evaluate each integral?

Question

∫² ∫²4xydxdy ***(on the bottom of the first ∫ there’s a 0, the second ∫ there’s a y)***

Answer 1

To do the nner integral, you have to take the integral with respect to 'x' because one of your bounds is the variable 'y'.

So, you have int(4xydx) from (y,2). Taking the integral with respect to 'x' means you treat everything else but the x as constants. In this case, the 4 and y can be treated as constants. Using the power rule for anti-derivatives, you get:

(2x^2y) evaluated from (y,2). Putting in the limits of integration, you get:

2(2^2)y - 2(y^2)y ---------> simplifying, you get 8y-2y^3. Now you the following integral:

int([8y-2y^3]dy) evaluated from (0,2). Now this is a simple integral with one variable. Use the power rule again to solve:

4y^2 - (2/4)y^4 [evaluated from (0,2)]. Evaluating at the bounds, you get:

4(2^2) - 0.5(2^4) - [4(0)^2 - 0.5(0)^4] ---------> = 8

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Hope this helps

Answer 2

Since I am not sure about the limits of integration, I will explain how to do it.

Do the inside integral first. To do this, just forget about the first sign and the 'dy' Then you get:

∫² (4xy) dx

In this integral, the 4 and the y are just constants, and the x term is your only variable. The antiderivative is just 4*y*(x^2/2) = 2*y*x^2

You need to evaluate this from your lower limit (y) to your upper limit (is this 2?). If so, you get:

2*y*(2^2) - 2*y*(y^2) = 8*y - 2*y^3

Then you need to do the outer integral, which is just an integral of a simple polynomial.