fundamental theorem of calculus – Mathemafrica http://www.mathemafrica.org All about maths in Africa Mon, 20 Apr 2020 06:44:50 +0000 en-US hourly 1 https://wordpress.org/?v=4.4.2 The Fundamental Theory of Calculus part 2 (part ii) http://www.mathemafrica.org/?p=15183 http://www.mathemafrica.org/?p=15183#respond Thu, 11 Jul 2019 10:31:19 +0000 http://www.mathemafrica.org/?p=15183 OK, get ready for some Calculus-Fu!

We have now said that rather than taking pesky limits of Riemann sums to calculate areas under curves (ie. definite integrals), all we need is to find an antiderivative of the function that we are looking at.

As a reminder, to calculate the definite integral of a continuous function, we have:

 

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

 

where F is any antiderivative of f

Remember that to calculate the area under the curve of f(x)=x^4 from, let’s say 2 to 5, we had to write:

 

\int_2^5 x^4 dx=\lim_{n\rightarrow \infty}\sum_{i=1}^n f(x_i)\Delta x=\lim_{n\rightarrow \infty} f\left(2+\frac{3i}{n}\right)\frac{3}{n}=\lim_{n\rightarrow\infty}\frac{3}{n}\left(2+\frac{3i}{n}\right)^4

 

And at that point we had barely even started because we still had to actually evaluate this sum, which is a hell of a calculation…then we have to calculate the limit. What a pain.

 

Now, we are told that all we have to do is to find any antiderivative of f(x)=x^4 and we are basically done.

 

Can we find a function which, when we take its derivative gives us x^4?…

]]> OK, get ready for some Calculus-Fu!

We have now said that rather than taking pesky limits of Riemann sums to calculate areas under curves (ie. definite integrals), all we need is to find an antiderivative of the function that we are looking at.

As a reminder, to calculate the definite integral of a continuous function, we have:

 

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

 

where F is any antiderivative of f

Remember that to calculate the area under the curve of f(x)=x^4 from, let’s say 2 to 5, we had to write:

 

\int_2^5 x^4 dx=\lim_{n\rightarrow \infty}\sum_{i=1}^n f(x_i)\Delta x=\lim_{n\rightarrow \infty} f\left(2+\frac{3i}{n}\right)\frac{3}{n}=\lim_{n\rightarrow\infty}\frac{3}{n}\left(2+\frac{3i}{n}\right)^4

 

And at that point we had barely even started because we still had to actually evaluate this sum, which is a hell of a calculation…then we have to calculate the limit. What a pain.

 

Now, we are told that all we have to do is to find any antiderivative of f(x)=x^4 and we are basically done.

 

Can we find a function which, when we take its derivative gives us x^4?…

]]> http://www.mathemafrica.org/?feed=rss2&p=15183 0 The Fundamental Theory of Calculus part 2 (part i) http://www.mathemafrica.org/?p=15177 http://www.mathemafrica.org/?p=15177#respond Thu, 11 Jul 2019 10:15:30 +0000 http://www.mathemafrica.org/?p=15177 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.

————-

 

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.…

]]> 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.

————-

 

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.…

]]> http://www.mathemafrica.org/?feed=rss2&p=15177 0 The Fundamental Theorem of Calculus part 1 (part iii) http://www.mathemafrica.org/?p=15170 http://www.mathemafrica.org/?p=15170#respond Thu, 11 Jul 2019 09:52:53 +0000 http://www.mathemafrica.org/?p=15170 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.…

]]> 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.…

]]> http://www.mathemafrica.org/?feed=rss2&p=15170 0 The Fundamental Theorem of Calculus part 1 (part ii) http://www.mathemafrica.org/?p=15151 http://www.mathemafrica.org/?p=15151#respond Thu, 11 Jul 2019 08:45:31 +0000 http://www.mathemafrica.org/?p=15151 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).…

]]> 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).…

]]> http://www.mathemafrica.org/?feed=rss2&p=15151 0 The Fundamental Theorem of Calculus, part 1 (part i) http://www.mathemafrica.org/?p=15139 http://www.mathemafrica.org/?p=15139#comments Tue, 09 Jul 2019 14:15:56 +0000 http://www.mathemafrica.org/?p=15139 We’ve seen some intriguing things in this course so far, and we’ve developed some clever tricks, from how to find the gradient of just about any function we can throw at you, to proving statements to be true for an infinite number of cases.

To some extent, this is what we have looked at so far (at least in terms of calculus, and building up to calculus):

Screen Shot 2019-07-09 at 16.06.58

However, we’re about to see some magic. We’re about to see the most important thing yet on this course, and indeed one of the most important moments in all of mathematical history.

We are going to see…actually, we are going to prove, that there is a relationship between rates of change and the area under a graph. This doesn’t sound that amazing, but its consequences have essentially allowed for the development of much of modern mathematics over the last 350 years.

The link that we are going to prove will allow us to find the area under graphs of functions for which taking the Riemann sum would be really hard.…

]]> We’ve seen some intriguing things in this course so far, and we’ve developed some clever tricks, from how to find the gradient of just about any function we can throw at you, to proving statements to be true for an infinite number of cases.

To some extent, this is what we have looked at so far (at least in terms of calculus, and building up to calculus):

Screen Shot 2019-07-09 at 16.06.58

However, we’re about to see some magic. We’re about to see the most important thing yet on this course, and indeed one of the most important moments in all of mathematical history.

We are going to see…actually, we are going to prove, that there is a relationship between rates of change and the area under a graph. This doesn’t sound that amazing, but its consequences have essentially allowed for the development of much of modern mathematics over the last 350 years.

The link that we are going to prove will allow us to find the area under graphs of functions for which taking the Riemann sum would be really hard.…

]]> http://www.mathemafrica.org/?feed=rss2&p=15139 1