Posted by Paul on Friday, February 19, 2010 at 3:57am.
I have the function f(x)=e^x*sinNx on the interval [0,1] where N is a positive integer. What does it mean describe the graph of the function when N={whatever integer}? And what happens to the graph and to the value of the integral as N approaches infinity? Does the graph confirm the limiting behavior of the integral's value?
* Calculus - Damon, Friday, February 19, 2010 at 10:07am
well, e^0 is 1
and e^.5 = 1.64
and e^1 is 2.72
so it is a sine wave with increasing amplitude as you approach 1 and frequency increasing with N
The integral of e^ax sin bx dx is
[e^ax/(a^2+b^2)] [a sin bx -b cos bx}here a = 1 and b = N
so
[e^x/(1+N^2)] [sin Nx - N cos Nx]
as N gets big
this looks like
e^x (-N cos Nx)/N^2
or
(-e^x/N)(cos Nx)
e^x is that small constant and cos Nx ranges between -1 and + 1 so as N gets big this goes to zero like 1/N
* Calculus - Paul, Wednesday, February 24, 2010 at 11:12am
But how did you arrive at "as N gets big this looks like e^x(-NcosNx)/N^2?????
Professor Damon is describing the integral, not the function.
Have you graphed this function? For various N?
Yes I have, but I don't understand why the integral would go to zero. And the description of the interval.
is it because every interval of one the integrals approach 1?
I meant to say zero
But then would this confirm the limiting behavior of the graph because at every interval up to infinity it would approaching zero?
To understand how the expression "as N gets big this looks like e^x(-NcosNx)/N^2" is derived, let's break it down step by step:
1. Start with the integral of the given function: ∫( e^x * sin(Nx) ) dx
2. Apply integration by parts, which states that the integral of u * dv is equal to uv minus the integral of v * du. Choose u = sin(Nx) and dv = e^x dx.
Applying integration by parts gives: ∫( e^x * sin(Nx) ) dx = [ (e^x)(-cos(Nx))/N ] - ∫( -cos(Nx)/N * e^x ) dx
3. Simplify the remaining integral: ∫( -cos(Nx)/N * e^x ) dx
4. Since e^x is a constant with respect to x, it can be taken outside of the integral: e^x * ∫( -cos(Nx)/N ) dx
5. The integral of -cos(Nx)/N dx can be computed as -1/N * ∫( cos(Nx) ) dx. This integral is evaluated as sin(Nx)/N: -sin(Nx)/N
6. Combining the results, the original integral becomes:
[ (e^x)(-cos(Nx))/N ] - e^x * ( -sin(Nx)/N )
Simplifying this further, we get:
-e^x * cos(Nx)/N + e^x * sin(Nx)/N
7. Factoring out e^x/N gives:
e^x * ( sin(Nx) - cos(Nx) ) / N
8. Finally, as N gets large, cos(Nx) and sin(Nx) oscillate rapidly between -1 and 1, resulting in the limiting behavior of the expression:
(-e^x/N) * ( cos(Nx) ) → 0
Therefore, as N approaches infinity, the graph of the function approaches y = 0, confirming the limiting behavior of the integral's value.