integration – 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 Fundamental theorem of calculus example http://www.mathemafrica.org/?p=14657 http://www.mathemafrica.org/?p=14657#respond Mon, 23 Jul 2018 16:14:26 +0000 http://www.mathemafrica.org/?p=14657 We did an example today in class which I wanted to go through again here. The question was to calculate

 

\frac{d}{dx}\int_a^{x^4}\sec t dt

 

We spot the pattern immediately that it’s an FTC part 1 type question, but it’s not quite there yet. In the FTC part 1, the upper limit of the integral is just x, and not x^4. A question that we would be able to answer is:

 

\frac{d}{dx}\int_a^{x}\sec t dt

 

This would just be \sec x. Or, of course, we can show that in exactly the same way:

 

\frac{d}{du}\int_a^{u}\sec t dt=\sec u

 

That’s just changing the names of the variables, which is fine, right? But that’s not quite the question. So, how can we convert from x^4 to u? Well, how about a substitution? How about letting x^4=u and seeing what happens. This is actually just a chain rule. It’s like if I asked you to calculate:

 

\frac{d}{dx} g(x^4).

 

You would just say: Let x^4=u and then we have:

 

\frac{d}{dx} g(x^4)=\frac{du}{dx}\frac{d}{du}g(u)=4x^3 g'(u).…

]]> We did an example today in class which I wanted to go through again here. The question was to calculate

 

\frac{d}{dx}\int_a^{x^4}\sec t dt

 

We spot the pattern immediately that it’s an FTC part 1 type question, but it’s not quite there yet. In the FTC part 1, the upper limit of the integral is just x, and not x^4. A question that we would be able to answer is:

 

\frac{d}{dx}\int_a^{x}\sec t dt

 

This would just be \sec x. Or, of course, we can show that in exactly the same way:

 

\frac{d}{du}\int_a^{u}\sec t dt=\sec u

 

That’s just changing the names of the variables, which is fine, right? But that’s not quite the question. So, how can we convert from x^4 to u? Well, how about a substitution? How about letting x^4=u and seeing what happens. This is actually just a chain rule. It’s like if I asked you to calculate:

 

\frac{d}{dx} g(x^4).

 

You would just say: Let x^4=u and then we have:

 

\frac{d}{dx} g(x^4)=\frac{du}{dx}\frac{d}{du}g(u)=4x^3 g'(u).…

]]> http://www.mathemafrica.org/?feed=rss2&p=14657 0 Some more volume visualisations http://www.mathemafrica.org/?p=13671 http://www.mathemafrica.org/?p=13671#comments Thu, 14 Sep 2017 13:44:24 +0000 http://www.mathemafrica.org/?p=13671 Here is an animation which may help you imaging a shape which has a circular base, with parallel slices perpendicular to the base being equilateral triangles:movie3

 

The same thing, where the slices are squares.movie4

 

And here is the region in the (x,y) plane between y=\sqrt{x}, the x-axis and the line x=1. rotated about the y-axis. Here a thin shell is drawn in the volume, then pulled out. Then it is replaced, then the volume is filled with shells, and each of them is pulled out of the volume vertically. This is to give you an idea about how to visualise the method of cylindrical shells.

 

movie5

How clear is this post?
…]]> Here is an animation which may help you imaging a shape which has a circular base, with parallel slices perpendicular to the base being equilateral triangles:movie3

 

The same thing, where the slices are squares.movie4

 

And here is the region in the (x,y) plane between y=\sqrt{x}, the x-axis and the line x=1. rotated about the y-axis. Here a thin shell is drawn in the volume, then pulled out. Then it is replaced, then the volume is filled with shells, and each of them is pulled out of the volume vertically. This is to give you an idea about how to visualise the method of cylindrical shells.

 

movie5

How clear is this post?
…]]> http://www.mathemafrica.org/?feed=rss2&p=13671 1 Guidelines for visualising and calculating volumes of revolution http://www.mathemafrica.org/?p=13659 http://www.mathemafrica.org/?p=13659#comments Wed, 13 Sep 2017 11:49:32 +0000 http://www.mathemafrica.org/?p=13659 I have seen some people try to blindly use the formulae for volumes of revolution by cylindrical cross-sections and by cylindrical shells, and I thought that I would write a guide as to how I would recommend tackling such problems, as generally just using the formulae will lead you down blind alleys.

I’ve created an example, with an animation, which I hope will help to master this technique.

So, here is a relatively fool-proof strategy:

  1. Draw the region which you are going to have to rotate around some axis. This will generally be a matter of:
    • Drawing the curves that you have been given
    • Finding where they intersect
  2. Draw the line about which you are supposed to rotate the region
  3. Draw the reflection of the region about the line of rotation: This gives you a slice through the volume that will be formed
  4. Now you have to decide which method to use:
    • Take a slice through the volume perpendicular to the axis of rotation.
…]]> I have seen some people try to blindly use the formulae for volumes of revolution by cylindrical cross-sections and by cylindrical shells, and I thought that I would write a guide as to how I would recommend tackling such problems, as generally just using the formulae will lead you down blind alleys.

