What is #f(x) = int (3x)/(x-4) # if #f(2)=1 #?
1 Answer
Explanation:
Integrating this is a little tricky:
First, bring out the constant
#f(x)=3intx/(x-4)dx#
From here, either long divide
#=3int(x-4+4)/(x-4)dx=3int(x-4)/(x-4)+4/(x-4)dx#
#=3int1+4/(x-4)dx#
Split up the integrals:
#=3int1dx+3int4/(x-4)dx#
Here, remember that the
#=3x+3int4/(x-4)dx#
Bring out the
#=3x+12int1/(x-4)dx#
Notice that we have an integral in the form
Since
#f(x)=3x+12lnabs(x-4)+C#
Now, we can use the original condition
#1=3(2)+12lnabs(2-4)+C#
#1=6+12lnabs(-2)+C#
#-5=12ln2+C#
#-5-12ln2=C#
Plug the value of
#f(x)=3x+12lnabs(x-4)-5-12ln2#