a6 - improper integrals (1 point) calculate $int_{0}^{1}\frac{dx}{x^{0.2}}$. write f if the integral doesnt…

Explanation:

Step1: Use 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 the integral $\int\frac{1}{x^{0.2}}dx=\int x^{-0.2}dx$. Applying the power - rule, we get $\frac{x^{-0.2 + 1}}{-0.2+1}+C=\frac{x^{0.8}}{0.8}+C$.

Step2: Evaluate the definite integral

We want to find $\int_{0}^{1}\frac{1}{x^{0.2}}dx=\lim_{a\rightarrow0^{+}}\int_{a}^{1}x^{-0.2}dx$. First, evaluate $\int_{a}^{1}x^{-0.2}dx=\left[\frac{x^{0.8}}{0.8}\right]{a}^{1}=\frac{1^{0.8}}{0.8}-\frac{a^{0.8}}{0.8}=\frac{1}{0.8}-\frac{a^{0.8}}{0.8}$. Then, take the limit as $a\rightarrow0^{+}$: $\lim{a\rightarrow0^{+}}\left(\frac{1}{0.8}-\frac{a^{0.8}}{0.8}\right)=\frac{1}{0.8}-0=\frac{1}{0.8}=\frac{10}{8}=\frac{5}{4}$.

Answer:

$\frac{5}{4}$