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January 7th, 2017, 12:38 PM   #1
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Area enclosed by the curve

Make a rough sketch of the curve y= (x-1)(x+2)(x-3) and find the area enclosed by this curve and the x axis.

Now I was thinking to myself ok - let's distribute this out

Leaving me - (hope this is correct) = $\displaystyle x^3 - 2x^2 - 5x + 6 $

So I choose random x values between -3 and 3.

However it resulted in some type of sine wave with some pretty low negative values.

So I'm thinking this (x-1)(x+2)(x-3) shouldn't be distributed. Maybe they are some type of coordinates.

Dunno. I can do the integration part by myself I think.

Can someone point me in the general right direction here?

Last edited by skipjack; January 7th, 2017 at 07:38 PM.
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January 7th, 2017, 12:52 PM   #2
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I just don't know what's going on in schools these days that you are at integration w/o having learning basic plotting tricks.

1) y has three x intercepts at 1, -2, 3

2) $x \to -\infty \Rightarrow y \to -\infty,~x \to \infty \Rightarrow y \to \infty$

3)
$y \in (-\infty, -3) \Rightarrow y < 0$
$y \in (-2, 1) \Rightarrow y > 0$
$y \in (1, 3) \Rightarrow y < 0$
$y \in (3, \infty) \Rightarrow y > 0$

this is all you need to make a rough sketch of your curve.

As far as find the area between the curve and the x-axis the question is ill defined. Strictly interpreted the answer is infinity. What they probably want is the integral from -2 to 3, but who knows.
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January 7th, 2017, 01:03 PM   #3
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General shape of a cubic function with a positive leading coefficient ...



$\displaystyle A = \int_{-2}^1 f(x) \, dx - \int_1^3 f(x) \, dx$
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January 7th, 2017, 01:16 PM   #4
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Lol Rom. Well I can explain that... You see I'm one of those thirty somethings that tried to return to college a lil late in life.
I completed a science diploma to get me in. It was applied so I've done a lil math but this stuff is like - whaaaooo!
A baptism of fire you could say - kind of like a scalded cat
Much like getting thrown in at the deep end. Mmmmhmmm
But I feel I've come a fair distance since then, believe it or not.
We all have to start somewhere. Everyone was a learner at one time.

Excellent Skeeter so I was on the right track at least.

Looks like this one in the middle here my graph I mean, which would be correct as that matches the structure of the trinomial.
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January 7th, 2017, 02:13 PM   #5
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So I calculated up the first area. From -2 to 1 came out at 77/12 - which looks about right.
Dang that takes a fair bit of time.

I didn't even do the second area.

That would be a killer in an exam.
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January 7th, 2017, 02:51 PM   #6
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$\bigg[\dfrac{x^4}{4}-\dfrac{2x^3}{3}-\dfrac{5x^2}{2}+6x\bigg]_{-2}^1 + \bigg[\dfrac{x^4}{4}-\dfrac{2x^3}{3}-\dfrac{5x^2}{2}+6x\bigg]_3^1$

$\bigg[\left(\dfrac{1}{4}-\dfrac{2}{3}-\dfrac{5}{2}+6\right)-\left(\dfrac{16}{4}+\dfrac{16}{3}-\dfrac{20}{2}-12\right)\bigg] + \bigg[\left(\dfrac{1}{4}-\dfrac{2}{3}-\dfrac{5}{2}+6\right)-\left(\dfrac{81}{4}-\dfrac{54}{3}-\dfrac{45}{2}+18\right)\bigg]$

$-\dfrac{95}{4} + \dfrac{34}{3} + \dfrac{55}{2} +6$

$-\dfrac{285}{12} + \dfrac{136}{12} + \dfrac{330}{12} + \dfrac{72}{12}$

$\dfrac{253}{12}$

don't you just love the "f" word? (fractions)
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January 7th, 2017, 03:35 PM   #7
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Quote:
Originally Posted by romsek View Post
3)
$y \in (-\infty, -3) \Rightarrow y < 0$
$y \in (-2, 1) \Rightarrow y > 0$
$y \in (1, 3) \Rightarrow y < 0$
$y \in (3, \infty) \Rightarrow y > 0$
jeeze, I'm just so sloppy. All those $y$'s on the left should of course be $x$'s

sorry for the confusion.
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January 8th, 2017, 06:26 AM   #8
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F the F word

Haha no problems Rom. I didn't know that method so no confusion.

Thanks Skeeter. Thought I was near ready for this exam, till I opened a different previous exam this morning and found the nightmare simplification problem, out the bowls of hell! It's not even funny.
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