ok, i didn't know i had to do two blogs either, so like a lot of other people, i am going to combine them on this blog.
we learned about substitution, substitution takes the place of the derivative rules for problems with a quotient rule and product rule, substitution has a few steps.
1. find u by looking inside the parentheses inside the problem
2. take the derivative of u to find du
3. go into the origional problem and switch out (substitute) the stuff
4. integrate
5. plug in
ok, so ln i kinda understand, ln integration happens when the bottom of a fraction is the u and the top is the derivative of u. so since the derivative of ln(x) is 1/x, the answer is just ln(u)+c like everytime.
e^x integration is pretty decent too, all you have to do for this kind of integration is set your u equal to the exponent of the e. So the derivative of e^x is just e^x times the derivative of x.
So there are four different methods of integration, LRAM, RRAM, MRAM, and trapezoidal.
The first formula you need to know is x=(b-a)/n [a,b] with n subintervals. You will need to know this because each of the next formulas require that you know what x is.
LRAM- left hand approximation. (this puts the rectangles used to find the area on the left side of the curve) x[f(a)+f(a+x)+...f(b)]
RRAM- right hand approximation. (this puts the rectangles used to find the area on the right side of the curve) x[f(a+x)+...f(b)]
MRAM- approximation from the middle. (this puts the rectangles right on top of the curve, so that the curve goes through the middle of each one) x[f(mid)+f(mid)+...]
Trapezoidal- this does not use squares, instead it uses trapezoids to eliminate most of the empty space inside the curve, and I think this is the most accurate. x/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]
i suck at integration, that's my only problem.
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integration will get easier with practice. that's pretty much all you can do to help yourself. just remember to plus c behind all indefinite integrals and that you don't need it behind definite integrals because you will get a number for your answer. also, if you have anything besides an equation for an answer with indefinite integrals, something is wrong. just remember, it's like taking the derivative backwards! :-)
ReplyDeleteIntegration is pretty much the same as derivative but backwards with sometimes adding substitution to get the derivative to wokr
ReplyDeleteintegration is basically setting the derivative back to the origional equation you add one to the exponent and then multiply the term by the reciprocal exponent
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