The problem involves evaluating the integral of sin(t^2) with variable limits of integration, specifically from x to x^3. The context is within the subject area of calculus, particularly focusing on the application of the fundamental theorem of calculus.
The discussion is ongoing, with various interpretations being explored regarding the nature of the problem. Some participants suggest using the fundamental theorem of calculus, while others question the original problem's wording and intent.
There is uncertainty about the exact requirements of the problem, with some participants questioning whether the integral should be evaluated directly or if the derivative is being sought. Additionally, there are concerns about the complexity of integrating sin(t^2) by hand.
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)