# Definite Integration

• Apr 28th 2012, 12:16 PM
Bashyboy
Definite Integration
The problem is $\int_1^e\frac{(\1+lnx)^2}{x}dx$

So, I figured that $u\equiv\1+lnx$,

then $\frac{du}{dx} = \frac{1}{x}\rightarrow{dx} = {x}{du}$

coming upon this step I am not entirely confident with what I am suppose to be substituting.
• Apr 28th 2012, 12:31 PM
Plato
Re: Definite Integration
Quote:

Originally Posted by Bashyboy
The problem is $\int_1^e\frac{(\1+lnx)^2}{x}dx$
So, I figured that $u\equiv\1+lnx$,
then $\frac{du}{dx} = \frac{1}{x}\rightarrow{dx} = {x}{du}$
coming upon this step I am not entirely confident with what I am suppose to be substituting.

$\mathop {\left. {\frac{{{{\left( {1 + \ln (x)} \right)}^3}}}{3}} \right|}\nolimits_{x = 1}^{x = e}$
• Apr 28th 2012, 12:37 PM
MathoMan
Re: Definite Integration
Setting $u=1+\ln x$ then differential of the left side of the equation should equal the differential of the right side so $du=d(1+\ln x)=(1+\ln x)'dx=\frac{dx}{x}$ and you should do the following:

$\int\limits_1^c\frac{(1+\ln x)^2}{x}dx=\int\limits_1^c (1+\ln x)^2\frac{dx}{x}=\left\vert\begin{array}{l}1+\ln x=u \\ \frac{dx}{x}=du\\ 1\to 1+\ln 1=1\\ c\to 1+\ln c \end{array}\right\vert=\int\limits_1^{1+\ln c}u^2 du$
• Apr 28th 2012, 12:41 PM
Bashyboy
Re: Definite Integration
Okay, so I was on the correct path. I ended up with $\int_1^e\frac{u^2}{x}{x}{du}$,
but wasn't sure if it was it was correct; now, obviously, it is correct, but I can't seem to figure out why it is correct. I am just trying to justify every step, but it just seems unusual that the x's cancel out and stuff like that.
• Apr 28th 2012, 12:49 PM
Bashyboy
Re: Definite Integration
Mathoman, I am not exactly sure what is meant by "then differential of the left side of the equation should equal the differential of the right side."
• Apr 28th 2012, 12:57 PM
Plato
Re: Definite Integration
Quote:

Originally Posted by Bashyboy
Mathoman, I am not exactly sure what is meant by "then differential of the left side of the equation should equal the differential of the right side."

Try to get away from using u-substitution. Learn the forms differentiation takes on.
• Apr 28th 2012, 01:03 PM
Bashyboy
Re: Definite Integration
• Apr 28th 2012, 01:06 PM
Plato
Re: Definite Integration
Quote:

Originally Posted by Bashyboy

What is the derivative of $y=\frac{(1+\ln(x))^3}{3}~?$

Learn the basic derivative forms.
• Apr 28th 2012, 01:18 PM
BrownianMan
Re: Definite Integration
Quote:

Originally Posted by Bashyboy
Mathoman, I am not exactly sure what is meant by "then differential of the left side of the equation should equal the differential of the right side."

He simply means that given u = 1 + ln x, the differential of u must be equal to the differential of 1 + ln x. The differential u is du, and the differential of 1 + ln x is dx/x. So when you do the substitution, you end up with the integral of u^2 from 1 to 1+ln e.
• Apr 28th 2012, 01:21 PM
skeeter
Re: Definite Integration
$\int_1^e \frac{(1+\ln{x})^2}{x} \, dx = \int_1^e \color{red}{(1+\ln{x})^2} \cdot \color{blue}{\frac{1}{x} \, dx} \, \color{black}{= \int_1^2} \color{red}{u^2} \, \color{blue}{du}$
• Apr 28th 2012, 08:15 PM
Prove It
Re: Definite Integration
Quote:

Originally Posted by Plato
Try to get away from using u-substitution. Learn the forms differentiation takes on.

Plato, you should not discourage a student from using a correct method (especially one that is very straightforward, such as a u substitution), especially if it's the method the student has been told to use. Yes, it would be nice if a student can recognise what particular derivatives look like, and therefore what their antiderivatives look like, but that's an unrealistic expectation, as it requires students to remember far too much. It's much better to have students remember a single process than an uncountable number of derivatives.
• Apr 29th 2012, 03:16 AM
Bashyboy
Re: Definite Integration
I figured by doing the u-substitution I would have one sure-fire way of doing, and that I would also see how the derivative and integrals relate in the process; so, now, I am just trying to make sense of the u-substitution in each integration case.
• Apr 29th 2012, 04:19 AM
Prove It
Re: Definite Integration
Quote:

Originally Posted by Bashyboy
I figured by doing the u-substitution I would have one sure-fire way of doing, and that I would also see how the derivative and integrals relate in the process; so, now, I am just trying to make sense of the u-substitution in each integration case.

Well like I said, a good strategy is to look for an inner function and see if its derivative is a multiple.

In this case, you have \displaystyle \begin{align*} \frac{\left(1 + \ln{x}\right)^2}{x} = \left(1 + \ln{x}\right)^2\frac{1}{x} \end{align*}

Can you see that if we use \displaystyle \begin{align*} 1 + \ln{x} \end{align*} as the inner function, its derivative is \displaystyle \begin{align*} \frac{1}{x} \end{align*}? So the integrand is of the form we require.

So we'd let \displaystyle \begin{align*} u = 1 + \ln{x} \implies \frac{du}{dx} = \frac{1}{x} \implies du = \frac{1}{x}\,dx \end{align*}, and note that when \displaystyle \begin{align*} x = 1, u = 1 \end{align*} and when \displaystyle \begin{align*} x = e, u = 2 \end{align*}, then the integral becomes

\displaystyle \begin{align*} \int_1^e{\left(1 + \ln{x}\right)^2\frac{1}{x}\,dx} &= \int_1^2{u^2\,du} \end{align*}

which is very easy to solve :)
• Apr 30th 2012, 10:13 AM
Ivanator27
Re: Definite Integration
Another way to go about it:

http://latex.codecogs.com/gif.latex?...lnx)^{2}}{x}dx

let http://latex.codecogs.com/gif.latex?\large u = 1 + lnx
http://latex.codecogs.com/gif.latex?... \frac{1}{x}dx
http://latex.codecogs.com/gif.latex?\large xdx = du

Integrating indefinitely yields:

http://latex.codecogs.com/gif.latex?...rac{1}{3}u^3+c

So...
http://latex.codecogs.com/gif.latex?...lnx)^{2}}{x}dx
http://latex.codecogs.com/gif.latex?...ight ]_{1}^{e}
And because the Cs cancel anyway:
http://latex.codecogs.com/gif.latex?...ight ]_{1}^{e}

This is how I've been taught to approach an example like this, I only change the upper and lower limits if substituting x back into the expression would make it unmanageable messy. Also, it's not so uncommon for all the Xs to cancel out of an example - when I learned how to integrate by substitution all the examples simplified greatly with the correct substitution.
• Apr 30th 2012, 01:28 PM
tom@ballooncalculus
Re: Definite Integration
And, just in case a picture helps...

http://www.ballooncalculus.org/draw/...twentynine.png

... where (key in spoiler) ...

Spoiler:
http://www.ballooncalculus.org/asy/chain.png

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

The general drift is...

http://www.ballooncalculus.org/asy/maps/intChain.png

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!