# Integration Involving Definite Integral

• Mar 17th 2014, 02:01 PM
nycmath
Integration Involving Definite Integral
I know members on this site expect work or effort to be posted with question. I included my work here. I want to know where I went wrong. See picture.

• Mar 17th 2014, 02:20 PM
Plato
Re: Integration Involving Definite Integral
Have a look at this.
• Mar 17th 2014, 02:35 PM
romsek
Re: Integration Involving Definite Integral
I get $\dfrac{1}{25-x^2}=\dfrac{1}{10} \left(\dfrac{1}{x+5}-\dfrac{1}{x-5} \right)$

which integrates to

$\dfrac{1}{10}\ln\left(\left|\dfrac{x+5}{x-5}\right|\right)$ and evaluating

$\dfrac{1}{10}\ln\left(\left|\dfrac{x+5}{x-5}\right|\right) \left|^4_0\right.=\\ \\ \dfrac{1}{10}\ln\left(\left|\dfrac{4+5}{4-5}\right|\right)-\dfrac{1}{10}\ln\left(\left|\dfrac{5}{-5}\right|\right)=\\ \\ \dfrac{1}{10}\ln(9)=\\ \\ \dfrac{1}{10}\ln(3^2)= \dfrac{1}{5}\ln(3)$

you should have been able to figure this out
• Mar 17th 2014, 02:35 PM
SlipEternal
Re: Integration Involving Definite Integral
$\displaystyle \int \dfrac{dx}{a^2-x^2} = \ln\left|\dfrac{a+x}{a-x} \right| + C$

You flipped the fraction on the RHS.

$\dfrac{1}{10}\ln\left(\dfrac{5+4}{5-4}\right) - \dfrac{1}{10}\ln\left(\dfrac{5+0}{5-0}\right) = \dfrac{1}{10}\ln 9$

(Since all values are positive, I am ignoring the absolute value symbols).

Next, using logarithm rules:

$\dfrac{1}{10}\ln 9 = \dfrac{1}{5}\cdot \dfrac{1}{2} \ln 9 = \dfrac{1}{5} \ln 9^{1/2} = \dfrac{1}{5}\ln 3$

Romsek beat me to it.
• Mar 17th 2014, 03:04 PM
nycmath
Re: Integration Involving Definite Integral
Thanks everyone.

However,where did 1/5 come from?

How does (1/10)*ln 9 become (1/5)(1/2)ln9, which then becomes (1/5)*ln(sqrt{9}) leading to the answer (1/5)*ln 3?

What led you to decide that 1/10 should be broken into two fractions (1/5)(1/2)?

The 1/2 infront of ln becomes the exponent of 9, which is the same as the square root of 9.
• Mar 17th 2014, 03:19 PM
nycmath
Re: Integration Involving Definite Integral
(1/5)*ln3, then my original answer is also correct, right? The 3 authors of the book decided to apply log rules to further simplify the answer, right?
• Mar 17th 2014, 03:22 PM
romsek
Re: Integration Involving Definite Integral
yes

though your derivation was a mess. It's a simple application of partial fractions. No substitutions were needed.
• Mar 17th 2014, 05:51 PM
SlipEternal
Re: Integration Involving Definite Integral
Quote:

Originally Posted by romsek
yes

though your derivation was a mess. It's a simple application of partial fractions. No substitutions were needed.

No, the OP's answer was wrong:

$\dfrac{1}{10}\ln\left|\dfrac{1}{9}\right| = \dfrac{1}{10}\ln 9^{-1} = -\dfrac{1}{10}\ln 3^2 = -\dfrac{1}{5}\ln 3 \neq \dfrac{1}{5} \ln 3$

nycmath: You used the wrong integral, as I pointed out in my post. That is why you got the wrong answer.

As for why $\dfrac{1}{10}\ln 9 = \dfrac{1}{5}\ln 3$, I broke up the fraction, but that is not needed. You can also do the following:

$\dfrac{1}{10}\ln 9 = \dfrac{1}{10}\ln 3^2 = \dfrac{2}{10}\ln 3 = \dfrac{1}{5}\ln 3$
• Mar 17th 2014, 06:08 PM
nycmath
Re: Integration Involving Definite Integral
Quote:

Originally Posted by SlipEternal
No, the OP's answer was wrong:

$\dfrac{1}{10}\ln\left|\dfrac{1}{9}\right| = \dfrac{1}{10}\ln 9^{-1} = -\dfrac{1}{10}\ln 3^2 = -\dfrac{1}{5}\ln 3 \neq \dfrac{1}{5} \ln 3$

nycmath: You used the wrong integral, as I pointed out in my post. That is why you got the wrong answer.

As for why $\dfrac{1}{10}\ln 9 = \dfrac{1}{5}\ln 3$, I broke up the fraction, but that is not needed. You can also do the following:

$\dfrac{1}{10}\ln 9 = \dfrac{1}{10}\ln 3^2 = \dfrac{2}{10}\ln 3 = \dfrac{1}{5}\ln 3$

I thank you for your guidance through this questions and others here. I know my integration skills have improved since joining the site.
• Mar 17th 2014, 06:11 PM
nycmath
Re: Integration Involving Definite Integral
Thank you everyone. I have learned so much here. I could not be where I am in my self-study of calculus without your help and guidance.