# Integration: General Power Formula

• May 18th 2013, 12:37 PM
togo
Integration: General Power Formula
Could you explain where I went wrong with this one?

28-1-16

(int) 0.8((3 + 2 ln(u))^3) du/u

Attempt:
u = 3 + 2 ln u
du = 1 / u

(.8 [3 + 2 ln u]^3) / 4

the ti-89 comes up with another answer when this integration is performed, which is long and bizarre.
• May 18th 2013, 03:32 PM
x3bnm
Re: Integration: General Power Formula
\begin{align*}\int \frac{0.8(3 + 2 \ln(u))^3}{u} du =& \int 0.8\left(\frac{27}{u} + \frac{54 \ln(u)}{u} + \frac{36 \ln^{2}(u)}{u} + \frac{8\ln^{3}(u)}{u} \right) \,\,du.....\text{[by expanding cube]}\\=& 0.8\left(\int \frac{27}{u}\,\,du +\int \frac{54 \ln(u)}{u}\,\,du + \int \frac{36 \ln^{2}(u)}{u}\,\,du + \int \frac{8\ln^{3}(u)}{u}\,\,du \right).....(1) \end{align*}

Let $m = \ln(u)\,\, \therefore dm = \frac{1}{u}\,\,du$

So (1) becomes:

\begin{align*}0.8\left(\int \frac{27}{u}\,\,du +\int \frac{54 \ln(u)}{u}\,\,du + \int \frac{36 \ln^{2}(u)}{u}\,\,du + \int \frac{8\ln^{3}(u)}{u}\,\,du \right) =& 0.8\left( 27\ln|u| + \int 54m\,\, dm + 12\ln^{3}|u|+ 2\ln^{4}|u| \right) + C\\ =& 0.8\left( 27\ln|u| + 27m^2 + 12\ln^{3}|u|+ 2\ln^{4}|u| \right) + C\\=& 0.8\left( 27\ln|u| + 27\ln^{2}|u| + 12\ln^{3}|u|+ 2\ln^{4}|u| \right) + C....\text{[because }m = \ln(u)\text{]}\\=& 21.6\ln|u| + 21.6\ln^{2}|u| + 9.6 \ln^{3}|u| + 1.6 \ln^{4}|u| + C........\text{[Answer]}\end{align*}

You can check the answer here to make sure.
int (0.8(3 + 2 ln(u))^3)/u - Wolfram|Alpha
• May 19th 2013, 10:27 AM
x3bnm
Re: Integration: General Power Formula
Quote:

Originally Posted by togo
Could you explain where I went wrong with this one?

28-1-16

(int) 0.8((3 + 2 ln(u))^3) du/u

Attempt:
u = 3 + 2 ln u
du = 1 / u

(.8 [3 + 2 ln u]^3) / 4

the ti-89 comes up with another answer when this integration is performed, which is long and bizarre.

You are almost on the right track.

$\int \frac{0.8(3 + 2 \ln(u))^3}{u} du$

Let $m = 3 + 2 \ln(u) \,\, \therefore dm = \frac{2}{u}\,\,du$

So plugging in $m$ and $dm$ we get:

\begin{align*}\frac{0.8(3 + 2 \ln(u))^3}{u} du =& \int \frac{0.8m^3}{2}\,\,dm\\=& \frac{0.8m^4}{8} + C\\=& \frac{0.8(3 + 2 \ln(u))^4}{8} + C....\text{[by plugging back the value of } m\text{]}\end{align*}

So, $\int \frac{0.8(3 + 2 \ln(u))^3}{u} du = \frac{0.8(3 + 2 \ln(u))^4}{8} + C....\text{[Answer]}$

You can check the answer by taking the derivative of $\frac{0.8(3 + 2 \ln(u))^4}{8} + C$ which
is equal to $\frac{0.8(3 + 2 \ln(u))^3}{u}$

derivative 0.8((3 + 2 ln(u))^4}/8) + C - Wolfram|Alpha

• May 19th 2013, 01:53 PM
togo
Re: Integration: General Power Formula
Hi how did 0.8m^3 / 2 turn into 0.8m^4 / 8?

how did the division by 2 turn into a division by 8?
• May 19th 2013, 05:50 PM
Prove It
Re: Integration: General Power Formula
Surely you know that $\displaystyle \int{ x^n \,dx} = \frac{x^{n+1}}{n+1} + C$...
• May 19th 2013, 08:54 PM
x3bnm
Re: Integration: General Power Formula
Quote:

Originally Posted by togo
Hi how did 0.8m^3 / 2 turn into 0.8m^4 / 8?

how did the division by 2 turn into a division by 8?

Integration is the opposite of differentiation hence the other name anti-derivative.

So by integrating what you'll get is when differentiated will be what you're integrating.

So:

$\int \frac{0.8m^3}{2}\,\,dm$ you're integrating and you'll get $\frac{0.8m^4}{8} + C$

When you differentiate $\frac{0.8m^4}{8} + C$ you'll get what you're integrating. So when you integrate you've to rearrange variables and constants in such a way that when you differentiate the rearrangement you'll end up with what you're integrating.

So if we differentiate $\frac{0.8m^4}{8} + C$ we get:
Let $f(x) = \frac{0.8m^4}{8} + C\,\, \therefore \frac{df}{dx} = \frac{0.8* 4* m^{(4-1)}}{8} + 0= \frac{0.8m^3}{2}$ which is what you're integrating.

And that's why $\int \frac{0.8m^3}{2}\,\,dm = \frac{0.8m^4}{8} + C$
• May 20th 2013, 12:06 PM
togo
Re: Integration: General Power Formula
Quote:

Originally Posted by Prove It
Surely you know that $\displaystyle \int{ x^n \,dx} = \frac{x^{n+1}}{n+1} + C$...

what confuses me is according to the general power rule, you add 1 to the exponent found on the integral, which is 3, so ends up being 4 on the bottom. While I understand it differentiates properly (post #6), seems like something other than general power rule was used to come up with a denominator of 8.
• May 20th 2013, 12:23 PM
x3bnm
Re: Integration: General Power Formula
Quote:

Originally Posted by togo
what confuses me is according to the general power rule, you add 1 to the exponent found on the integral, which is 3, so ends up being 4 on the bottom. While I understand it differentiates properly (post #6), seems like something other than general power rule was used to come up with a denominator of 8.

You first separate the constants from variable. So it will be:

\begin{align*}\int \frac{0.8m^3}{2}\,\,dm =& \frac{0.8}{2} \int {m^3}\,\,dm \\=& \frac{0.8}{2}\left( \frac{m^{(3+1)}}{3+1}\right) + C\\=& \frac{0.8}{2}\left(\frac{m^4}{4}\right)+C\\=& \frac{0.8m^4}{8}+ C\end{align*}