The integral to evaluate is ∫(x to x^3) sin(t^2) dt, which requires the application of the second fundamental theorem of calculus. The correct approach involves differentiating the integral, yielding the expression 3x^2 sin(x^6) - sin(x^2). However, the solution also involves the Fresnel integral, represented as sqrt(π/2)[S(sqrt(2/π)x^3) - S(sqrt(2/π)x)], where S(x) denotes the Fresnel integral of sin(t^2) dt from 0 to x. The discussion clarifies that the problem is not about evaluating the integral directly but rather finding its derivative.
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Frostbytez said:I think they want the derivative not to evaluate the actual integral. Can I use the second fundamental theorem then?
Frostbytez said: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.
brydustin said:====
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)