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 \(f\).…

So, we are now ready to prove the FTC part 1. We’re going to follow the proof in Stewart and add in some discussion as we go along to motivate what we are doing. What we are going to prove is that:

\[\frac{d}{dx} \int_a^x f(t) dt=f(x)\]

for \(x\in [a,b]\) when \(f\) is continuous on \([a,b]\).

Proof:

we define \(g(x)=\int_a^x f(t)dt\) and we want to find the derivative of \(g\). We will do this by using the fundamental definition of the derivative, so let’s look at calculating this function at \(x\) and \(x+h\) – ie. how much does it change when we change \(x\) by a little bit?

\[g(x+h)-g(x)=\int_a^{x+h}f(t) dt-\int_a^x f(t) dt\]

But remember that the definite integral is just the area, so this difference is the area between a and x+h minus the area between a and x. Which is just the area between x and x+h. Using the properties of integrals, we can write this formally as:

\[g(x+h)-g(x)=\int_a^{x+h}f(t) dt-\int_a^x f(t) dt=\left(\int_a^{x}f(t)+\int_x^{x+h}f(t)\right)-\int_a^{x}f(t)=\int_x^{x+h}f(t)dt\]

and we can write, for \(h\ne 0\):

\[\frac{g(x+h)-g(x)}{h}=\frac{1}{h}\int_x^{x+h}f(t)dt\]

Restated, we can think of this as the area between x and x+h divided by h.…

OK, so up to now we can’t actually use the FTC (Fundamental Theorem of Calculus) to calculate any areas. That will come from the FTC part 2.

For now, let’s take some examples and see what the FTC is saying. I’ll restate it here:

The Fundamental Theorem of Calculus, part 1

If \(f\) is continuous on \([a,b]\) then the function \(g\) defined by:

\(g(x)=\int_a^x f(t) dt\),     for \(a\le x\le b\)

is continuous on \([a,b]\) and differentiable on \((a,b)\), and \(g'(x)=f(x)\).

——

Let’s look at some examples. We’re going to take an example that we can calculate using a Riemann sum. Let’s choose \(f(x)=x^2\).

If we integrate this from \(0\) to some point \(x\) – ie. calculate the area under the curve, we get:

\(\int_0^x t^2 dt=\frac{x^3}{3}\).

Make sure that you can indeed get this by calculating the Riemann sum.

So, what does the FTC part 1 tell us? It says that if we take the derivative of this area, with respect to the upper limit, \(x\), then we get back \(f(x)\).…