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

In summary, the conversation discusses the relationship between two functions, f and F, and their corresponding integrals. The integral of F, which represents the area under the curve, is initially larger than the integral of f, but as x increases, the growth of f outpaces that of F. This seems counterintuitive since the integral is supposed to represent the area under the curve. However, it is important to note that area and distance (or speed) are two different types of quantities and cannot be directly compared. The conversation ends with the understanding that while it may seem strange, the integral of a positive function can indeed be smaller than the function itself.
  • #1
MoarGrades
15
0
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
MoarGrades said:
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?
 
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  • #3
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
MoarGrades said:
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?
 
  • #5
HallsofIvy said:
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 traveled a distance that is less than his speed after a large enough t?

Is it just because f grows so fast?
 
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  • #6
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
HallsofIvy said:
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?
 
  • #8
MoarGrades said:
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.
 
  • #9
MoarGrades said:
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.
 
  • #10
Hurkyl said:
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.
 

1. Why does the graph of the integral of x^3e^x^2 have a steep slope?

The steep slope on the graph of the integral of x^3e^x^2 is due to the exponential function e^x^2. As x increases, e^x^2 grows at a much faster rate than x^3, resulting in a steep increase in the integral.

2. Why does the graph of the integral of x^3e^x^2 have a large positive value?

The large positive value on the graph of the integral of x^3e^x^2 is a result of the positive nature of both x^3 and e^x^2. When these two functions are integrated, the resulting value will also be positive.

3. Why does the graph of the integral of x^3e^x^2 have a smaller slope at smaller x values?

The smaller slope at smaller x values on the graph of the integral of x^3e^x^2 is due to the fact that the exponential function e^x^2 grows at a much faster rate than x^3. As x gets smaller, the effect of x^3 becomes less significant in comparison to e^x^2, resulting in a smaller slope.

4. Why does the graph of the integral of x^3e^x^2 have a positive value even when x is negative?

The positive value on the graph of the integral of x^3e^x^2 for negative x values is due to the fact that the integral of a function measures the area under the curve, regardless of the sign of x. Therefore, even when x is negative, the area under the curve can still be positive.

5. Why does the graph of the integral of x^3e^x^2 approach infinity as x approaches infinity?

The graph of the integral of x^3e^x^2 approaches infinity as x approaches infinity because both x^3 and e^x^2 are positive and increasing functions. As x gets larger, the integral will continue to increase at a faster rate, ultimately approaching infinity.

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