Compute the area bounded by the curve y = x 3 − 6 x 2 + 9 x and the line y = x .
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...Ugghh.... Why am I so careless.... :(
absolutely perfect!
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 :)
First, let's note that x 3 − 6 x 2 + 9 x = x ( x − 3 ) 2 which tells us that the zeros of that function are 0 (multiplicity 1) and 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 − 3 ) 2 − 1 = 0 ⇒ ( x − 3 ) 2 = 1 ⇒ x − 3 = ± 1 , x = 0 , 2 , 4 Now all we need to know is which function is above the other over the intervals of [ 0 , 2 ] and [ 2 , 4 ] , which we can do by plugging in an x -value that is between the bounds of each interval, for example; 1 3 − 6 ( 1 2 ) + 9 ( 1 ) = 4 > 1 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 ) ( 6 − 2 ) + ( 6 − 2 ) = 4 + 4 = 8
Calculation problems
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
So the two lines intersect when x = 0 , x = 2 and 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 , and vice versa for 2 ≤ x ≤ 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 = [ 4 x 4 − 3 6 x 3 + 2 8 x 2 ] 0 2 + [ 4 − 4 x 4 + 3 6 x 3 − 2 8 x 2 ] 2 4 = [ 4 x 4 − 2 x 3 + 4 x 2 ] 0 2 + [ 4 − 4 x 4 + 2 x 3 − 4 x 2 ] 2 4 = ( ( 4 − 1 6 + 1 6 ) − ( 0 − 0 + 0 ) ) + ( ( − 6 4 + 1 2 8 − 6 4 ) − ( − 4 + 1 6 − 1 6 ) ) = ( 4 − 0 ) + ( 0 − 4 ) = 4 + 4 = 8
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Let's get the intersection points of both functions: x 3 − 6 x 2 + 9 x = x = >
x ( x − 2 ) ( x − 4 ) = 0 => x = 0 , 2 o r 4
I used graphical solution for ease:
imgur
(Used Wolfram Alpha)
We can see the the curve lies above the line in the interval 0 < = x < = 2 while the line lies above the curve in the interval 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 =
( 4 x 4 − 2 x 3 + 4 x 2 ) ∣ 0 2 + ( 4 − x 4 + 2 x 3 − 4 x 2 ) ∣ 2 4 = 8 .