Yankel said:
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 $\displaystyle \begin{align*} -\frac{1}{4} \left( e^2 + 3e^{-2} \right) \,\textrm{units}^2 \approx 1.95 \, \textrm{units}^2 \end{align*}$.
