Area between the curves

Calculus Level 2

Compute the area bounded by the curve y = x 3 6 x 2 + 9 x y = x^3 - 6x^2 + 9x and the line y = x y = x .


The answer is 8.

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3 solutions

Mohamed Abdelaaty
Dec 26, 2013

Let's get the intersection points of both functions: x 3 6 x 2 + 9 x = x = > x^3 - 6x^2 + 9x = x =>

x ( x 2 ) ( x 4 ) = 0 x(x-2)(x-4) = 0 => x = 0 , 2 o r 4 0, 2 or 4

I used graphical solution for ease:

imgur imgur

(Used Wolfram Alpha)

We can see the the curve lies above the line in the interval 0 < = x < = 2 0 <= x <= 2 while the line lies above the curve in the interval 2 < = x < = 4 2 <= x <= 4

So our integration is:

0 2 ( x 3 6 x 2 + 9 x x ) d x + 2 4 ( x ( x 3 6 x 2 + 9 x ) d x = \int_0^2 (x^3 - 6x^2 + 9x - x)dx + \int_2^4 (x - (x^3 - 6x^2 + 9x)dx =

( x 4 4 2 x 3 + 4 x 2 ) 0 2 (\frac{x^4}{4} - 2x^3 + 4x^2)|_0^2 + ( x 4 4 + 2 x 3 4 x 2 ) 2 4 = 8 . (\frac{-x^4}{4} + 2x^3 -4x^2)|_2^4 = \boxed{8}.

...Ugghh.... Why am I so careless.... :(

敬全 钟 - 7 years, 5 months ago

absolutely perfect!

Sharad Gaikwad - 7 years, 2 months ago

We don't have to know which function lies above the other, only the interval matters and we only count the absolute of the integration of the intervals.

I used this method as it's much clearer though :)

Mohamed Abdelaaty - 7 years, 5 months ago
Ian Tillman
Dec 18, 2013

First, let's note that x 3 6 x 2 + 9 x = x ( x 3 ) 2 x^3-6x^2+9x=x(x-3)^2 which tells us that the zeros of that function are 0 0 (multiplicity 1) and 3 3 (multiplicity 2). To find where the functions intersect, we simply need to set them equal to each other, giving us x ( x 3 ) 2 = x x ( x 3 ) 2 x = 0 x ( ( x 3 ) 2 1 ) = 0 x(x-3)^2=x\Rightarrow x(x-3)^2-x=0\Rightarrow x((x-3)^2-1)=0 ( x 3 ) 2 1 = 0 ( x 3 ) 2 = 1 x 3 = ± 1 , x = 0 , 2 , 4 (x-3)^2-1=0\Rightarrow (x-3)^2=1 \Rightarrow x-3=\pm1, x=0,2,4 Now all we need to know is which function is above the other over the intervals of [ 0 , 2 ] [0,2] and [ 2 , 4 ] [2,4] , which we can do by plugging in an x x -value that is between the bounds of each interval, for example; 1 3 6 ( 1 2 ) + 9 ( 1 ) = 4 > 1 1^3-6(1^2)+9(1)=4 > 1 3 3 6 ( 3 2 ) + 9 ( 3 ) = 0 < 3 3^3-6(3^2)+9(3)=0 < 3 Now that we know which function is above the other during each of the intervals, we can now calculate the area that they bound. ( 0 2 x 3 6 x 2 + 9 x d x 0 2 x d x ) + ( 2 4 x d x 2 4 x 3 6 x 2 + 9 x d x ) (\int_0^2 x^3-6x^2+9x\ \mathrm{d}x-\int_0^2 x\ \mathrm { d}x)+(\int_2^4x\ \mathrm { d}x-\int_2^4 x^3-6x^2+9x\ \mathrm { d}x) ( 6 2 ) + ( 6 2 ) = 4 + 4 = 8 (6-2)+(6-2)=4+4=\boxed{8}

Calculation problems

Calvin David - 7 years, 4 months ago
J Thompson
Feb 7, 2014

To find the area between the lines, we need to know where they intersect.

x 3 6 x 2 + 9 x = x x 3 6 x 2 + 8 x = 0 x ( x 2 6 x + 8 ) = 0 x ( x 2 ) ( x 4 ) = 0 \begin{aligned}x^3 - 6x^2 + 9x = x &\implies x^3 - 6x^2 + 8x = 0\\ &\implies x(x^2 - 6x + 8) = 0\\ &\implies x(x - 2)(x - 4) = 0\end{aligned}

So the two lines intersect when x = 0 x = 0 , x = 2 x = 2 and x = 4 x = 4 .

By considering the shape of the cubic graph (and the fact that we have three intersections), it should be apparent that the cubic function is greater than the linear function for 0 x 2 0 \le x \le 2 , and vice versa for 2 x 4 2 \le x \le 4 . Thus the required area is given by:

0 2 ( ( x 3 6 x 2 + 9 x ) x ) d x + 2 4 ( x ( x 3 6 x 2 + 9 x ) ) d x = 0 2 ( x 3 6 x 2 + 8 x ) d x + 2 4 ( x 3 + 6 x 2 8 x ) d x = [ x 4 4 6 x 3 3 + 8 x 2 2 ] 0 2 + [ 4 x 4 4 + 6 x 3 3 8 x 2 2 ] 2 4 = [ x 4 4 2 x 3 + 4 x 2 ] 0 2 + [ 4 x 4 4 + 2 x 3 4 x 2 ] 2 4 = ( ( 4 16 + 16 ) ( 0 0 + 0 ) ) + ( ( 64 + 128 64 ) ( 4 + 16 16 ) ) = ( 4 0 ) + ( 0 4 ) = 4 + 4 = 8 \begin{aligned} \int_0^2{\!\Big( (x^3 - 6x^2 + 9x) - x\Big)}\; \mathrm{d}x + \int_2^4{\!\Big(x - (x^3 - 6x^2 + 9x)\Big)}\; \mathrm{d}x &= \int_0^2{\!(x^3 - 6x^2 + 8x)} \; \mathrm{d}x + \int_2^4{\!(-x^3 + 6x^2 - 8x)} \; \mathrm{d}x \\ &= \left[\frac{x^4}{4} - \frac{6x^3}{3} + \frac{8x^2}{2}\right]_0^2 + \left[ \frac{-4x^4}{4} + \frac{6x^3}{3} - \frac{8x^2}{2}\right]_2^4\\ &= \left[\frac{x^4}{4} - 2x^3 + 4x^2\right]_0^2 + \left[ \frac{-4x^4}{4} + 2x^3 - 4x^2\right]_2^4 \\ &= \Big( (4 - 16 + 16) - (0 - 0 + 0)\Big) + \Big( (-64 + 128 - 64) - (-4 + 16 - 16)\Big)\\ &= (4 - 0) + (0 - 4)\\ &= 4 + 4\\ &= \boxed{8}\end{aligned}

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