Does the integral of x^(3/2) sin(2x) converge over the range of 0 to 1?

[iva]

Homework Statement

Discuss the convergence of

Integral(x^3/2 sin 2x dx) range is from 0 to 1

Homework Equations

|sin x| =< 1

The Attempt at a Solution

The sine function converges absolutely. It is also increasing from 0 to 90 degrees, decreases until 27 decrease and it is negative from 180 to 270 before it starts increasing again becoming positive at 360 degrees again.

sin 2x will always be less than or equal to 1 and greater or equal to -1. But for this interval of 0 to 1 it is increasing and therefore the function x^3/2 sin 2x dx over the given range is increasing and therefore diverging

(My lack of understanding here is that my textbook says that this test should only be used where f(x) is positive, decreasing and continuous. Yet I am asked to discuss this integral which over the given range of 0 to 1 is increasing, hence I'm compelled to say that it increases and therefore diverges.)

 

[micromass]

Firstly, you should not work with degrees. I'm sure your textbook works with radians. So you should work with them as well!

Secondly, I do not understand what you want to test for convergence! The function $$x^{3/2}\sin(2x)$$ clearly exists and is continuous on the entire interval [0,1], thus the integral with ranges 0 and 1 exists.
You should only test for convergence when working with open or infinite intervals.

Am I misunderstanding the question??

 

[iva]

I agree, but the question specifically asks us to discuss the convergence of this function. If there was no integral and it was a series i would say that the series is monotone,bounded by 0 and 1 and divergent as it never decreases but oscilates; but as it's a function I'm not sure. Clearly it is a wave graph, but what do I say about its convergence, specifically over the interval [0 1]? That it is bounded divergent?

Thanks

 

[micromass]

Well, the integral converges trivially, since the function is continuous on a closed interval. That's how I would answer that...

I still think there's something wrong with the question though...

 

[iva]

Actually, would this reasoning make sense:

since the |sin t| =< 1 as the value of x increases the x^3/2 part will inflate the wave increasingly, so that it continuous to oscilate but gets bigger and bigger, therefore being monotone diverging? I found an image for what I'm trying to describe http://www.google.co.za/imgres?imgu...age=1&ndsp=24&ved=1t:429,r:0,s:0&tx=63&ty=23"

 

[iva]

having said this , that function converges towards the origin.. guess maybe i just need to mention all these properties that I believe the function to have.