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March 30, 2015

Posted by **Paul** on Wednesday, February 24, 2010 at 11:13am.

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?????

- calculus -
**bobpursley**, Wednesday, February 24, 2010 at 11:36amProfessor Damon is describing the integral, not the function.

Have you graphed this function? For various N?

- calculus -
**Paul**, Wednesday, February 24, 2010 at 12:11pmYes I have, but I don't understand why the integral would go to zero. And the description of the interval.

- calculus -
**Paul**, Wednesday, February 24, 2010 at 12:18pmis it because every interval of one the integrals approach 1?

- calculus -
**Paul**, Wednesday, February 24, 2010 at 12:18pmI meant to say zero

- calculus -
**Paul**, Wednesday, February 24, 2010 at 12:20pmBut then would this confirm the limiting behavior of the graph because at every interval up to infinity it would approaching zero?

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