Print Page - Area bound by Mod graphs? (Question)

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: This-is-not-me on October 18, 2013, 12:29:20 pm

Title: Area bound by Mod graphs? (Question)
Post by: This-is-not-me on October 18, 2013, 12:29:20 pm

Hey guys I'm having some trouble with part b of this question can someone help me understand it?

Title: Re: Area bound by Mod graphs? (Question)
Post by: b^3 on October 18, 2013, 02:02:51 pm

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 $x=1$ . So if we draw $x=1$ on the graph we obtain the following.
https://www.desmos.com/calculator/u6os20tebi
Now if we look at what area is bounded by $x=0$ , $y=0$ , $x=1$ and our curve we get the following.

(http://i1082.photobucket.com/albums/j373/mclaren200800/region_zpsf1a449f2.png)

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/1fqtcirpgn
When we applied the modulus, we reflected this curve across the axis.
i.e. Our curve is $\begin{alignedat}{1}2x-3 & \geq0 \\ x & \geq\frac{3}{2} \\ f(x) & =\begin{cases} \frac{-\left(-6\right)}{2x-3} & x<\frac{3}{2} \\ \frac{-6}{2x-3} & x\geq\frac{3}{2} \end{cases} \\ & =\begin{cases} \frac{6}{2x-3} & x<\frac{3}{2} \\ \frac{-6}{2x-3} & x\geq\frac{3}{2} \end{cases} \end{alignedat} $
Now we are only interested in the cure on the left, $y=\frac{6}{2x-3}$ , but we notice that our area is below the $x$ -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 $y$ axis horizontally, our first terminal is $x=0$ , our second terminal will be given by $x$ value at the intersection of the two lines, which we know is $x=1$ .

Spoiler

$\begin{alignedat}{1}-\int_{0}^{1}\frac{6}{2x-3}dx & =-\left[6\times\frac{1}{2}\log_{e}|2x-3|\right]_{0}^{1} \\ & =-3\log_{e}\left(1\right)+3\log_{e}\left(3\right) \\ & =3\log_{e}\left(\frac{3}{1}\right) \\ & =3\log_{e}\left(3\right) \end{alignedat}$
Remember, you need to divide by the coefficient in front of the $x$ 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).

Title: Re: Area bound by Mod graphs? (Question)
Post by: This-is-not-me on October 18, 2013, 03:27:27 pm

Wow thanks for the detailed response I finally understand it, but just a clarification

Quote from: b^3 on October 18, 2013, 02:02:51 pm

i.e. Our curve is $\begin{alignedat}{1}2x-3 & \geq0 \\ x & \geq\frac{3}{2} \\ f(x) & =\begin{cases} \frac{-\left(-6\right)}{2x-3} & x<\frac{2}{3} \\ \frac{-6}{2x-3} & x\geq\frac{2}{3} \end{cases} \\ & =\begin{cases} \frac{6}{2x-3} & x<\frac{2}{3} \\ \frac{-6}{2x-3} & x\geq\frac{2}{3} \end{cases} \end{alignedat} $

this part here the domain is meant to be 3/2 not 2/3 right? Or am I missing something?

And as for the region being negative it basically means we have to do upper curve-lower curve where the upper curve is y=0 correct?

Title: Re: Area bound by Mod graphs? (Question)
Post by: b^3 on October 18, 2013, 04:22:39 pm

Yeah it should be $\frac{3}{2}$ , my bad, I'll fix it up now.

Quote from: This-is-not-me on October 18, 2013, 03:27:27 pm

And as for the region being negative it basically means we have to do upper curve-lower curve where the upper curve is y=0 correct?

Well yeah, I guess you can think of it that way.

Title: Re: Area bound by Mod graphs? (Question)
Post by: This-is-not-me on October 18, 2013, 08:32:34 pm

Sweet thanks :)

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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: This-is-not-me on October 18, 2013, 12:29:20 pm