Question

determine the following indefinite integral. check your work by differentiation.
intsqrt5{r^{3}}dr
intsqrt5{r^{3}}dr=square

Explanation:

Step1: Rewrite the radical as a power

We know that $\sqrt[5]{r^{3}}=r^{\frac{3}{5}}$. So the integral becomes $\int r^{\frac{3}{5}}dr$.

Step2: Apply the power - rule for integration

The power - rule for integration is $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$). Here $n=\frac{3}{5}$, so $\int r^{\frac{3}{5}}dr=\frac{r^{\frac{3}{5}+1}}{\frac{3}{5}+1}+C$.

Step3: Simplify the exponent and the denominator

$\frac{3}{5}+1=\frac{3 + 5}{5}=\frac{8}{5}$. Then $\frac{r^{\frac{3}{5}+1}}{\frac{3}{5}+1}+C=\frac{r^{\frac{8}{5}}}{\frac{8}{5}}+C=\frac{5}{8}r^{\frac{8}{5}}+C$.

Step4: Check by differentiation

Differentiate $\frac{5}{8}r^{\frac{8}{5}}+C$ using the power - rule for differentiation $\frac{d}{dr}(x^{n})=nx^{n - 1}$. We have $\frac{d}{dr}(\frac{5}{8}r^{\frac{8}{5}}+C)=\frac{5}{8}\times\frac{8}{5}r^{\frac{8}{5}-1}=r^{\frac{3}{5}}=\sqrt[5]{r^{3}}$.

Answer:

$\frac{5}{8}r^{\frac{8}{5}}+C$