(int_{0}^{1}\frac{1}{x^{6}}+sqrt4{x^{3}}dx)
(int_{0}^{1}\frac{1}{x^{6}}+sqrt4{x^{3}}dx)
Answer
Explanation:
Step1: Rewrite the integrand
Rewrite $\frac{1}{x^{6}}+\sqrt[4]{x^{3}}$ as $x^{-6}+x^{\frac{3}{4}}$.
Step2: Integrate term - by - term
The integral of $x^{-6}$ is $\frac{x^{-6 + 1}}{-6+1}=-\frac{1}{5x^{5}}$ and the integral of $x^{\frac{3}{4}}$ is $\frac{x^{\frac{3}{4}+1}}{\frac{3}{4}+1}=\frac{x^{\frac{7}{4}}}{\frac{7}{4}}=\frac{4}{7}x^{\frac{7}{4}}$. So, $\int(x^{-6}+x^{\frac{3}{4}})dx=-\frac{1}{5x^{5}}+\frac{4}{7}x^{\frac{7}{4}}+C$.
Step3: Evaluate the definite integral
$\left(-\frac{1}{5x^{5}}+\frac{4}{7}x^{\frac{7}{4}}\right)\big|{0}^{1}=\left(-\frac{1}{5\times1^{5}}+\frac{4}{7}\times1^{\frac{7}{4}}\right)-\lim{x\rightarrow0^{+}}\left(-\frac{1}{5x^{5}}+\frac{4}{7}x^{\frac{7}{4}}\right)$. The second term $\lim_{x\rightarrow0^{+}}\left(-\frac{1}{5x^{5}}+\frac{4}{7}x^{\frac{7}{4}}\right)=-\infty$. So the integral $\int_{0}^{1}\left(\frac{1}{x^{6}}+\sqrt[4]{x^{3}}\right)dx$ diverges.
Answer:
The integral diverges.