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 April 12th, 2012, 09:23 AM #1 Member   Joined: Apr 2012 Posts: 47 Thanks: 0 surface Integration need to calculate surface between fx and gx. not sure i used correct word surface. $fx= x^{3}-x$ $gx=1-x^{2}$ fx=gx $x^{3}-x+1-x^{2}=0$ im stock, hope someone can show me how to do it.
 April 12th, 2012, 11:02 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: surface Integration While surface isn't an incorrect term, typically the word area is used. For finding the intersections, write the equation as: $f(x)=g(x)$ $x^3-x=1-x^2$ $x^3-x+x^2-1=0$ $x$$x^2-1$$+$$x^2-1$$=0$ $(x+1)$$x^2-1$$=0$ $(x+1)^2(x-1)=0$ $x=-1,\,1$ We may determine that on (-1,1) using a test point (x = 0) that we have $g(x)>f(x)$ and so the area A bounded by the two functions is: $A=\int_{-1}\,^1 g(x)-f(x)\,dx=\int_{-1}\,^1 -x^3-x^2+x+1\,dx$
 April 12th, 2012, 11:45 AM #3 Member   Joined: Apr 2012 Posts: 47 Thanks: 0 Re: surface Integration is possible to use long division methode? if yes how, because teacher say is possible to use long division, but i dont see fx degree is higher then gx and this problem is not in $\frac{a}{b}$ form $x(x^{2}-1)+(x^{2}-1)=0$ can you do one step more, dont understand this one. $(x+1)+(x^{2}-1)=0$
 April 12th, 2012, 11:51 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: surface Integration I wouldn't use division, just factoring because it's simpler in this case. We have: $x$$x^2-1$$+$$x^2-1$$=0$ $x$$x^2-1$$+1$$x^2-1$$=0$ The two terms on the left have a common factor of $x^2-1$ so we have: $(x+1)$$x^2-1$$=0$ $(x+1)(x+1)(x-1)=0$ $(x+1)^2(x-1)=0$
 April 12th, 2012, 12:01 PM #5 Member   Joined: Apr 2012 Posts: 47 Thanks: 0 Re: surface Integration can you tell me how to use division in this case. i know how to use division methode but dont know $x^{3}+x^{2}-x-1$ division by what? i still missing something...
 April 12th, 2012, 12:16 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: surface Integration Well, we could state: $\frac{f(x)}{g(x)}=\frac{x^3-x}{1-x^2}=-\frac{x$$1-x^2$$}{1-x^2}=-x$ So we may state: $f(x)=-xg(x)$ So now, when we equate the two functions to find the intersections, we have: $g(x)=f(x)$ $g(x)=-xg(x)$ $(x+1)g(x)=0$ $(x+1)$$1-x^2$$=0$ $(x+1)$$x^2-1$$=0$ $(x+1)(x+1)(x-1)=0$ $(x+1)^2(x-1)=0$ Now you see why I recommend just factoring, it is simpler.
 April 12th, 2012, 02:05 PM #7 Member   Joined: Apr 2012 Posts: 47 Thanks: 0 Re: surface Integration aaaaaaa now i see, thanks alot but dont know why im very bad in calculate area .... like this 2 problems. a) y=fx= $\sqrt{4x-1}$ lines are x=0 and x=2 b) y=fx$x^{3}+x^{2}-2x$ so this one $x(x^{2}+x-2)$ $x(x+2)(x-1)$ so $x=0,-2,1$ i miss gx?
April 13th, 2012, 02:58 AM   #8
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Re: surface Integration

not sure im doing right, hope someone can help me out. thanks!
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April 13th, 2012, 03:12 AM   #9
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Re: surface Integration

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April 13th, 2012, 04:28 AM   #10
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Re: surface Integration

confused
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