question 15 · 1 point evaluate the definite integral given below. $int_{0}^{3}(-2x^{\frac{1}{3}}-2x^{\frac{2}…

Explanation:

Step1: Apply power - rule for integration

The power - rule for integration is $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). For $\int(-2x^{\frac{1}{3}}-2x^{\frac{2}{3}})dx=-2\int x^{\frac{1}{3}}dx-2\int x^{\frac{2}{3}}dx$. $-2\times\frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}-2\times\frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1}=-2\times\frac{x^{\frac{4}{3}}}{\frac{4}{3}}-2\times\frac{x^{\frac{5}{3}}}{\frac{5}{3}}=-\frac{3}{2}x^{\frac{4}{3}}-\frac{6}{5}x^{\frac{5}{3}}+C$.

Step2: Evaluate the definite integral

$\left[-\frac{3}{2}x^{\frac{4}{3}}-\frac{6}{5}x^{\frac{5}{3}}\right]_0^3$. First, substitute $x = 3$: $-\frac{3}{2}(3)^{\frac{4}{3}}-\frac{6}{5}(3)^{\frac{5}{3}}=-\frac{3}{2}\times3^{\frac{4}{3}}-\frac{6}{5}\times3^{\frac{5}{3}}$. $=-\frac{3^{1+\frac{4}{3}}}{2}-\frac{6\times3^{\frac{5}{3}}}{5}=-\frac{3^{\frac{7}{3}}}{2}-\frac{6\times3^{\frac{5}{3}}}{5}$. $=-\frac{3^{2}\times3^{\frac{1}{3}}}{2}-\frac{6\times3\times3^{\frac{2}{3}}}{5}=-\frac{9\sqrt[3]{3}}{2}-\frac{18\sqrt[3]{9}}{5}$. Then substitute $x = 0$: $-\frac{3}{2}(0)^{\frac{4}{3}}-\frac{6}{5}(0)^{\frac{5}{3}} = 0$. The result of the definite integral is $-\frac{9\sqrt[3]{3}}{2}-\frac{18\sqrt[3]{9}}{5}$.

Answer:

$-\frac{9\sqrt[3]{3}}{2}-\frac{18\sqrt[3]{9}}{5}$