$\\int_0^{\\frac{7}{3}} \\sqrt{9 + 3w}dw$

Explanation:

Step1: Use substitution

Let (u = 9+3w), then (du=3dw), (dw=\frac{1}{3}du). When (w = 0), (u=9); when (w=\frac{7}{3}), (u=9 + 3\times\frac{7}{3}=16). The integral becomes (\frac{1}{3}\int_{9}^{16}\sqrt{u}du).

Step2: Integrate (\sqrt{u})

Since (\int u^{n}du=\frac{u^{n + 1}}{n+1}+C) ((n\neq - 1)), for (n=\frac{1}{2}), (\int\sqrt{u}du=\int u^{\frac{1}{2}}du=\frac{2}{3}u^{\frac{3}{2}}+C). So (\frac{1}{3}\int_{9}^{16}\sqrt{u}du=\frac{1}{3}\times\frac{2}{3}\left[u^{\frac{3}{2}}\right]_{9}^{16}).

Step3: Evaluate the definite - integral

(\frac{2}{9}\left(u^{\frac{3}{2}}\right)\big|_{9}^{16}=\frac{2}{9}(16^{\frac{3}{2}}-9^{\frac{3}{2}})). We know that (a^{\frac{m}{n}}=\sqrt[n]{a^{m}}), so (16^{\frac{3}{2}}=\left(\sqrt{16}\right)^{3}=4^{3} = 64), (9^{\frac{3}{2}}=\left(\sqrt{9}\right)^{3}=3^{3}=27). Then (\frac{2}{9}(64 - 27)=\frac{2}{9}\times37=\frac{74}{9}).

Answer:

(\frac{74}{9})