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Given

  • $f(x) = x^3 - ax$
  • $g(x) = 2x^2 - bx$
  • $a \ne b$
  • $f(x)$ and $g(x)$ meet in two points, one of them is a tangent.

I am supposed to find the area enclosed between the two curves.

I've managed to find the points: $x = 0, x = 1$, the thing I'm having trouble with is finding the area enclosed by those two function from $x = 0$ to $x = 1$.

I tried to integrate like this: $\int_{0}^{1}(x^3-2x^2+bx-ax)dx$, but I can't get rid of $a$ and $b$.

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Why do you want to get rid of $a$ and $b$? –  Thomas Apr 5 '12 at 21:58
    
They do meet where $x=0$. I do not see the $x=1$. I also do not see a question. And do you mean that the two curves are tangent to each other at one point where they meet? It could be useful if you quote the full problem. –  André Nicolas Apr 5 '12 at 21:59
    
@ThomasM Because the this is the requirement. The answer should be a pure number. –  Lior Apr 5 '12 at 22:05
    
@André Nicolas The question is: What is the area enclosed by the the curves from x = 0 to x = 1? They are not tangent to each other, they meat at x = 0 and share a tangent line in x = 1. I did my best translating it to English :) –  Lior Apr 5 '12 at 22:05
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2 Answers

up vote 1 down vote accepted

To find the intersection points of $f$ and $g$, notice that $f(x)-g(x)=$ $x^3-2x^2-(a-b)x=$ $x\left(x^2-2x+(b-a)\right)$ has roots at $0$ and at $1\pm\sqrt{a-b+1}$ for $a-b+1\ge0$. This is three points for $a-b>-1$ (or $ba+1$). Since we are given that there are exactly two intersection points, we therefore conclude that $a-b=-1$, and that $f-g=x(x-1)^2$.

Next, $f'(x)-g'(x)=3x^2-4x-(a-b)$ has roots at $x=\frac{2\pm\sqrt{3(a-b)+4}}{3}$ $=\frac{2\pm1}{3}=\frac13,1$. So the curves enclose a region for $x\in[0,1]$ and are tangent at the left endpoint. Since $b-a=1$, you can get rid of them in the integral because they only occur as a difference:

$$ A %=\int_0^1 x(x^2-2x+1)\,dx =\int_0^1 x(x-1)^2\,dx =\int_{-1}^0 (x+1)x^2\,dx %=\int_{-1}^0 (x^3+x^2)\,dx =\left[\frac{x^4}{4}+\frac{x^3}{3}\right]_{-1}^0 %=\frac{0-1^4}{4}+\frac{0+1^3}{3} =\frac13-\frac14 =\frac1{12} $$ with a few algebraic manipulations (and one substitution -- see the source for more steps). Of course, you can also do the integral in the more routine way as $$ A =\int_0^1 x(x^2-2x+1)\,dx =\int_0^1 (x^3-2x^2+x)\,dx =\left[\frac14x^4-\frac23x^3+\frac12x^2\right]_0^1 =\frac34-\frac23 =\frac1{12} $$ and the two together help (as in my case!) to vet errors.

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If they meet at x=1, then b=a+1. Use that and the terms with a and b will cancel.

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