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Calculate the double integral.

3xy2
x2 + 1dA, R = {(x, y) | 0 ≤ x ≤ 1, −2 ≤ y ≤ 2}
iintegral.gif
R

Sagot :

the double integral.3xy2 x2 + 1dA, R = {(x, y) | 0 ≤ x ≤ 1, −2 ≤ y ≤ 2}

∫∫3xy2 dx dy = 16

We start by calculating the limits of integration. The given region, R, is bounded by 0 ≤ x ≤ 1 and −2 ≤ y ≤ 2. This means that the double integral is:

∫∫3xy2 dx dy = ∫0→1∫-2→2 3xy2 dx dy

Next, we will calculate the integral with respect to x. We can easily do this by integrating 3xy2 with respect to x. This gives us the following:

∫0→1 3xy2 dx = x3y2 + c

Now, we can substitute this expression into the double integral, giving us:

∫∫3xy2 dx dy = ∫0→1 (x3y2 + c) dy

Finally, we can integrate this expression with respect to y, giving us the following:

∫∫3xy2 dx dy = x3y3 + cy + d

Now, we can substitute the limits of integration into this expression, giving us the following:

∫∫3xy2 dx dy = 1(32) + c(-2) + d = 16

therefore, the double integral is equal to 16.

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