How do you evaluate the definite integral int dx/(4-x) from [0,2]?

Sep 18, 2016

$\ln \left\mid 2 \right\mid$

Explanation:

We have: ${\int}_{0}^{2} \frac{\mathrm{dx}}{4 - x}$

First, let's evaluate $\int \frac{\mathrm{dx}}{4 - x}$.

Let $u = 4 - x \implies \frac{\mathrm{du}}{\mathrm{dx}} = - 1 \implies \mathrm{du} = - \mathrm{dx}$:

$= \int - \frac{1}{u} \mathrm{dx}$

$= - \int \frac{1}{u} \mathrm{du}$

$= - \ln \left\mid u \right\mid$

We can now replace $u$ with $4 - x$:

$= - \ln \left\mid 4 - x \right\mid$

Now, let's evaluate the definite integral with the interval $\left[0 , 2\right]$:

$= \left(- \ln \left\mid 4 - \left(2\right) \right\mid\right) - \left(- \ln \left\mid 4 - \left(0\right) \right\mid\right)$

$= - \ln \left\mid 2 \right\mid + \ln \left\mid 4 \right\mid$

$= - \ln \left\mid 2 \right\mid + \ln \left\mid {2}^{2} \right\mid$

Using the laws of logarithms:

$= - \ln \left\mid 2 \right\mid + 2 \ln \left\mid 2 \right\mid$

$= \ln \left\mid 2 \right\mid$

Sep 18, 2016

$\ln 2$.

Explanation:

Let $I = {\int}_{0}^{2} \frac{\mathrm{dx}}{4 - x}$

We subst. $4 - x = t \therefore - \mathrm{dx} = \mathrm{dt} , \mathmr{and} , \mathrm{dx} = - \mathrm{dt}$

Next, we change the limits of Integral :

$4 - x = t , x = 0 \Rightarrow t = 4 , \mathmr{and} , x = 2 \Rightarrow t = 2$. Hence,

$I = {\int}_{4}^{2} - \frac{\mathrm{dt}}{t} = {\int}_{2}^{4} \frac{\mathrm{dt}}{t} = {\left[\ln | t |\right]}_{2}^{4} = \ln 4 - \ln 2$.

$\therefore I = \ln \left(\frac{4}{2}\right) = \ln 2$.