This week we were off of school. I got way to used to it and I definately do not want to go to school tomorrow. The integration packet was pretty difficult, but we got through the most of it. The packet included things we've learned over the past weeks with integration. The packet included problems that delt with definite integrals, indefinite integrals, Riemann Sums, substitution, substitution with e, natural logs, and space between a curve and a line.
First of all, an integral is the opposite of a derivative. Instead of multiplying the exponent to the coefficient in front then subtracting from the exponent, you will first add one to the exponent then divide by it.
A indefinite integral is the most basic type of integral. It is just doing the opposite of a derivitive, and the answer comes out in the form of an expression.
A definite integral is an integral with a range. For definite integrals, you first integrate the problem, then you plug in for your range. The number at the top is b and the number at the bottom is a. Once plugged in, the sum of a is subtracted from the sum of b, leaving you with a number for an answer. If there is ever an absolute value in the problem, the integral must be split. This can be done by solving for x.
Substitution is used when there seems to be a chain rule, a product rule, or a quotent rule. These rules do not apply in integration, so substitution must be used. Substitution is basically forcing a problem to work, using one rule for all problems. In the problem, a derivative and non derivitive must be located. The nonderivitive will be noted with u and the derivitive with du.
Substitution with e is the same. The derivite of e is e, so the answer to any integral with e will basically be e with it's exponent + c
Tonight working on the packet, I had a few problems. One of them delt with finding the area between a parabola and a line. I just don't understand these problems. Also, in the packet there were natural log problems with ln in the integral. I don't know what to do with these.
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Most of the time there is a ln left over, it is ln of a number. So it is a constant. So you can just pull it out in front.
ReplyDeleteFor example
1
over
ln|6|(x-1)
The integral would be
ln|x-1|
over
ln|6|
because you just take the 1/ln|6|