How do you evaluate the definite integral $\int \frac{d x}{x}$ $int dx/x$ from $[\frac{1}{e}, e]$ $[1/e,e]$ ?

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Answer 1 · 2016-10-28T12:58:42

2

$\int_{\frac{1}{e}}^{e} \frac{d x}{x} = {[ln x]}_{\frac{1}{e}}^{e}$ $int_(1/e)^e(dx)/x=[lnx]_(1/e)^e$

$= ln e - ln (\frac{1}{e})$ $=lne-ln(1/e)$

$= ln e - [ln 1 - ln e]$ $=lne-[ln1-lne]$

$= 1 - 0 + 1$ $=1-0+1$

$= 2$ $=2$

Answer 2 · 2016-10-28T13:00:34

The answer is $2$ $2$

The integral is $\int \frac{d x}{x} = ln x + C$ $intdx/x=lnx + C$

So the integral is $\int \frac{d x}{x} = (ln x)$ $intdx/x=(lnx)$

So from $(\frac{1}{e}, e)$ $(1/e,e)$ ,
we have $\int \frac{d x}{x} = ln e - ln (\frac{1}{e}) = ln e - ln 1 + ln e$ $intdx/x=lne-ln(1/e)=lne-ln1+lne$

As $ln e = 1$ $lne=1$ and $ln 1 = 0$ $ln1=0$
So we have finally $\int \frac{d x}{x} = 1 - 0 + 1 = 2$ $intdx/x=1-0+1=2$