# Area under a curve

#### Yankel

##### Active member
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).

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

Staff member
What is the domain of the given curve?

x>0 ?

#### MarkFL

Staff member
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
(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

Staff member
If the origin is not on the curve, then can we say that area to which you refer is truly bounded?

#### Yankel

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

#### Prove It

##### Well-known member
MHB Math Helper
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*}.