Definite integral question

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January 6th 2009, 04:10 PM
Pringles

Definite integral question

Quote:

Originally Posted by Chris L T521

For this to be a pdf, $\int_{-\infty}^{\infty}P\left(x\right)\,dx=1$

Thus, solve the equation $\int_{-\infty}^{\infty} k\left(1-x^3\right)\,dx=1\implies \int_0^1 k\left(1-x^3\right)\,dx=1$ for k.

Can you take it from here?

see this question below(integration)They are asking that the integral be evaluated.
is the procedure just like what we did(integration in the above question?)

$\int_{0}^{1}\frac{1}{x^{2}+3x+2}dx=\int_{0}^{1}\fr ac{1}{x+1}dx-\int_{0}^{1}\frac{1}{x+2}dx$
January 6th 2009, 06:56 PM
mr fantastic

Quote:

Originally Posted by Pringles

Quote:

Originally Posted by Chris L T521

For this to be a pdf, $\int_{-\infty}^{\infty}P\left(x\right)\,dx=1$

Thus, solve the equation $\int_{-\infty}^{\infty} k\left(1-x^3\right)\,dx=1\implies \int_0^1 k\left(1-x^3\right)\,dx=1$ for k.

Can you take it from here?

see this question below(integration)They are asking that the integral be evaluated.
is the procedure just like what we did(integration in the above question?)

$\int_{0}^{1}\frac{1}{x^{2}+3x+2}dx=\int_{0}^{1}\fr ac{1}{x+1}dx-\int_{0}^{1}\frac{1}{x+2}dx$

It has nothing to do with that.

You're expected to know and apply the standard form $\int \frac{1}{x + a} \, dx = \ln |x + a|$ (I have omitted the arbitrary constant).
January 6th 2009, 07:40 PM
Mush

Quote:

Originally Posted by mr fantastic

It has nothing to do with that.

You're expected to know and apply the standard form $\int \frac{1}{x + a} \, dx = \ln |x + a|$ (I have omitted the arbitrary constant).

No disrespect meant, but I am wondering why you don't just include the arbitrary constant in your posts about integration? Surely it's easier and more mathematically rigorous to write "+ C" rather than "(I have omitted the arbitrary constant)".
January 6th 2009, 09:40 PM
Mush

Quote:

Originally Posted by Pringles

Evaluate the integral

$\int_{0}^{1}\frac{1}{x^{2}+3x+2}dx=\int_{0}^{1}\fr ac{1}{x+1}dx-\int_{0}^{1}\frac{1}{x+2}dx$

I would appreciate it if someone would be kind enough to give me the full workings with some explanation.

Thankyou(Crying)

Do you know how expression was split into two?

$\displaystyle \int_{0}^{1}\frac{1}{x^{2}+3x+2}dx=\int_{0}^{1}\fr ac{1}{x+1}dx-\int_{0}^{1}\frac{1}{x+2}dx$

The general rule for such integrals is $\displaystyle \int \frac{1}{x+a}dx = \ln{|x+a|} + C$ . Hence:

$\displaystyle \int_{0}^{1}\frac{1}{x+1}dx-\int_{0}^{1}\frac{1}{x+2}dx = [\ln{|x+1|}]^1_0 -[\ln{|x+2|}]^1_0$

$\displaystyle = [\ln{|1+1|}-\ln{|0+1|}]-[\ln{|1+2|}-\ln{|0+2|}]$

$=\displaystyle \ln{|2|}-\ln{|1|}-\ln{|3|}+\ln{|2|}$

$=\displaystyle 2\ln{|2|}-\ln{|3|}$ since $\displaystyle \ln{|1|} = 0$

$=\displaystyle \ln{|2^2|}-\ln{|3|}$ since $\displaystyle a\ln{|b|} = \ln{|b^a|}$

$=\displaystyle \ln{|4|}-\ln{|3|}$

$=\displaystyle \ln{|\frac{4}{3}|}$ since $\displaystyle \ln{|a|}-\ln{|b|}=\ln{|\frac{a}{b}|}$
January 6th 2009, 11:09 PM
mr fantastic

Quote:

Originally Posted by Mush

No disrespect meant, but I am wondering why you don't just include the arbitrary constant in your posts about integration? Surely it's easier and more mathematically rigorous to write "+ C" rather than "(I have omitted the arbitrary constant)".

If the result is to be used by the OP in a definite integral then I omit the constant (and say so).

If the result is part of an indefinite integral that the OP is trying to solve then I include the constant.

My reasons for doing this are not too difficult to fathom I think ....