OK, now we come onto the part of the FTC that you are going to use most. We are finally going to show the direct link between the definite integral and the antiderivative. I know that you’ve been holding your breaths until this moment. Get ready to breath a sign of relief:

The Fundamental Theorem of Calculus, Part 2 (also known as the Evaluation Theorem)

If $f$ is continuous on $[a,b]$ then

$\int_a^b f(x) dx=F(b)-F(a)$

where $F$ is any antiderivative of $f$. Ie any function such that $F'=f$.

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This means that, very excitingly, now to calculate the area under the curve of a continuous function we no longer have to do any ghastly Riemann sums. We just have to find an antiderivative!

OK, let’s prove this one straight away.

We’ll define:

$g(x)=\int_a^x f(t)dt$

and we know from the FTC part 1 how to take derivatives of this. It’s just $g'(x)=f(x)$. This says that $g$ is an antiderivative of $f$.

If $F$ is some other antiderivative of $f$ on $[a,b]$ then we know that $F$ and $g$ are the same thing, up to a constant, $C$. ie, on $(a,b)$

$F(x)=g(x)+C$

We know though that $F$ and $g$ are continuous on [a,b] so we can take the limit of $F(x)=g(x)+C$ to $a$ or $b$ and indeed the expression holds true there too, so it is true on $[a,b]$.

OK, now take our expression for $g$ and calculate it at $x=a$. Well, that’s just the area between $a$ and $a$…which is 0, duh:

$g(a)=\int_a^a f(t)dt=0$.

Now we can calculate the difference between $F$ at $a$ and $b$:

$F(b)-F(a)=(g(b)+C)-(g(a)+C)=(g(b)+C)-(0+C)=g(b)=\int_a^b f(t)dt$

And we’re done. Phew…that wasn’t too painful, was it?

This is probably the most powerful mathematical statement we have made in in this course. Use this tool wisely!

 How clear is this post?