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Friday, May 1, 2009

3rd Last Class Before the Exam

WOW! Just 2 more classes until the big exam. That was very fast.

Today we reviewed Techniques of integration.
The first one we looked at was the U-Substitution technique. This can be used when you are antidifferentiating a function that is composed of a factor that is a composite function and the other factor is the derivative of the inner function of the composite function.

I did this question on the smartboard. First I chose a part of the function to be substituted for U. I saw that the denominator and the exponent of e was root x, so I chose that one.
So I let u equal root x and differentiated u as shown in the image above. I also solved for solved for x using the substitution, just in case I needed it. So I substituted all values necessary and I got an antiderivative. Great we found an antiderivative, how do we know if it correct? Just differentiated the antiderivative that you found and if you get the function that was to be antidifferentiated, then you got the right antiderivative.

This question is similar to the previous one but this one is an integral. What is the difference you ask? An antiderivative will give you a function, an integral will give you a number. Even though they are different, they are also related.
To answer this question, just antidifferentiate the function and substitute the limits into the antiderivative to find the answer. This is the definition of the second part of the Fundamental Theorem of Calculus.

Oh no! There is an antiderivative that we've never seen before. What is the antiderivative of cosecant, let alone the antiderivative of cosecant squared. I can't remember how we thought of this but Mr K got us to differentiate Cotangent. Using the Quotient Rule because cotangent can be written as cosine divided by sine as shown in the image above using abbreviations. Don't try to memorize the antiderivative/derivative relationship. Just understand that you can easily build it when needed.

Mr. K began talking about the antiderivative of x to the power of some number times e to the x. He said that it was a telescopic function which means as you evaluate it, the function will get longer, but things can be reduced or factored in such a way that it will shrink like a telescope.

I was playing around with some examples and I discovered the pattern. It is such a cool antiderivative. I'll give an example:

I got to this part by using integration by parts. So f = x³, my f' = 3x², g' = e^x, and g = e^x. Notice that the second term is still an antiderivative that we cannot find easily, but it is a factor of 2 functions, so let's use integration by parts again. So we'll get:

Once again we cant antidifferentiate one of the functions so let's integrate by parts once more:

Now we can finally evaluate that antiderivative. Once you do that you'll notice that there is a common factor of e^x. So let's factor that out and you might see something completely amazing. I'll show you:

Okay look at the polynomial in the brackets. When have you seen those specific terms before? If you think and look carefully the consecutive terms are derivatives of the previous term. I tried this again with another power of x and I got the same pattern. So when you are asked to find the antiderivative of x to the power of some number times e^x, the antiderivative will be e^x times the power of x in the function being antidifferentiated minus the derivative of the previous function until you reach a constant. Don't forget the plus c.

tn-1 means the previous term.

Sorry if my algebra was bad on the image above. But if you try it yourself, it really is a beautiful thing :P

The next few slides on the slide show posted by Mr K is for homework. Study for the exam on Wednesday. Next Scribe will be Francis.

Here is the video of the Scribe:

These guys have very funny podcasts.

That's all folks.

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