What is int_1^e (lnx)/(2x) dx?

Aug 7, 2016

$= \frac{1}{4}$

Explanation:

${\int}_{1}^{e} \frac{\ln x}{2 x} \mathrm{dx}$

$= {\int}_{1}^{e} \frac{d}{\mathrm{dx}} \left(\frac{1}{4} {\ln}^{2} x\right) \mathrm{dx}$

$= \frac{1}{4} {\left[{\ln}^{2} x\right]}_{1}^{e}$

$= \frac{1}{4} {\left[{1}^{2} - 0\right]}_{1}^{e} = \frac{1}{4}$

Aug 7, 2016

$\frac{1}{4}$

Explanation:

Can do this in a number of ways, here are two of them. The first is to use a substitution:

$\textcolor{red}{\text{Method 1}}$

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

Let $u = \ln \left(x\right) \implies \mathrm{du} = \frac{\mathrm{dx}}{x}$

Transforming the limits:

$u = \ln \left(x\right) \implies u : 0 \rightarrow 1$

Integral becomes:

$\frac{1}{2} {\int}_{0}^{1} u \mathrm{du} = \frac{1}{2} {\left[\frac{1}{2} {u}^{2}\right]}_{0}^{1} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$

This is the simpler way, but you might not always be able to make a substitution. An alternative is integration by parts.

$\textcolor{red}{\text{Method 2}}$

Use integration by parts:

For functions $u \left(x\right) , v \left(x\right)$:

$\int u v ' \mathrm{dx} = u v - \int u ' v \mathrm{dx}$

$u \left(x\right) = \ln \left(x\right) \implies u ' \left(x\right) = \frac{1}{x}$

$v ' \left(x\right) = \frac{1}{2 x} \implies v \left(x\right) = \frac{1}{2} \ln \left(x\right)$

$\int \frac{\ln \left(x\right)}{2 x} \mathrm{dx} = \frac{1}{2} \ln \left(x\right) \ln \left(x\right) - \int \frac{\ln \left(x\right)}{2 x} \mathrm{dx}$

Grouping like terms:

$2 \int \frac{\ln \left(x\right)}{2 x} \mathrm{dx} = \frac{1}{2} \ln \left(x\right) \ln \left(x\right) + C$

$\therefore \int \frac{\ln \left(x\right)}{2 x} \mathrm{dx} = \frac{1}{4} \ln \left(x\right) \ln \left(x\right) + C$

We are working with a definite integral though, so applying limits and removing the constant:

${\int}_{1}^{e} \frac{\ln \left(x\right)}{2 x} \mathrm{dx} = {\left[\frac{1}{4} \ln \left(x\right) \ln \left(x\right)\right]}_{1}^{e}$

$= \frac{1}{4} \ln \left(e\right) \ln \left(e\right) - \frac{1}{4} \ln \left(1\right) \ln \left(1\right)$

$\ln \left(e\right) = 1 , \ln \left(1\right) = 0$

$\implies {\int}_{1}^{e} \frac{\ln \left(x\right)}{2 x} \mathrm{dx} = \frac{1}{4}$