Heidi T. answered • 10/27/19

Experienced tutor/teacher/scientist

To solve this, let z = f(x,y). This makes it clear that you are integrating with respect to x and y. The order of integration is determined by the relationships between x and y. In this case, we can write y = f(x), so would want to do the integration with respect to y first. If it were done in the reverse, you would have to consider the fact that y is a function of x, so would be much more difficult.

You are told that the function y = x^{2} is the lower limit and y = 0 is the upper limit (one is below, the other above). The values of x are between 0 and 1; so these are the lower and upper limits of x, respectively.

This leads to the integral:

∫_{0}^{1} ∫^{0}_{x^2} (2x siny) dy dx

since x does not depend on y, can separate it out:

∫_{0}^{1} 2x [∫^{0}_{x^2} ( sin (y}) dy] dx

The integral of ∫ sin(u) du = - cos{u) + C

so ∫^{0}_{x^2} ( sin (y}) dy = ( - cos(y)|_{x^2}^{0} = - cos (0) - (- cos(x^{2})) = cos(x^{2}) - 1

You now have an integral in x only: ∫_{0}^{1} 2x [ cos(x^{2}) - 1 ] dx, which can be separated into: ∫_{0}^{1} cos(x^{2}) 2x dx - ∫_{0}^{1} 2x dx,

The second part of this is a simple integral you should be able to solve easily. For the first part, apply the identity:

∫ cos(u) du = sin{u) + C, where u = x^{2} and du = 2x the limits are unchanged because u = 1^{2} = 1 when x = 1 and u = 0^{2} = 0 when x = 0

Once you have solved both parts, combine them and that should give you the answer.