What is the total area of the region between the curves y=6x^2-18x and y=-6x from x=1 to x=3?

To find the area between the curves y = 6x² - 18x and y = -6x from x = 1 to x = 3, we need to integrate the absolute difference of the two functions over the given interval.

Find the difference between the functions: (6x² - 18x) - (-6x) = 6x² - 18x + 6x = 6x² - 12x

Determine where the functions intersect (if they do within the interval): Set 6x² - 18x = -6x 6x² - 12x = 0 6x(x - 2) = 0 x = 0 or x = 2

Since x = 2 is within our interval [1, 3], we need to consider the intervals [1, 2] and [2, 3] separately.

Integrate the absolute difference of the functions over the intervals:

Interval [1, 2]: In this interval, -6x is greater than or equal to 6x^2 - 18x.

So we integrate: ∫(from 1 to 2) |-6x - (6x² - 18x)| dx = ∫(from 1 to 2) |-6x² + 12x| dx = ∫(from 1 to 2) (-6x² + 12x) dx = [-2x³ + 6x²] (from 1 to 2) = (-16 + 24) - (-2 + 6) = 8 - 4 = 4

Interval [2, 3]: In this interval, 6x² - 18x is greater than or equal to -6x.

So we integrate: ∫(from 2 to 3) |(6x² - 18x) - (-6x)| dx = ∫(from 2 to 3) |6x² - 12x| dx = ∫(from 2 to 3) (6x² - 12x) dx = [2x³ - 6x²] (from 2 to 3) = (54 - 54) - (16 - 24) = 0 - (-8) = 8

Add the absolute values of the integrals: Total area = |4| + |8| = 4 + 8 = 12

Therefore, the total area of the region between the curves y = 6x² - 18x and y = -6x from x = 1 to x = 3 is 4.