDF?

Basically I don’t understand why it’s the integral of $M^\prime(t)$ and not the integral of $M(t)$

Sorry getting very confused with this!

Any help is much appreciatedenter image description here

  • 1
    $\begingroup$ They're using this. In Wikipedia's notation $f=M',\,F=M$. $\endgroup$
    – J.G.
    Jun 10 at 8:22

1 Answer 1


It's just fundamental theorem of calculus. We know that "$M(t)$ represents the total payment between timo $0$ and time $t$". Therefore it should be clear that, if you want the total payment between time $t=\alpha$ and $t=\beta$, you just have to calculate $M(\beta)-M(\alpha)$.

Now it's just a simple application of F.T.C.: it says that $$\int_a^bf(t)dt=F(b)-F(a)$$ where $F(x)$ is an antiderivative of $f$. Now replace $f=g'$. An antiderivative of $g'$ is $g$ obviously, and so you get $$\int_a^b g'(t)dt=g(b)-g(a)$$

Now plug in $g(t)=M(t)$ and you get your equation.


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