okay so these two i'm putting together because i went on vacation and was not able to do my second one when it was dueee.
let's go over the lram and that stuffff.
LRAM is left hand approximation
FORMULA: delta x [f(a) + f( delta x +a) .... + f( delta x - b)]
EXAMPLE: calculate the left Riemann Sum for -4x -5 on the interval [-3, -1] divided into 2 subintervals.
delta x equals: -1+3 /2 = 2/2 = 1
1[ f(-3) + f(-3 +1)]
1[ f( -3) + f(-2)]
then plug into your equation
RRAM is right hand approximation
FORMULA: delta x [ f(a + delta x) + .... + f(b)]
so using the same example:
1[ f( -2) + f(-1)] and then plug into your equation
MRAM is to calculate the middle
FORMULA: delta x [ f(mid) + f(mid) + .... ]
To find midpoints, you would add the two numbers together then divide by two
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DIFFERENCE BETWEEN INDEFINITE AND DEFINITE INTERGRALS
Indefinate - only an equation with the intergral symbol; you simply take the derivative backwards, you divide the exponent from the coefficient and add one to the exponent. Also, you MUST place a + c at the end of the equation. Why, you may ask, because you don't know if the beginning equation had a constant at the end of it, so you MUST mark it, or it will be wrong, for these you get an equation as an answerr.
Definate Integrals - almost the same thing except there will be a number at the top and bottom of the integral symbol; the end of your equation will be marked with a dx. You treat it the same as an indefinate integral except you plug in the top number, or the b, into the derivative and then you have to SUBTRACT the plugged in bottom number, a, from the derivative plugged in. You get a numerical answerrr!
EXAMPLE:
3
S x^2 dx
0
= x^3/3 ..... (3)^3/3 - (0)^3/3 ...... = 9-0
= 9
linearization:
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
VOLUME BY WASHERS
FORMULA: pie times the integral of the [top function] squared minus the [bottom function] squared times dx.
If you don't have the inbetween number you have to set the functions equal, but if you do, then it's worked the same way as above. square the formula's that were given and simplify. then take the integral of it and plug in the numbers they give you or you found by setting the formulas equal to each other and then solve like any other one by subracting them.
QUESTIONSS! i have no idea what e integration is, i mean when i see it i might, but just hearing people say it i don't have a clue! anyone up for explaining?
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e integration isn't terribly too hard.. you set your u to the exponent of the and more times than not, when you take the derivative of u, that will be what's in front of the e. for example...
ReplyDelete(2x+2) e^(x^2+2x)
u = x^2+2x du=2x+2
So the integral of e^u du is just e^u + c. So your answer would be
e^(x^2+2x) + c
e integration is actually really easy. If you're being asked to integrate e, you use substitution. U will always be the exponent e is raised to. You then integrate accordingly. Most of the time, your answer will be e raised to whatever it was previously raised to plus c.
ReplyDeletee integration:
ReplyDeletewhatever is raised to the e power will be your u and du will be the derivative of u. For example:
e^2x-1dx
u=2x-1 du=2
rewrite the function as:
1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.
when you see e raised to something, you will set up up with u and du. your u will be what e is raised to and your du will be u's derivative. Then if you need to manipulate the problem to show du, then you do that accordingly. Then you raise e to u.. your answer will be du the e integration +C... hope this helps you girl!
ReplyDeletee integration issomething i actually know. To integrate e, you use substitution. Your gonna always be the exponent e is raised to. Then you inteegrate. Most likely, your answer will be e raised to whatever it was previously raised to plus c.
ReplyDeletee integration is not too hard, it just gets me when i see it too. You just use substitution. u is equal to the exponent that the e is raised to. Then du is the derivative of u obviously. You just integrate it from there by using substitution and voila :)
ReplyDeleteeven though you have 6 people who explained to you what e integration is, i will also do it because i need a second comment and i actually understand e integration! haha
ReplyDeleteit's pretty much using substitution.
so you need to find your U and your DU
e^whatever
whatever will always be your U.
DU will be the derivative of whatever..always.
then simply follow all substitution rules until you get your answer. :)