Evaluate integral {x,-2,2} [sinx / (1 + x^6)]dx

[jwpaine]

Where do I begin with this one?
Evaluate \(\displaystyle \int _{-2}^{2} \frac{sinx}{1 + x^6}dx\)

I'm not sure where to begin, as I can't think of an antiderivitive to plug in the lower and upper bounds....

We haven't been on integration long, and the professor is not the best teacher.

John

 

[stapel]

Where do I begin with this one?
Evaluate \(\displaystyle \int _{-2}^{2} \frac{sinx}{1 + x^6}dx\)

Personally, I'd begin by asking the professor if maybe there's a typo, because "The Integrator" gives a really nasty result for the indefinite integral... :shock:

Eliz.

 

[jwpaine]

Where do I begin with this one?
Evaluate \(\displaystyle \int _{-2}^{2} \frac{sinx}{1 + x^6}dx\)

Personally, I'd begin by asking the professor if maybe there's a typo, because "The Integrator" gives a really nasty result for the indefinite integral... :shock:

Eliz.

Well my TI-89 gives me an answer of 0. But I know this can't be right, because if we graph the function, we can see plenty of "area" below and above the sin wave between -2 and 2

Maybe there is another way besides finding an anti-derivative?

 

[o_O]

The reason that the area is 0 is because the region below the x-axis takes into the account of the negative number. For example, if you integrated sinx from -pi/2 to 0, you would get -1. Thus, if you tried integrating sinx from -pi/2 to pi/2, you would get 0 as well. To solve this "problem", whenever you are evaluating the integrals, you would have to split it up into regions below and above the axis (helps if you graph it).

As for the integral the integrator gives ... Naaassty.

Edit: Just for kicks :mrgreen:

\(\displaystyle \int\frac{sinx}{1+x^{6}}dx\)

\(\displaystyle =\)

[attachment=0:2tvgnf06]MSP.gif[/attachment:2tvgnf06]

 

[stapel]

Well my TI-89 gives me an answer of 0. But I know this can't be right, because if we graph the function, we can see plenty of "area" below and above the sin wave between -2 and 2

I'm not sure what you mean by "below and above the sin[sic] wave", since you're not integrating "sin(x)"...?

Are you familiar with the "area" concept of the integral? If so, are you familiar with the idea of "negative" area (being area below, and thus subtracted from, the x-axis)?

Maybe there is another way besides finding an anti-derivative?

Are you allowed to use the concept of area? If so, you could prove that 1 + x[sup:a3f8yyoc]6[/sup:a3f8yyoc] is an even function, symmetric about the y-axis (and thus on the interval -2 < x [/u]<[/u] 2). Then note (or prove) that sin(x) is an odd function, with sin(-x) = -sin(x) for any given value of x.

Use these facts to compare the positive area (to the right of the y-axis) with the negative area (to the left). What must be the result?

But if you have to actually do the integration, I have no suggestions. Sorry!

Eliz.

 

[pka]

Oh, come on you all.
\(\displaystyle f(x) = \frac{{\sin (x)}}{{1 + x^6 }}\) is an odd function.

Therefore, \(\displaystyle \left( {\forall a} \right)\,\int\limits_{ - a}^a {f(x)dx} = 0\,\)

 

[galactus]

Hey JW:

For what my inputs worth, when you see horrible integrals like this there is mostly an observation to make.

Think of that first before you start trying to brutally solve it.

Here's the graph. You can see it's odd. Therefore, evaluates to 0.

 

[jwpaine]

Hey JW:

For what my inputs worth, when you see horrible integrals like this there is mostly an observation to make.

Think of that first before you start trying to brutally solve it.

Here's the graph. You can see it's odd. Therefore, evaluates to 0.

Thanks galactus. I PMed you something un-related. Just FYI in case the notification system is down. You can delete this when you get it.

Cheers,
John