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

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Answer 1 · 2016-09-18T19:31:42

#ln abs(2)#

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 · 2016-09-18T19:31:59

#ln2#.

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#.