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

by MoarGrades
Tags: graph, integral, x3ex2
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MoarGrades
#1
Feb5-11, 05:35 PM
P: 15
So if you graph f = (x^3)(e^x^2) and F = (1/2)(e^x^2)(x^2-1), 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.
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#2
Feb6-11, 03:16 AM
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Quote Quote by MoarGrades View Post
So if you graph f = (x^3)(e^x^2) and F = (1/2)(e^x^2)(x^2-1), 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.
Well in the same functions you got (e^x^2) so we just remove that from both and get:
f=(x^3)
F=(1/2)(x^2-1)

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?
arildno
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Feb6-11, 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.

HallsofIvy
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Feb6-11, 11:50 AM
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Why does the graph of the integral of x^3e^x^2 do this?

Quote Quote by MoarGrades View Post
So if you graph f = (x^3)(e^x^2) and F = (1/2)(e^x^2)(x^2-1), 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?
Can you explain why you think that would be true?

Why does F grow slower than f? It makes no sense to me.
Why does it not make sense? What does the one have to do with the other?
MoarGrades
#5
Feb6-11, 12:21 PM
P: 15
Quote Quote by HallsofIvy View Post
Can you explain why you think that would be true?


Why does it not make sense? What does the one have to do with the other?
Because isn't an integral supposed to represent the area under a curve? How can the area be smaller than the value of the function for a positive function?

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^2-1)? 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?
HallsofIvy
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Feb7-11, 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.
MoarGrades
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Feb7-11, 10:44 AM
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Quote Quote by HallsofIvy View Post
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.
What do you mean? Isn't the integral of a positive function from a to b the area under it? Or if you describe that function as velocity, isn't the integral the displacement?
Hurkyl
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Feb7-11, 10:50 AM
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Quote Quote by MoarGrades View Post
What do you mean? Isn't the integral of a positive function from a to b the area under it? Or if you describe that function as velocity, isn't the integral the displacement?
A better analogy would be to wonder how you could travel just one mile if your velocity is 60 miles per hour.
HallsofIvy
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Feb7-11, 02:39 PM
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Quote Quote by MoarGrades View Post
What do you mean? Isn't the integral of a positive function from a to b the area under it? Or if you describe that function as velocity, isn't the integral the displacement?
Yes, but I not saying that those "integrals" don't give those quantities. My point is that you cannot compare "area" to "distance" or "speed" to "distance". It simply doesn't make sense to say that a given area is larger than a given distance or that a given speed is greater tnan a given distance any more than you can say that a specific distance is larger than a given time.
MoarGrades
#10
Feb7-11, 03:47 PM
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Quote Quote by Hurkyl View Post
A better analogy would be to wonder how you could travel just one mile if your velocity is 60 miles per hour.
Got it. thanks.


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