And then we can antidifferentiate sinx and e^x easily, and after antidifferentiating it, it becomes...
Note: *Don't forget the + C. It was evident during the class that we constantly forgot to add the + C. It costs a mark!*
So we had the integral of the function (x^2 + 1)/x on the interval from 1 to e with respect to x. Because we don't know how to antidifferentiate quotients, we can massage that to separate terms so that they're both divided by x. The x's reduce on the first term and that leaves us with the integral of the function x + 1/x on the interval from 1 to e with respect to x. From here, we can antidifferentiate this, a LOT easier than before.
We now have x^2 /2 + lnx evaluated on the interval from 1 to e. Now we get...
After evaluating, we get the final answer of e^ + 1 / 2. This leads us to conclude to not differentiate too fast, and algebraically change it so that it's easier to differentiate it.
But on to the main dish. If we can differentiate multiplying terms, can we antidifferentiate multiplying terms as well? Well we tried that out with this question. Mr. K pretty much confused us when we asked if there was a product rule for antidifferentiating, and his answer was a "kind of, not really, definitely, maybe" answer. So, basically, yes, but mostly no. -__-"



So because there's no g'(x), we multiply by one using a full fraction using the coefficient number from g(x). From the fraction, remove the half from the integral so that we're left with F'(g(x)) * g'(x), which is sin(2x) * 1/2 + C <-- There it is again! OMG!
Anyways, that wasn't fun because I fell asleep for an hour and woke up really tired, so I'm happy that I'm finished, 'cause I have to study for Chemistry now.. Like, now. I forgot, I have to crown the next scribe. Hi, I'm Justus. No, I'm not really Justus, I'm saying that Justus is the next scribe for the last day of the semester. Have fun guy.

2 comments:
Hey Rence,
Just for fun I put your post into Wordle:
http://www.wordle.net/gallery/wrdl/484137/Anti-diff
What do you think?
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