1. Double Integral question

Hey everyone, I'm having some trouble in my Multivariable Calculus class. I hope someone can help! I'm sorry that I don't know how to properly make the integral look pretty! Thanks in advance.

$\displaystyle Integral(Integral(sin(x^2))dx)dy$

the limits are y < x < 3, 0 < y < 3.

Wolfram gives me an answer of 0.955565, but I'd love to know how it came up with this.

2. There isn't an elementary anti-derivative of $\displaystyle \sin(x^2)$ as far as I know. Wolfram may have used a numerical integration technique, such as the Trapezoidal Rule, to approximate the integral.

3. Originally Posted by NOX Andrew
There isn't an elementary anti-derivative of $\displaystyle \sin(x^2)$ as far as I know. Wolfram may have used a numerical integration technique, such as the Trapezoidal Rule, to approximate the integral.
True, but this is a double integral and we can change the bounds to find a possible solution.

$\displaystyle \int_0^3 \int_y^3 sin x^2 dxdy = \int_0^3 \int_0^x sin x^2 dydx = \int_0^3 sin x^2 dx \int_0^x dy = \int_0^3 xsin x^2 dx$

And we can now solve that without any trouble!

4. Originally Posted by AllanCuz
True, but this is a double integral and we can change the bounds to find a possible solution.

$\displaystyle \int_0^3 \int_y^3 sin x^2 dxdy = \int_0^3 \int_0^x sin x^2 dydx = \int_0^3 sin x^2 dx \int_0^x dy = \int_0^3 xsin x^2 dx$

And we can now solve that now without any trouble!
Right! Fubini's Theorem! Thanks!