Hi
Forgot Sunday, was going to submit Monday, but sick…so…yay BLOG!
Integration!!!! (exclamation marks) (did I spell that right?...oh wll)
INDEFINITE: So I actually got this: you add one to the exponent and divide by your new exponent and voila!!!!
And for the trig ones…you’re going to be seriously out of luck if you don’t know your DERIVATIVE FORMULAS!!! Muy importante.
So, examples, shall we? (I’m just going to use my “S” for integral in lack of an actual cool integral button on my stupid computer…wait..I just found it!!!!)
∫ 3x2 dx
Add 1 to the exponent giving you 3, and divide by, you guessed it, 3!
3x3
3
Which then simplifies to :
x3 + C
*only put the +C when doing indefinite integration, NOT DEFINITE (thought I’d stress that).
You know an integral is DEFINITE when it has these little numbers on the side of ∫. You basically do the same thing as in indefinite, but you have to subtract whatever your top number plugged in is minus your bottom number plugged in. Make sense? Not really. Not coherent at the moment.
So example!!!!! (again with the exclamation marks!..gah!...no!...it’s like an instant reflex!!!.....jeez us!)
0∫3 x3
Integrate:
¼ x4 l30
Plug in 3 and you get 81/4, and 0 gives you 0
So your answer is 81/4!! (Exclamation points..)
This past week we also did LRAM and MRAM and RRAM and all that other good stuff and I got it, but I just need to actually MEMORIZE the formulas, you know?
So, I do believe that is about it for everything I know. Now for the questions:
1. Can someone explain to me what Substitution is since I missed that?
2. Also, I need a good example of a definite integration with absolute value…kind of confusing.
THANKS MUCHO!!!! (exclamation points…from now on I’m using *)
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Substitution is integration pretty much instead you pull out some variables and then plug back in... so lets say you have(the square root) of 3x-1dx. your u=3x-1 and du=3
ReplyDeletethen you have to put a 1/3 in front to balance the three.
so you have 1/2{(3x-1)^1/2 (3)dx
so now by integrating you get..
1/2(2/3u^3/2)+c
so your final answer would be..
1/3 (2x-1)^3/2+c
hope this helps!
For the absolute value...
ReplyDeleteYou set the inside = to 0. So lets say absolute value of 2x-1 on the interval from 0 to 4.
Set the 2x-1=0
Solve for x to get x=1/2
Now, set up two integrals.
From 0 to 1/2 and from 1/2 to 4.
Make the first one (the one from 0 to 1/2) negative.
So it would be
from 0 to 1/2 (integrate -(2x-1))
from 1/2 to 4 (integrate (2x-1))
and whatever you get for those two, just add them. that's your answer.
in the subsititution, to find du you just take the derivative of whatever's in the inside right?
ReplyDeletewhat about when you have like 2 things being multiplied
If you have two things being multiplied either one thing is the du or you have to actually multiply it out.
ReplyDelete