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Evaluate the following iterated integrals. $$\int_{1}^{4} \int_{0}^{4} \sqrt{u v} d u d v$$

Short Answer

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Question: Evaluate the iterated integral $$\int_{1}^{4} \int_{0}^{4} \sqrt{u v} d u d v$$. Answer: $$\frac{496}{5}$$

Step by step solution

01

Integrate with respect to u

To evaluate the integral with respect to \(u\), we have to calculate: $$\int_{0}^{4} \sqrt{u v} d u$$ Recall that \(\sqrt{u v} = (u v)^{1/2}\). Therefore, we rewrite the integral as follows: $$\int_{0}^{4} (u v)^{1/2} d u$$ Now, integrate the integral with respect to \(u\): $$\int_{0}^{4} (u v)^{1/2} d u = \frac{2}{3}(u v)^{\frac{3}{2}}\Big|_0^4$$ Evaluate the antiderivative at the limits of integration: $$\frac{2}{3}(u v)^{\frac{3}{2}}\Big|_0^4 = \frac{2}{3}(4v)^{\frac{3}{2}} - \frac{2}{3}(0v)^{\frac{3}{2}} = \frac{2}{3}(4^{\frac{3}{2}}v^{\frac{3}{2}})$$
02

Integrate with respect to v

Now that we have the antiderivative with respect to \(u\), we can integrate with respect to \(v\). The new integral is as follows: $$\int_{1}^{4} \frac{2}{3}(4^{\frac{3}{2}}v^{\frac{3}{2}}) d v$$ Integrate the integral with respect to \(v\): $$\int_{1}^{4} \frac{2}{3}(4^{\frac{3}{2}}v^{\frac{3}{2}}) d v = \frac{2}{5}(4^{\frac{3}{2}}v^{\frac{5}{2}})\Big|_1^4$$ Evaluate the antiderivative at the limits of integration: $$\frac{2}{5}(4^{\frac{3}{2}}v^{\frac{5}{2}})\Big|_1^4 = \frac{2}{5}(4^{\frac{3}{2}}4^{\frac{5}{2}}) - \frac{2}{5}(4^{\frac{3}{2}}1^{\frac{5}{2}}) = \frac{2}{5}(4^4) - \frac{2}{5}(4^{\frac{3}{2}})$$
03

Simplify and Calculate

The result can be simplified as follows: $$\frac{2}{5}(4^4) - \frac{2}{5}(4^{\frac{3}{2}}) = \frac{2}{5}(4^4 - 4^{\frac{3}{2}}) = \frac{2}{5}(256-8) = \frac{2}{5}(248)$$ The final answer is: $$\frac{496}{5}$$

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