integration

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January 21st 2009, 10:46 PM
sss

integration

1.A particle that moves along a straight line has velocity
http://webwork2.math.nau.edu/webwork...1cc5bac1b1.png
meters per second after t seconds. How many meters will it travel during the first t seconds?

I used integration by parts but i didnt get the write answer.

2.

http://webwork2.math.nau.edu/webwork...07e7018e31.png
January 21st 2009, 11:27 PM
o_O

(1) You're going to need to use integration by parts twice.

First round requires: $u = t^2$ and $dv = e^{-2t} dt$

You will get: $\int t^2 e^{-2t} = -\frac{1}{2}t^2e^{-2t} + 2 {\color{red}\int te^{-2t}dt}$

Use integration by parts again on the red: $u = t$ and $dv = e^{-2t}dt$

Can you carry on from here?

________________________

(2) Again, you'll need integration by parts: $u = \arctan (3x)$ and $dv = xdx$

You'll get: $\int x\arctan (3x) dx = \frac{x^2}{2}\arctan (3x) - \frac{3}{2} {\color{blue}\int \frac{x^2}{1+9x^2} dx}$

The blue integral should be easier to deal with now. You'll need long division or you can notice that: $\frac{x^2}{1+9x^2} \cdot \frac{9}{9} = \frac{9x^2 \ {\color{red}+ 1 - 1}}{9 (1+9x^2)} = \frac{1}{9} - \frac{1}{9} \left( \frac{1}{1+9x^2}\right)$
January 21st 2009, 11:44 PM
DeMath

Quote:

Originally Posted by sss

I used integration by parts but i didnt get the write answer.
2.

http://webwork2.math.nau.edu/webwork...07e7018e31.png

$\int {x\arctan \left( {3x} \right)} dx = \frac{{{x^2}}}{2}\arctan \left( {3x} \right) - \frac{1}{2}\int {\frac{{3{x^2}}}{{{{\left( {3x} \right)}^2} + 1}}dx} =$

$= \frac{{{x^2}}}{2}\arctan \left( {3x} \right) - \frac{1}{6}\int {\frac{{{{\left( {3x} \right)}^2} + 1 - 1}}{{{{\left( {3x} \right)}^2} + 1}}dx} =$

$= \frac{{{x^2}}}{2}\arctan \left( {3x} \right) - \frac{x}{6} + \frac{1}{6}\int {\frac{{dx}}{{{{\left( {3x} \right)}^2} + 1}}} =$

$= \frac{{{x^2}}}{2}\arctan \left( {3x} \right) - \frac{x}{6} + \frac{1}{{18}}\int {\frac{{d\left( {3x} \right)}}{{{{\left( {3x} \right)}^2} + 1}}} =$

$= \frac{{{x^2}}}{2}\arctan \left( {3x} \right) - \frac{x}{6} + \frac{1}{{18}}\arctan \left( {3x} \right) + C =$

$= \frac{1}{{18}}\left( {9{x^2} + 1} \right)\arctan \left( {3x} \right) - \frac{x}{6} + C.$