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Old Apr 16, 2009 | 02:34 PM
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Default Integration Help

f(x) = 3x+2 from x =1 to x =3
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Old Apr 16, 2009 | 02:40 PM
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integrate f(x)=3x+2

we get
3(x^3)/3 +2x/1
simplify
x^3 +2x

now evaluate at x=3, and x=1

3^3 +2*3 =15
1^3 =2*1 =2
subtract them
15-2=13

answer is 13


this is a super simple problem, did you just start integration?
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Old Apr 16, 2009 | 02:45 PM
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Yeah we just started...I thought we have to find the anti-derivative...

3/2x^2 + 2x
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Old Apr 16, 2009 | 02:50 PM
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Originally Posted by F22B Prelude
Yeah we just started...I thought we have to find the anti-derivative...

3/2x^2 + 2x
anti-derivative just means an indefinite integral, same shit

where is the 2 coming from in the 3/2x^2?

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Old Apr 16, 2009 | 02:51 PM
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anti-derivative of 3x is 3/2x^2...isn't it
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Old Apr 16, 2009 | 02:53 PM
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Originally Posted by F22B Prelude
anti-derivative of 3x is 3/2x^2...isn't it
ah god damn it, i wrote it down wrong

lets try that again give me a minute to type shit out
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Old Apr 16, 2009 | 02:55 PM
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well yeah, so you get 3/2x^2 +2x
then evaluate
so 3/2(3) +2(3) = 9/2 +6 = 21/2
3/2(1) +2(1)= 3/2 +2 = 7/2

21/2 - 7/2 = 14/2 = 7
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Old Apr 16, 2009 | 02:58 PM
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Explain this to me then

Approximate the area under the graph of f(x) and above the xaxis using the following methods with n =4.
1.) Use left end points
2.) Use right end points
3.)Average the answers in parts a and b
4.) Use midpoints
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Old Apr 16, 2009 | 03:02 PM
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oh shit you are starting from the beginning

did you attend class the day they did this crap?
its a visual representation of what integration means

integration tells you the area under a graph to the x-axis, positive aread
doing the endpoint thing what they mean is that to calculate the area under a graph physically they break the area into rectangles that have their tops with either the right top corner, left top corner, or midpoint touching the eqaution on the graph
then you use the area of all these little rectangles to estimate the area under the graph
using the 3 different placements of the points give you slightly different answers, and you will see that the more rectangles you use the closer to a perfect answer you get

thus what integration really does is an infinite number of rectangles, giving a perfect slution
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Old Apr 16, 2009 | 03:03 PM
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He threw 5 sections at us this week, and he started discussing this with 5 minutes left and gave us homework on it...I know what it means, but how the fuck am i supposed to fucking do it
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