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Find the integral $\int(2x^2-3sin\: x+5\sqrt x)\;dx$

$\begin{array}{1 1} \frac{2x^3}{3}+3\cos x+\frac{10}{3}x^\frac{3}{2}+c \\ \frac{2x^3}{3}-3\cos x+\frac{10}{3}x^\frac{3}{2}+c \\ \frac{2x^3}{3}-3\sin x+\large\frac{10}{3}x^\frac{3}{2}+c \\ \frac{2x^3}{3}-3\cos x+\large\frac{10}{3}x^\frac{5}{2}+c. \end{array}$

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1 Answer

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  • $(i)\;\int x^ndx=\frac{x^{n+1}}{n+1}+c$.
  • $(ii)\;\int \sin xdx=-\cos x+c.$
$\int(2x^2-3\sin x+5\sqrt x)dx.$
 
$\;\;\;=\int 2x^2dx-3\int \sin xdx+5\int \sqrt xdx.$
 
$\;\;\;=2\big(\frac{x^3}{3}\big)-3(-\cos x)+5\bigg(\frac{x^\frac{1}{2}+1}{\frac{1}{2}+1}\bigg)+c$.
 
$\;\;\;=\frac{2x^3}{3}+3\cos x+5\times\frac{2}{3}x^\frac{3}{2}+c.$
 
$\;\;\;=\frac{2x^3}{3}+3\cos x+\frac{10}{3}x^\frac{3}{2}+c.$
 
Hence $\int(2x^2-3\sin x+5\sqrt x)dx=\frac{2x^3}{3}+3\cos x+\frac{10}{3}x^\frac{3}{2}+c.$

 

answered Jan 27, 2013 by sreemathi.v
 
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