Question

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

Answer 1

`ln abs(2)`

Explanation:

We have: `int_(0)^(2) (dx) / (4 - x)`

First, let's evaluate `int (dx) / (4 - x)`.

Let `u = 4 - x => (du) / (dx) = - 1 => du = - dx`:

`= int - (1) / (u) dx`

`= - int (1) / (u) du`

`= - ln abs(u)`

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

`= - ln abs(4 - x)`

Now, let's evaluate the definite integral with the interval `[0, 2]`:

`= (- ln abs(4 - (2))) - (- ln abs(4 - (0)))`

`= - ln abs(2) + ln abs(4)`

`= - ln abs(2) + ln abs(2^(2))`

Using the laws of logarithms:

`= - ln abs(2) + 2 ln abs(2)`

`= ln abs(2)`

Answer 2

`ln2`.

Explanation:

Let `I=int_0^2 dx/(4-x)`

We subst. `4-x=t :. -dx=dt, or, dx=-dt`

Next, we change the limits of Integral :

`4-x=t, x=0 rArr t=4, and, x=2 rArr t=2`. Hence,

`I=int_4^2 -dt/t=int_2^4 dt/t=[ln|t|]_2^4=ln4-ln2`.

`:. I=ln(4/2)=ln2`.