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Why does the graph of the integral of x^3e^x^2 do this? 
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#1
Feb511, 05:35 PM

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So if you graph f = (x^3)(e^x^2) and F = (1/2)(e^x^2)(x^21), its integral, from 0 to 1, F starts out much larger than f, and then f become starts growing much much faster than F. Shouldn't F be the one growing the fastest, as it's supposed to be the area under f?
Why does F grow slower than f? It makes no sense to me. 


#2
Feb611, 03:16 AM

P: 49

f=(x^3) F=(1/2)(x^21) We can just multiply the numbers in the second function(F) and thus get: f=(x^3) F=0.5(x^2)0.5 The 0.5 is a very small value so if we subtract it from a big number(for example 10) then it doesn't make a big difference. For that reason we just take it away and get: f=x^3 F=0.5(x^2) As you see now when you put any value for x in these functions f is always bigger!! What is bigger? 10^3 or the half of 10^2? 


#3
Feb611, 04:28 AM

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The fraction f/F can be written as:
[tex]\frac{f}{F}=\frac{2x^{3}}{x^{2}2}[/tex] Suppose that x is a lot less than 1. Then, the denominator can be simplified as: [tex]\frac{2x^{3}}{x^{2}2}\approx\frac{2x^{3}}{2}=x^{3}[/tex] which is much less than 1 in absolute vale. That is, when x is small, we have that f/F is much less than 1, i.e, f is much less than F. 


#4
Feb611, 11:50 AM

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Why does the graph of the integral of x^3e^x^2 do this?



#5
Feb611, 12:21 PM

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For example, wouldn't this integral mean that someone running at a speed of (t^3)(e^t^2), at time t, travels a distance of (1/2)(e^t^2)(t^21)? How does it make sense that he travelled a distance that is less than his speed after a large enough t? Is it just because f grows so fast? 


#6
Feb711, 08:27 AM

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What makes no sense to me is to think that it is possible to compare two different kinds of quantities! Area is measured in "distance^2" and cannot be compared to "distance". Similarly, speed is measured in "distance/time" and it makes no sense to compare speed to distance.



#7
Feb711, 10:44 AM

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#8
Feb711, 10:50 AM

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#9
Feb711, 02:39 PM

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#10
Feb711, 03:47 PM

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