+0  
 
0
245
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avatar+1828 

I think it must be 1 converge 

because the term that I circled it equal to zero ! 

I'm I right ? 

 

difficulty advanced
xvxvxv  Mar 17, 2015

Best Answer 

 #3
avatar+26399 
+10

According to Mathcad:

 

 integral:

 

and Wolfram Alpha won't evaluate it.

.

Alan  Mar 18, 2015
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3+0 Answers

 #1
avatar+91435 
+5

Here is the graph.

https://www.desmos.com/calculator/orgvwef1o3

 

Why does what you circled =0 ??

What does diverge and converge mean?

Thinking about what they 'may' mean this doesn't really do either does it?

 

I know I am out of my depth here. 

I guess I am hoping someone will teach me something that is easy to learn. (Well one can dream)  :)  

Melody  Mar 18, 2015
 #2
avatar+80956 
+10

Like Melody, I'm on shaky ground here....so.....you might have to take this with a grain of salt....

By the Integral Test, the function in the integrand must be eventually positive and decreasing on the interval of interest - in this case, 0 to ∞.

Here's the graph of  y = cos(x)*e^(-sin(x))

.....https://www.desmos.com/calculator/ho13rj5sjr

Note that this function is not eventually positive and decreasing on the interval (it is periodic).....thus.....the integral diverges

That's my two cents worth.....

Alan???  heureka???....to the rescue???

 

  

CPhill  Mar 18, 2015
 #3
avatar+26399 
+10
Best Answer

According to Mathcad:

 

 integral:

 

and Wolfram Alpha won't evaluate it.

.

Alan  Mar 18, 2015

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