# Question #94b54

Jan 4, 2018

$f ' \left(x\right) = \ln \left(x\right) + 1$

${\int}_{e}^{2 e} \ln \left(x\right) \mathrm{dx} = 2 e \ln \left(2\right) \approx 3.76834$

#### Explanation:

We have

$f \left(x\right) = x \ln \left(x\right)$

Its derivative can be found with the product rule:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left\{x\right\} \ln \left(x\right) + x \frac{d}{\mathrm{dx}} \left\{\ln \left(x\right)\right\}$

$= 1 \cdot \ln \left(x\right) + x \cdot \frac{1}{x}$

$= \ln \left(x\right) + 1$

Now that we have $f ' \left(x\right)$ we can use this to evaluate the integral:

${\int}_{e}^{2 e} \ln \left(x\right) \mathrm{dx}$

Now we use a simple trick of adding $1$ and subtracting $1$ like this:

$= {\int}_{e}^{2 e} \ln \left(x\right) + 1 - 1 \mathrm{dx}$

We can now break this integral down like so:

$= {\int}_{e}^{2 e} \ln \left(x\right) + 1 \mathrm{dx} + {\int}_{e}^{2 e} - 1 \mathrm{dx}$

$= {\int}_{e}^{2 e} \ln \left(x\right) + 1 \mathrm{dx} - {\int}_{e}^{2 e} 1 \mathrm{dx}$

We already have the answer to the first integral from the first part of the question: the expression inside the integral is what we found for $f ' \left(x\right)$ so the integral will simply be $f \left(x\right)$. The second integral will of course just be $- x$. So evaluating the integral and its limits yiels:

$= {\left[x \ln \left(x\right)\right]}_{e}^{2 e} - {\left[x\right]}_{e}^{2 e}$

$= \left\{\left(2 e\right) \ln \left(2 e\right)\right\} - \left\{e \ln \left(e\right)\right\} - \left(\left\{2 e\right\} - \left\{e\right\}\right)$

$2 e \ln \left(2 e\right) - e - 2 e + e$

$2 e \left(\ln \left(2 e\right) - 1\right) = 2 e \left(\ln 2 + \ln e - 1\right)$

$2 e \ln \left(2\right) \approx 3.76834$