I’ve created an example, with an animation, which I hope will help to master this technique.

So, here is a relatively fool-proof strategy:

  1. Draw the region which you are going to have to rotate around some axis. This will generally be a matter of:
    • Drawing the curves that you have been given
    • Finding where they intersect
  2. Draw the line about which you are supposed to rotate the region
  3. Draw the reflection of the region about the line of rotation: This gives you a slice through the volume that will be formed
  4. Now you have to decide which method to use:
    • Take a slice through the volume perpendicular to the axis of rotation.
…]]> http://www.mathemafrica.org/?feed=rss2&p=13659 2 MAM1000W 2017 semester 2, lecture 1 (part i) http://www.mathemafrica.org/?p=13535 http://www.mathemafrica.org/?p=13535#comments Sun, 13 Aug 2017 09:57:32 +0000 http://www.mathemafrica.org/?p=13535 I wanted to put up a little summary of some of the most important things to remember from the end of last semester. There was a sudden input of new concepts, so let’s put some of them down here to get a clear reminder of what we need to know. A few things in this post:

  • The antiderivative
  • Sigma notation
  • Areas under curves

Antiderivatives

An antiderivative of a function f on an open interval I is a function F such that:

 

F'(x)=f(x) for every x\in I

 

Note that we say an antiderivative, not the antiderivative. There can be many functions whose derivatives give the same thing. While we know that:

 

\frac{d}{dx}\sin x=\cos x

 

and therefore  \sin x is an antiderivative of \cos x, we can also say that:

 

\frac{d}{dx}(\sin x+3)=\cos x

 

So \sin x+3 is also an antiderivative of \cos x. In fact for any constant c it is true that \sin x+c is an antiderivative of \cos x. We will come up with some clever notation for the antiderivative soon.…

]]> I wanted to put up a little summary of some of the most important things to remember from the end of last semester. There was a sudden input of new concepts, so let’s put some of them down here to get a clear reminder of what we need to know. A few things in this post:

  • The antiderivative
  • Sigma notation
  • Areas under curves

Antiderivatives

An antiderivative of a function f on an open interval I is a function F such that:

 

F'(x)=f(x) for every x\in I

 

Note that we say an antiderivative, not the antiderivative. There can be many functions whose derivatives give the same thing. While we know that:

 

\frac{d}{dx}\sin x=\cos x

 

and therefore  \sin x is an antiderivative of \cos x, we can also say that:

 

\frac{d}{dx}(\sin x+3)=\cos x

 

So \sin x+3 is also an antiderivative of \cos x. In fact for any constant c it is true that \sin x+c is an antiderivative of \cos x. We will come up with some clever notation for the antiderivative soon.…

]]> http://www.mathemafrica.org/?feed=rss2&p=13535 1 Picturing volumes of revolution http://www.mathemafrica.org/?p=13128 http://www.mathemafrica.org/?p=13128#comments Thu, 04 Aug 2016 13:25:51 +0000 http://www.mathemafrica.org/?p=13128 One of the homework questions this week was the following:

Let R=\{(x,y)\in \mathbb{R}^2:y\ge 0, \cos x\le y\le \sin x\,\,and\,\,0\le x\le\pi\}.

a) Sketch the region R and find its area.
b) Find the volume of the solid obtained by rotating the region R around the y-axis.

The first thing to do is to sketch the graphs of y=\cos x and y=\sin x. Once you’ve done that, the other parts of the inequalities should be clear. It should look like the red region in the following plot:

 

pic1Now we have to imagine bringing out a third axis, perpendicular to the picture above, ie. coming out towards us. We then want to rotate the red form here about the vertical axis. This we can imagine doing in the following animation:

 

trialGiven this form we can then think about either taking horizontal cross-sections through it, which will give us thin annuli, or we can take vertical, circular slices to give us thin shells. Adding these together and integrating should give us the same answer whichever way we choose to slice it, but one way will be considerably easier.…

]]> One of the homework questions this week was the following:

Let R=\{(x,y)\in \mathbb{R}^2:y\ge 0, \cos x\le y\le \sin x\,\,and\,\,0\le x\le\pi\}.

a) Sketch the region R and find its area.
b) Find the volume of the solid obtained by rotating the region R around the y-axis.

The first thing to do is to sketch the graphs of y=\cos x and y=\sin x. Once you’ve done that, the other parts of the inequalities should be clear. It should look like the red region in the following plot:

 

pic1Now we have to imagine bringing out a third axis, perpendicular to the picture above, ie. coming out towards us. We then want to rotate the red form here about the vertical axis. This we can imagine doing in the following animation:

 

trialGiven this form we can then think about either taking horizontal cross-sections through it, which will give us thin annuli, or we can take vertical, circular slices to give us thin shells. Adding these together and integrating should give us the same answer whichever way we choose to slice it, but one way will be considerably easier.…

]]> http://www.mathemafrica.org/?feed=rss2&p=13128 1