Answer to Question #349961 in Calculus for zalea

Question #349961

Evaluate the following integrals. (Show your solution)


1)2xx1) \intop 2x√x dxdx

2) \intop 2/x³2/√x³ dxdx


1
Expert's answer
2022-06-14T00:05:35-0400

We'll use the next basic integration formula:

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


So we have:

1)

2xxdx=2x1x12dx=2x32dx=\int 2x\sqrt{x}dx=\int2x^1x^{\frac{1}{2}}dx=\int2x^{\frac{3}{2}}dx=


=2x32+132+1+C=2x5252+C=45x52+C==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=


=45x2+12+C=45x2x12+C=45x2x+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) 2x3dx=2x32dx=2x32dx=\int\frac{2}{\sqrt{x^3}}dx=\int\frac{2}{x^{\frac{3}{2}}}dx=\int2x^{-\frac{3}{2}}dx=


=2x32+132+1+C=2x1212+C=4x12+C=4x+C.=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) 2xxdx=45x2x+C,\int 2x\sqrt{x}dx=\frac{4}{5}x^2\sqrt{x}+C,


2) 2x3dx=4x+C.\int\frac{2}{\sqrt{x^3}}dx=-\frac{4}{\sqrt{x}}+C.


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