Question

Let $f\left(x\right),(x\ge 1)$ be a differentiable function satisfying $f\left(x\right)=(\ln x{)}^2-\underset{1}{\overset{e}{\int }}\frac{f\left(t\right)}{t}dt.$ If the area bounded by the tangent line of $y=f\left(x\right)$ at point $(e,f(e\left)\right),$ the curve $y=f\left(x\right)$ and the line $x=1$ is $A$. Then the value of $\left[A\right]$ is ,
where $[.]$ denotes the greatest integer function.

Answer 1

Let
$C=\underset{1}{\overset{e}{\int }}\frac{f\left(t\right)}{t}dt$
Then
$f\left(x\right)=(\ln x{)}^2-C$
$\therefore C=\underset{1}{\overset{e}{\int }}\frac{(\ln t{)}^2-C}{t}dt$
$=\underset{1}{\overset{e}{\int }}\frac{(\ln t{)}^2}{t}dt-C\underset{1}{\overset{e}{\int }}\frac{1}{t}dt$
$=\underset{0}{\overset{1}{\int }}{y}^{2}dy-C[\ln t{]}_1^e$
$\Rightarrow C=\frac{1}{3}-C$
$\Rightarrow C=\frac{1}{6}$
$\therefore f\left(x\right)=(\ln x{)}^2-\frac{1}{6}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{2\ln x}{x}$
Slope of tangent at point $(e,f(e\left)\right)$ is,
$m=\frac{2\ln e}{e}=\frac{2}{e}$
$x=e\Rightarrow y=(\ln e{)}^2-\frac{1}{6}=\frac{5}{6}$
Therefore, equation of tangent is,
$y-\frac{5}{6}=\frac{2}{e}(x-e)$
$\Rightarrow y=\frac{2x}{e}-\frac{7}{6}$


$\therefore A=\underset{1}{\overset{e}{\int }}[\left(\right(\ln x{)}^2-\frac{1}{6})-(\frac{2x}{e}-\frac{7}{6})]dx$
$\Rightarrow A=\underset{1}{\overset{e}{\int }}\left(\right(\ln x{)}^2-\frac{2x}{e}+1)dx$
$\Rightarrow A=\underset{1}{\overset{e}{\int }}(\ln x{)}^{2}dx-\left[\frac{e^2-1}{e}\right]+e-1\Rightarrow A=\left[\right[x(\ln x{)}^2{]}_1^e-\underset{1}{\overset{e}{\int }}2\ln xdx]+\frac{1}{e}-1\Rightarrow A=e+\frac{1}{e}-1-2\underset{1}{\overset{e}{\int }}\ln xdx\Rightarrow A=e+\frac{1}{e}-1-2\left[x\right(\ln x-1){]}_1^e$
$\Rightarrow A=e+\frac{1}{e}-3$

Therefore,
$\left[A\right]=0$