I don't even know what to call this - and I am bit baffled by the question:
Q) explain why the graph of the function f(x)=(4-x)In(1/2*x) lies above the x- aixs for (2<x<4) and below the x-axis for 4<x<6.
does this mean x= 3 and 5 and you substitute this in equation to see what you get?
Now I know this- so sorry for posting it. But I have to use this fact that graph is positive 2<x<4 and negative for 4<x<6 and find the area enclosed by the graph anf the x- axis between x=2 and x=6
You posted this in the Calculus section so you should understand that the area under the graph of y= f(x), above the x-axis, for a< x< b is given by .
To integrate (4- x)ln(x/2), let u= x/2 so that x= 2u and you have (4- 2u)ln(u)= 4ln(u)- 2uln(u).
You should be able to integrate each of those by parts.
It would be better to update this thread with your working.
You can cancel an x in the integral though because we know it cannot be 0 and we're not integrating from 0.
I'm not sure where 7/4 comes from and you will need to take care with your signs when you subtract it with the first part of the expression