INSTRUCTOR: Part a. Use the properties of the double integral and Fubini's theorem to evaluate the integral from 0 to 1 with respect to x of the interval from negative 1 to 3 with respect to y of the function 3 minus x plus 4y.

First we'll take the integral with respect to y of 3 minus x plus 4y, which would be 3y minus xy plus 2y squared. And we will evaluate that from negative 1 to 3, and those are y values. So that this becomes integral from 0 to 1 with respect to x of that inner function.

Now, when we evaluate this from negative 1 to 3, this becomes 28 minus 4x. So our integral is from 0 to 1 with respect to x of our function 28 minus 4x. Taking the integral with respect to x. This is 28 x minus 2x squared evaluated from 0 to 1, which equals 26.

Part b. Show that 0 is less than or equal to the double integral over the region R with respect to A of the function sine parentheses pi x close parentheses, multiplied by cosine open parentheses pi y close parentheses. That integral is less than or equal to 1 over 32, where R is equal to the integral from 0 to 1/4 by the interval 1/4 to 1/2.

Now, looking at the x values 0 to 1/4 and the y values 1/4 to 1/2, we need to notice that sine of parentheses pi x close parentheses is bounded between 0 and square root of 2 over 2.

Similarly, cosine parentheses pi y close parentheses is bounded by 0 and the square root of 2 over 2. Which means the product of these two, which is actually the integrand, is between-- and I'll just read off my inequality. 0 is less than or equal to sine parentheses pi close parentheses multiplied by cosine open parentheses pi y close parentheses is less than or equal to 1/2.

Now also, the area of R, A of R, is equal to 1/4 multiplied by 1/4, which is 1/16. So what we will consider is the double integral for our lower bound from 0 to 1/4 with respect to x of the integral from 1/4 to 1/2 with respect to y of the function 0. That was our lower bound.

And that double integral will be equal to 0, giving us the lower bound of that interview we were trying to find. To find the upper bound, we want the integral from 0 to 1/4 with respect to x of the integral from 1/4 to 1/2 with respect to y of the function 1/16.

Sorry. Of 1/2. The function 1/2 because that was our upper bound of our inequality. Now this will be equal to integral from 0 to 1/4 of 1/2 y evaluated from 1/4 to 1/2 and all of that with respect to x, which is equal to 1/8.

All right. So that will be equal to the integral from 0 to 1/4 with respect to x of 1/8, which will be equal to 1 over 32 giving us the upper bound that we wanted.