Week #14 Calculus
So this week in Calculus we learned a few odd and in things about integration and how to use it etc.
So far my two favorite things about Calculus are ln integration and e integration.
ln is so easy because most of the time, whenever there is a fraction, the bottom will be your u and the top will be the derivative of that. Even if the top isn't the derivative, most of the time you only have to add a number in front of the integration symbol to balance it out.
So let's say... what is the equation that models the area under the curve 1/(x-1). Well, you have to take the integral of this but you run into a problem whenever you can't split it up into two fractions. So, you go to ln. The derivative of ln(x) is simply 1/x. So in this case, set u to the bottom and du is 1dx. Now its the integral of 1/u du which is simply ln(u). Substituting back in for u will give you ln(x-1) + c as your answer.
These can get much more fun but they are always so simple. That's why I like them :-).
e^x integration is just as simple. Whenever working with e^x integration, your u is always set to the exponent of the e. The derivative of e^x is simply e^x times the derivative of x. So when working with these, really all you have to worry about it balancing it out with the derivative of x.
For example, e^(2x). This would become (1/2)e^(2x) + c, The reason behind the (1/2) is because if you take the derivative of e^(2x) it is 2e^(2x). However, we don't want that 2. So we take it out by putting a (1/2) in front, just like we did for other integrals.
Natural log and e integration are really easy and easy to identify... I hope everyone else likes it as much as me.
The only issue I've had with these types of integration is doing really well with the simplifying...but Brandi said its okay. I suppose we will get better with it.
-John
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Okay, John, since you're amazing at these, would you mind helping me out a bit? :)
ReplyDeleteFor ln integration, can you always manipulate the top to match the bottom? Like in here:
(x^2+x+1)/(x^2+1)
The top can't match the bottom in this case. The top has to be one exponent less than the bottom. You would have to do the division for this problem.
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