Determine the area of the region enclosed by the graphs

Question

Determine the area of the region enclosed by the graphs of f(x) = x² - 4x and g(x) = x³ - 6x² + 8x.

Answer 1

First, set functions equal to each other: f(x) = g(x)

x² - 4x = x³ - 6x² + 8x

0 = x³ -7x² + 12x

0 = x*(x²-7x+12)

Has roots: x = 0, 3 and 4

Next, lets look at the interval from x = 0 to x = 3:

g(x) is above f(x) on the graph, but to also determine this algebraically we can pick a point between x = 0 to x = 3, for instance x = 1.

f(x=1) = -3 and g(x=1) = 1 - 6 + 8 = 3. So g(x) is the upper function.

So the area between the two curves from x = 0 to x = 3 is equal to the definite integral of the upper function minus the lower function from 0 to 3.

That is, ∫ x³ - 6x² + 8x - (x² - 4x) from 0 to 3

= ∫ x³ - 7x² + 12x from 0 to 3

= x⁴/4 - (7/3)x³ + (12/2)*x^2 from 0 to 3

= 3⁴/4 - (7/3)3³ + (12/2)*3^2 - (0⁴/4 - (7/3)0³ + (12/2)*0^2)

= 81/4 - 63 + 54

= 45/4

Now lets look at the integral from x = 3 to x = 4:

From the graph, we see that f(x) is above g(x), this can also be determined algebraically as explained above.

So the area between the two curves from x = 3 to x = 4 is equal to the definite integral of the upper function minus the lower function from 3 to 4.

That is, ∫ (x² - 4x) - (x³ - 6x² + 8x) from 0 to 3

= ∫ -x³ + 7x² - 12x from 3 to 4

= -x⁴/4 + (7/3)x³ - (12/2)*x^2 from 3 to 4

= -4⁴/4 + (7/3)4³ - (12/2)*4^2 - (-3⁴/4 + (7/3)3³ - (12/2)*3^2)

= 7/12

Finally, the total area is the sum of these two results, from x = 0 to x =4.

That is 45/4 + 7/12 = 71/6 ≈ 11.83.