91Ó°ÊÓ

First, we need to integrate the function with respect to u. The integral looks like this: $$\int_{1}^{4} \left( \int_{0}^{4} \sqrt{u v} d u \right) d v$$ We need to evaluate the inner integral first: $$\int_{0}^{4} \sqrt{u v} d u$$ To make this easier, let's rewrite the function as: $$\int_{0}^{4} u^{\frac{1}{2}} v^{\frac{1}{2}} d u$$ Now integrate with respect to u using the power rule: $$\int u^{\frac{1}{2}} v^{\frac{1}{2}} d u = \frac{2}{3} u^{\frac{3}{2}} v^{\frac{1}{2}} + C(v)$$ Now, we will evaluate the integral at the given bounds (0 and 4) and subtract: $$\left[\frac{2}{3} u^{\frac{3}{2}} v^{\frac{1}{2}}\right]_{0}^{4} = \frac{2}{3} \cdot 4^\frac{3}{2} v^{\frac{1}{2}} - \frac{2}{3} \cdot 0^\frac{3}{2} v^{\frac{1}{2}}$$ $$= \frac{8}{3}v^\frac{1}{2}$$

Now that we have integrated with respect to u, we will now integrate with respect to v: $$\int_{1}^{4} \frac{8}{3}v^\frac{1}{2} d v$$ Using power rule again, we get: $$\int v^{\frac{1}{2}} d v = \frac{2}{3}v^{\frac{3}{2}} + C$$ Now multiply by the constant, so we have: $$\frac{8}{3} \int v^{\frac{1}{2}} d v = \frac{16}{9}v^{\frac{3}{2}} + C$$ Now we need to evaluate the integral at the given bounds (1 and 4) and subtract: $$\left[\frac{16}{9}v^{\frac{3}{2}}\right]_{1}^{4} = \frac{16}{9} \cdot 4^\frac{3}{2} - \frac{16}{9} \cdot 1^\frac{3}{2}$$ $$= \frac{64}{3} - \frac{16}{9} = \frac{176}{9}$$ So, the value of the iterated integral is: $$\int_{1}^{4} \int_{0}^{4} \sqrt{u v} d u d v = \frac{176}{9}$$