Integrate sin(t²) from t = x to t = x³

[Frostbytez]

Homework Statement

Evaluate the integral (upper limit x^3 and lower limit x) sin (t^2) dt

Homework Equations

The Attempt at a Solution

I tried using the second fundamental theorem to do: 3x^2 sin (x^6) - sin (x^2). I'm not sure if this is right though.

 

[Dickfore]

Is it some kind of a double integral?

 

[Frostbytez]

what do you mean by double integral?

 

[Galgenstrick]

You have to use Fresnel integrals. Integrating the sin(t^2) by hand is not easy. What class is this for?

 

[Frostbytez]

It's AP Calculus AB.

 

[Galgenstrick]

That question shouldn't be there, are you sure you read the problem right? could it have been (sin(t))^2 instead?The answer to the problem posted is as follows:

sqrt(Pi/2)[(S(sqrt(2/Pi)*x^3)-S(sqrt(2/Pi)*x)], where S(x) is the Fresnel integral of sin(t^2)dt from 0 to x.

 

[Frostbytez]

I think they want the derivative not to evaluate the actual integral. Can I use the second fundamental theorem then?

 

[Galgenstrick]

Would you type in the instructions for the problem? I am not entirely sure what they are asking.

 

[Dick]

I think they want the derivative not to evaluate the actual integral. Can I use the second fundamental theorem then?

The question does ask for the derivative of that integral, yes? And yes, you can use a fundamental theorem. And yes, you did it correctly.

 

[brydustin]

Homework Statement

Evaluate the integral (upper limit x^3 and lower limit x) sin (t^2) dt

Homework Equations

The Attempt at a Solution

I tried using the second fundamental theorem to do: 3x^2 sin (x^6) - sin (x^2). I'm not sure if this is right though.

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Yeah, I think that you are just over-complicating this... what is the derivative of
y = (-1/(t^2))cos(t^2) ...

Okay, the derviative of that is the integrand of your problem.
evalulating y from the top and bottom (of the integral domain) yields:
(-1/x^6)cos(x^6) +(1/x^2)cos(x^2)

 

[Dick]

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Yeah, I think that you are just over-complicating this... what is the derivative of
y = (-1/(t^2))cos(t^2) ...

Okay, the derviative of that is the integrand of your problem.
evalulating y from the top and bottom (of the integral domain) yields:
(-1/x^6)cos(x^6) +(1/x^2)cos(x^2)

Wrong. The derivative of that is DEFINITELY not sin(t^2).