My Math Forum Area enclosed by the curve

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 January 7th, 2017, 12:38 PM #1 Member   Joined: Nov 2016 From: Ireland Posts: 64 Thanks: 2 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.
 January 7th, 2017, 12:52 PM #2 Senior Member     Joined: Sep 2015 From: CA Posts: 933 Thanks: 504 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.
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 Member   Joined: Nov 2016 From: Ireland Posts: 64 Thanks: 2 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.
 January 7th, 2017, 02:13 PM #5 Member   Joined: Nov 2016 From: Ireland Posts: 64 Thanks: 2 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.
 January 7th, 2017, 02:51 PM #6 Math Team   Joined: Jul 2011 From: Texas Posts: 2,311 Thanks: 1137 $\bigg[\dfrac{x^4}{4}-\dfrac{2x^3}{3}-\dfrac{5x^2}{2}+6x\bigg]_{-2}^1 + \bigg[\dfrac{x^4}{4}-\dfrac{2x^3}{3}-\dfrac{5x^2}{2}+6x\bigg]_3^1$ $\bigg[\left(\dfrac{1}{4}-\dfrac{2}{3}-\dfrac{5}{2}+6\right)-\left(\dfrac{16}{4}+\dfrac{16}{3}-\dfrac{20}{2}-12\right)\bigg] + \bigg[\left(\dfrac{1}{4}-\dfrac{2}{3}-\dfrac{5}{2}+6\right)-\left(\dfrac{81}{4}-\dfrac{54}{3}-\dfrac{45}{2}+18\right)\bigg]$ $-\dfrac{95}{4} + \dfrac{34}{3} + \dfrac{55}{2} +6$ $-\dfrac{285}{12} + \dfrac{136}{12} + \dfrac{330}{12} + \dfrac{72}{12}$ $\dfrac{253}{12}$ don't you just love the "f" word? (fractions)
January 7th, 2017, 03:35 PM   #7
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Quote:
 Originally Posted by romsek 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.

 January 8th, 2017, 06:26 AM #8 Member   Joined: Nov 2016 From: Ireland Posts: 64 Thanks: 2 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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