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 $f(x)$. That is, if we ask:

How quickly is the area increasing as we increase the upper limit? The answer is just the value of the function at that point. As an example, if we ask how quickly the area under the curve is increasing at the point $x=4$, the answer is $\left.\frac{d}{dx}\int_0^x t^2 dt\right|_{x=4}=\left.x^2\right|_{x=4}=4^2=16$.

Again, in slightly different words: The rate at which an area under a curve increases as you change the upper bound of the area is just the value of the function itself at that point:

The derivative (with respect to x) of the integral of a function (from a to x) is the function itself.

This means that so long as a function is continuous on a closed interval, then I can always calculate how fast the area is increasing. For instance, if I want to know how quickly the area under the graph of $f(x)=\sin(\cos\sqrt{x})^4$ is increasing at the point $x=5$, the answer is just $\sin(\cos\sqrt(5))^4$.

Let’s take the same animation that we looked at before: The red line is the area under the curve. The rate of change of the area is just the gradient of the red curve, and indeed its value is simply the value of the function $f(x)$ itself: $\frac{d}{dx}\int_a^x f(t) dt=f(x)$,    for $a\le x\le b$ where the function is continuous on $[a,b]$.

This gives us some tools for calculating rather complicated things which you may never have imagined. I can now solve equations where there are integrals in the equations themselves! For instance:

Find the function $f(x)$ for which: $\int_0^x f(t)dt=x^2+\sin x$

That is, find a function whose area from 0 to some point $x$ is simply equal to $x^2+\sin x$. The FTC part 1 gives us a way to do this. Just take the derivative of both side. Taking the derivative of the left hand side just returns $f(x)$ by the FTC part 1, and the derivative of the right hand side is $2x+\cos x$. This means that we have: $\frac{d}{dx}\int_0^x f(t)dt=\frac{d}{dx}\left(x^2+\sin x\right)$

giving us: $f(x)=2x+\cos x$.

When you calculate the area of this function from 0 to x, the value of the area is equal to $x^2+\sin x$.

The FTC allows to calculate some really strange things now. Imagine before if I’d asked you to calculate: $\frac{d}{dx}\int_2^\frac{1}{x}\sin^4 t dt$

Here we have defined a function which is the area of some trig function from 2 up to a value $\frac{1}{x}$ and we want to know how quickly this changes as we change $x$.

Well, we can perform a substitution here. Let’s let $\frac{1}{x}=u$. Then for any function p(x) $\frac{d p(x)}{dx}=\frac{du}{dx}\frac{d p(x(u))}{du}=-\frac{1}{x^2}\frac{d p(x(u))}{du}$

So our original question is now: $\frac{d}{dx}\int_2^\frac{1}{x}\sin^4 t dt=-\frac{1}{x^2}\frac{d}{du} \int_2^u\sin^4 t dt$

and other than the fact that here we have $u$ rather than $x$, the latter part of this is exactly as in the FTC part 1, so it gives us: $-\frac{1}{x^2}\frac{d}{du} \int_2^u\sin^4 t dt=-\frac{1}{x^2}\sin^4 u=-\frac{1}{x^2}\sin^4\frac{1}{x}$

and we are done. Please don’t just look at this and nod your heads, or knot your brows, write it down and make sure that every step makes sense.

It’s not clear why any of this is useful yet, but think of it at the moment as another tool in your mathematical toolkit. This is a new type of object that you can (with practice) manipulate easily, and can use this to solve problems which otherwise would have been impossible.

In the next post we are actually going to prove the FTC part 1.

 How clear is this post?