Answer to Question #349961 in Calculus for zalea

Question

Evaluate the following integrals. (Show your solution)

$1) \intop 2x√x$ $dx$

2) $\intop$ $2/√x³$ $dx$

Answer 1

We'll use the next basic integration formula:

$\int x^ndx=\frac{x^{n+1}}{n+1}+C.$

So we have:

1)

$\int 2x\sqrt{x}dx=\int2x^1x^{\frac{1}{2}}dx=\int2x^{\frac{3}{2}}dx=$

$=2\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+C=2\frac{x^{\frac{5}{2}}}{\frac{5}{2}}+C=\frac{4}{5}x^{{\frac{5}{2}}}+C=$

$=\frac{4}{5}x^{2+\frac{1}{2}}+C=\frac{4}{5}x^2x^{\frac{1}{2}}+C=\frac{4}{5}x^2\sqrt{x}+C.$

2) $\int\frac{2}{\sqrt{x^3}}dx=\int\frac{2}{x^{\frac{3}{2}}}dx=\int2x^{-\frac{3}{2}}dx=$

$=2\frac{x^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}+C=2\frac{x^{-\frac{1}{2}}}{-\frac{1}{2}}+C=-\frac{4}{x^{\frac{1}{2}}}+C=-\frac{4}{\sqrt{x}}+C.$

Answer: 1) $\int 2x\sqrt{x}dx=\frac{4}{5}x^2\sqrt{x}+C,$

2) $\int\frac{2}{\sqrt{x^3}}dx=-\frac{4}{\sqrt{x}}+C.$