Firstly, draw the situation out, you'll have the graph from part a.
Now we need to look at what's bounded by the curve, the axes (so the x and y axis) and the line
. So if we draw
on the graph we obtain the following.
https://www.desmos.com/calculator/u6os20tebiNow if we look at what area is bounded by
,
,
and our curve we get the following.
Now we need to look at which curve, the positive or negative part of our mod we have for the area that we're interested in, the curve on the left hand side. Our original curve is the green curve here,
https://www.desmos.com/calculator/1fqtcirpgnWhen we applied the modulus, we reflected this curve across the axis.
i.e. Our curve is
 & =\begin{cases}<br />\frac{-\left(-6\right)}{2x-3} & x<\frac{3}{2}<br />\\ \frac{-6}{2x-3} & x\geq\frac{3}{2}<br />\end{cases}<br />\\ & =\begin{cases}<br />\frac{6}{2x-3} & x<\frac{3}{2}<br />\\ \frac{-6}{2x-3} & x\geq\frac{3}{2}<br />\end{cases}<br />\end{alignedat}<br /> )
Now we are only interested in the cure on the left,
, but we notice that our area is below the
-axis, so we would get a negative value if we integrate, instead we will integrate the negative of our curve.
Now we need to know our terminals, since our region starts at the
axis horizontally, our first terminal is
, our second terminal will be given by
value at the intersection of the two lines, which we know is
.
Spoiler
+3\log_{e}\left(3\right)<br />\\ & =3\log_{e}\left(\frac{3}{1}\right)<br />\\ & =3\log_{e}\left(3\right)<br />\end{alignedat})
Remember, you need to divide by the coefficient in front of the
when you integrate, and don't forget the modulus signs on the log.
The above kinda over complicates it, but is the process you'd normally use to find it when given a similar modulus function. It's just that in this case our region ends up negative, so we end up finding the negative of this, which gives back our original curve (this won't always happen).
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