Finding perimeter and area using integration

Philip K.

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Joined
Apr 28, 2020
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7
Hello,
I have the following problem:
Given the parabola y=x^2-4x+3 find the area and the parimeter of OAB
A=(1,0), B=(0,3), C=(3,0)

I calculated the area

\(\displaystyle
S oab= \int ^{1}_{0}\left( x^{2}-4x+3\right) dx
=\left| \left( \dfrac {x^{3}}{3}-\dfrac {4x^{2}}{2}+3x\right) \right|
\ = \left| \left( \dfrac {1}{3}-\dfrac {4}{3}+3\right) -0\right| \ = \dfrac {4}{3}

\)

However i am stuck into the parimeter.

\(\displaystyle
P oab=\left| OA\right| +\left| OB\right| +\left| AB\right|\\
= 1 +3 + \int ^{1}_{0}\sqrt {1+\left( \dfrac {d}{dy}\right) ^{2}dx}\\
\begin{aligned}=4+\int ^{1}_{0}\sqrt {1+\left( 2x-4\right) ^{2}}d_{x}\\
=4+\int ^{1}_{0}\sqrt {4x-16x+12}dx\\ =4+2\int ^{1}_{0}\sqrt {x^{2}-4x+\dfrac {17}{4}}dx\\
=4+2\int ^{1}_{0}\sqrt {\left( x-2\right) ^{2}+\dfrac {1}{4}}dx\end{aligned}

\)

Here i tried to substitute u= x-2, but i couldnt solve it.
 

HallsofIvy

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Jan 27, 2012
Messages
6,099
I don't understand this. You give the parabola y= x^2- 4x+ 3 but then you ask for the are and perimeter of OAB which is a right triangle (with area (1/2)( 1)(3)= 3/2 and perimeter \(\displaystyle 1+ 3+ \sqrt{1+ 9}= 4+ \sqrt{9}\)) but has nothing to do with the parabola. Further you give a point, C, that lies on the parabla but has nothing to do with the triangle! The area you calculated is the area between the x-axis and the parabola between x= 1 and x= 3 which, again, has nothing to do with the triangle!
 

Philip K.

New member
Joined
Apr 28, 2020
Messages
7
I don't understand this. You give the parabola y= x^2- 4x+ 3 but then you ask for the are and perimeter of OAB which is a right triangle (with area (1/2)( 1)(3)= 3/2 and perimeter \(\displaystyle 1+ 3+ \sqrt{1+ 9}= 4+ \sqrt{9}\)) but has nothing to do with the parabola. Further you give a point, C, that lies on the parabla but has nothing to do with the triangle! The area you calculated is the area between the x-axis and the parabola between x= 1 and x= 3 which, again, has nothing to do with the triangle!
Here is the graph.
1590280560956.png
 

HallsofIvy

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Jan 27, 2012
Messages
6,099
So you are NOT asked to find the "area and perimeter of OAB"! That was what was confusing me. You are asked find the area and perimeter of the region bounded by the x and y axes and the parabola.

Yes, the area is the integral \(\displaystyle \int_0^1 x^2- 4x+ 3 dx= \left[\frac{1}{3}x^3- 2x^2+ 3x\right]_0^1= \frac{1}{3}- 2+ 3= \frac{1- 6+ 9}{3}= \frac{4}{3}\) as you say.

To do the integral for arclength, \(\displaystyle \int_0^1 \sqrt{(x- 2)^2+ \frac{1}{4}} dx\), start by letting u= x- 2 and du= dx. When x= 0, u= -2 and when x= 1, u= -1 so the integral becomes \(\displaystyle \int_{-2}^{-1} \sqrt{u^2+ 1/4}du\). Factor out that \(\displaystyle \frac{1}{4}\): \(\displaystyle \frac{1}{2}\int_{-2}^{-1} \sqrt{4u^2+ 1}du\).

Do recognize that? If not, recall the identity \(\displaystyle sin^2(\theta)+ cos^2(\theta)= 1\(\displaystyle and divide through by \(\displaystyle cos^2(\theta)\) to get \(\displaystyle \frac{sin^2(\theta)}{cos^2(\theta)}+ 1= \frac{1}{cs^2(\theta)}\) or \(\displaystyle tan^2(\theta)+ 1= sec^2(\theta)\). So, if se let \(\displaystyle 4u= tan(\theta)\), \(\displaystyle \sqrt{4u^2+ 1}= \sqrt{tan^2(\theta)}= sec(\theta)\). When u= -2, \(\displaystyle tan(\theta)= -8\) so what is \(\displaystyle \theta\). When u= -1. \(\displaystyle tan(\theta)= -4\) so what is \(\displaystyle \theta\)?\)\)
 

Philip K.

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Joined
Apr 28, 2020
Messages
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I dont get that "When x= 0, u= -2 and when x= 1, u= -1 "
The rest i understand
 

HallsofIvy

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Jan 27, 2012
Messages
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You had \(\displaystyle (x- 2)^2- \frac{1}{4}\). I suggested letting u= x- 2 so that would become \(\displaystyle u^2-\frac{1}{4}\). Your original integral was from x= 0 to x= 1. When x= 0, u= 0- 2= -2 and when x= 1, u= 1- 2= -1.
 
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