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How do you solve:

$$\int_{7}^{10} \int_{11}^{14} x^{2} 4y ~ dx~dy$$

Not sure where to really start with this, so this is an example for the actual problem. Thanks!

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As the integration limits are constant and the integrand is separable, we may simply write the integral as

$$\int_7^{10} 4 y \, dy \int_{11}^{14} x^2 \, dx = 4 \frac{1}{2} (10^2-7^2) \frac{1}{3} (14^3-11^3) = 102 \cdot 471 = 48402$$

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we have $$\int_{7}^{10}\int_{11}^{14}x^2 4y\hspace 1mmdxdy=\int_{7}^{10}(\int_{11}^{14}x^2 4y\hspace 1mmdx)dy$$$$=\int_{7}^{10}4y(\int_{11}^{14}x^2 \hspace 1mmdx)dy$$$$=\int_{7}^{10}4y(\frac{14^3}{3}-\frac{11^3}{3})\hspace 1mmdy$$$$=\int_{7}^{10}4y(471)\hspace 1mmdy$$$$=\int_{7}^{10}4y(\frac{14^3}{3}-\frac{11^3}{3})\hspace 1mmdxdy$$$$=\int_{7}^{10}1884ydy$$$$=48042$$

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up vote -1 down vote accepted

Solving the first integral, we get 1/3x^3 + 4yx which gives us 471 + 12y. Next we can solve the next integral giving us 1719. Add this to both, giving us 1719. See here:

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This is the same problem and I didn't even realize it. Thank you! – mario1433 Feb 11 '13 at 2:42
Forgive me, but isn't the integrand as the OP wrote it $x^2 \cdot 4 y$, not $x^2 + 4 y$? – Ron Gordon Feb 11 '13 at 2:50

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