January 7th, 2017, 12:38 PM  #1 
Member Joined: Nov 2016 From: Ireland Posts: 54 Thanks: 2  Area enclosed by the curve
Make a rough sketch of the curve y= (x1)(x+2)(x3) 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 (x1)(x+2)(x3) 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: 749 Thanks: 398 
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 xaxis 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 
Math Team Joined: Jul 2011 From: Texas Posts: 2,197 Thanks: 1048 
General shape of a cubic function with a positive leading coefficient ... $\displaystyle A = \int_{2}^1 f(x) \, dx  \int_1^3 f(x) \, dx$ 
January 7th, 2017, 01:16 PM  #4 
Member Joined: Nov 2016 From: Ireland Posts: 54 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: 54 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: 2,197 Thanks: 1048 
$\bigg[\dfrac{x^4}{4}\dfrac{2x^3}{3}\dfrac{5x^2}{2}+6x\bigg]_{2}^1 + \bigg[\dfrac{x^4}{4}\dfrac{2x^3}{3}\dfrac{5x^2}{2}+6x\bigg]_3^1$ $\bigg[\left(\dfrac{1}{4}\dfrac{2}{3}\dfrac{5}{2}+6\right)\left(\dfrac{16}{4}+\dfrac{16}{3}\dfrac{20}{2}12\right)\bigg] + \bigg[\left(\dfrac{1}{4}\dfrac{2}{3}\dfrac{5}{2}+6\right)\left(\dfrac{81}{4}\dfrac{54}{3}\dfrac{45}{2}+18\right)\bigg]$ $\dfrac{95}{4} + \dfrac{34}{3} + \dfrac{55}{2} +6$ $\dfrac{285}{12} + \dfrac{136}{12} + \dfrac{330}{12} + \dfrac{72}{12}$ $\dfrac{253}{12}$ don't you just love the "f" word? (fractions) 
January 7th, 2017, 03:35 PM  #7 
Senior Member Joined: Sep 2015 From: CA Posts: 749 Thanks: 398  
January 8th, 2017, 06:26 AM  #8 
Member Joined: Nov 2016 From: Ireland Posts: 54 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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