Welcome to our community

Be a part of something great, join today!

Area under a curve

Yankel

Active member
Jan 27, 2012
398
Hello,

I am looking for the area between

\[f(x)=x\cdot ln^{2}(x)-x\]

and the x-axis.

I have a solution in hand, it suggests that the area is:

\[\int_{\frac{1}{e}}^{e}(x-x\cdot ln^{2}(x))dx\approx 1.95\]

I have a problem with this solution, I don't understand where the area between 0 and 1/e had gone to...

I plotted the function in maple, and an area is appearing very clearly (see photo).

Capture.PNG

So my question: Do you think like me, that the solution attached to this exercise is wrong? If not, where does this area gone to ?

thanks !
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
What is the domain of the given curve?
 

Yankel

Active member
Jan 27, 2012
398
x>0 ?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, and so even though we may state (and you may want to verify this with L'Hôpital's Rule):

\(\displaystyle \lim_{x\to0}f(x)=0\)

Can we say that the point $(0,0)$ is actually on the curve?
 

Yankel

Active member
Jan 27, 2012
398
(0,0) is not on the curve, I know. But what about the area from 0 to 1\e ? I cant ignore it because 0 is not on the graph.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
If the origin is not on the curve, then can we say that area to which you refer is truly bounded?
 

Yankel

Active member
Jan 27, 2012
398
A-ha, now I understand what you say...
Didnt see it coming.
Thanks !! Great help.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
Hello,

I am looking for the area between

\[f(x)=x\cdot ln^{2}(x)-x\]

and the x-axis.

I have a solution in hand, it suggests that the area is:

\[\int_{\frac{1}{e}}^{e}(x-x\cdot ln^{2}(x))dx\approx 1.95\]

I have a problem with this solution, I don't understand where the area between 0 and 1/e had gone to...

I plotted the function in maple, and an area is appearing very clearly (see photo).

View attachment 2029

So my question: Do you think like me, that the solution attached to this exercise is wrong? If not, where does this area gone to ?

thanks !
Mark is correct that the only bounded area is where $\displaystyle \begin{align*} \frac{1}{e} \leq x \leq e \end{align*}$. Notice that this region is BELOW the x axis, so you will get a negative answer. But since you are asked for the AREA, you must give the absolute value of this amount (as areas, being a physical quantity, are always nonnegative).

Now as for the actual integration...

$\displaystyle \begin{align*} \int_{\frac{1}{e}}^e{x\left[ \ln{(x)} \right] ^2 - x \, dx } &= \int_{\frac{1}{e}}^e{x\left\{ \left[ \ln{(x)} \right] ^2 - 1 \right\} \, dx } \\ &= \int_{\frac{1}{e}}^e{\frac{x^2 \left\{ \left[ \ln{(x)} \right] ^2 - 1 \right\} }{x}\, dx} \end{align*}$

Now make the substitution $\displaystyle \begin{align*} t = \ln{(x)} \implies dt = \frac{1}{x}\,dx \end{align*}$, and note that when $\displaystyle \begin{align*} x = \frac{1}{e}, t = -1 \end{align*}$ and when $\displaystyle \begin{align*} x = e, t = 1 \end{align*}$, the integral becomes

$\displaystyle \begin{align*} \int_{\frac{1}{e}}^e{\frac{x^2 \left\{ \left[ \ln{(x)} \right] ^2 - 1 \right\} }{x}\,dx} &= \int_{-1}^1{ \left( e^t \right) ^2 \left( t^2 - 1 \right) \, dt } \\ &= \int_{-1}^1{e^{2t} \left( t^2 - 1 \right) \, dt } \end{align*}$

Now applying integration by parts with $\displaystyle \begin{align*} u = t^2 - 1 \implies du = 2t\,dt \end{align*}$ and $\displaystyle \begin{align*} dv = e^{2t} \, dt \implies v = \frac{1}{2}e^{2t} \end{align*}$ we have

$\displaystyle \begin{align*} \int{ e^{2t} \left( t^2 - 1 \right) \, dt} &= \frac{1}{2} e^{2t} \left( t^2 - 1 \right) - \int{t\,e^{2t}\,dt} \end{align*}$

and applying integration by parts again with $\displaystyle \begin{align*} u = t \implies du = dt \end{align*}$ and $\displaystyle \begin{align*} dv = e^{2t} \, dt \implies v = \frac{1}{2} e^{2t} \end{align*}$ and we have

$\displaystyle \begin{align*} \frac{1}{2}e^{2t} \left( t^2 - 1 \right) - \int{t\,e^{2t}\,dt} &= \frac{1}{2}e^{2t} \left( t^2 - 1 \right) - \left( \frac{1}{2}t\,e^{2t} - \int{\frac{1}{2} e^{2t}\, dt} \right) \\ &= \frac{1}{2} e^{2t} \left( t^2 - 1 \right) - \frac{1}{2} t\, e^{2t} + \frac{1}{2} \int{ e^{2t} \, dt} \\ &= \frac{1}{2}e^{2t} \left( t^2 - 1 \right) - \frac{1}{2} t\, e^{2t} + \frac{1}{4}e^{2t} + C \\ &= \frac{1}{2} e^{2t} \left( t^2 - 1 - t + \frac{1}{2} \right) + C \\ &= \frac{1}{2} e^{2t} \left( t^2 - t - \frac{1}{2} \right) + C \end{align*}$

and so evaluating the definite integral we have

$\displaystyle \begin{align*} \int_{-1}^1{e^{2t} \left( t^2 - 1 \right) \, dt} &= \frac{1}{2} \left[ e^{2t} \left( t^2 - t - \frac{1}{2} \right) \right] _{-1}^1 \\ &= \frac{1}{2} \left( -\frac{1}{2}e^2 - \frac{3}{2}e^{-2} \right) \\ &= -\frac{1}{4} \left( e^2 + 3e^{-2} \right) \\ &\approx. -1.95 \end{align*}$

So the area you are looking for is $\displaystyle \begin{align*} -\frac{1}{4} \left( e^2 + 3e^{-2} \right) \,\textrm{units}^2 \approx 1.95 \, \textrm{units}^2 \end{align*}$.