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To find the area of the shaded region formed by the positive x-axis and the graphs of \( f(x) = -6x + 72 \) and \( g(x) = (x - 12)^2 \), we need to set up an integral that represents the area between these two curves from point A to point B. Point A is the x-intercept of \( f(x) \) and point B is the intersection of \( f(x) \) and \( g(x) \).

However, the integral provided in the image, \( \int(f(x - 12)^2) = -6x^2 + 144x - 792 \), is incorrect. To set up the integral correctly, we need to integrate the difference between \( f(x) \) and \( g(x) \) over the interval from A to B, assuming that A is to the left of B:

\( \text{Area} = \int_{A}^{B} (f(x) - g(x)) \, dx \)

First, we find the x-intercept of \( f(x) \):

\( -6x + 72 = 0 \)
\( x = 12 \)

So point A is at x = 12. To find point B, we need to solve the equation \( f(x) = g(x) \):

\( -6x + 72 = (x - 12)^2 \)
\( -6x + 72 = x^2 - 24x + 144 \)
\( x^2 - 18x + 72 = 0 \)

By factoring or using the quadratic formula, we find the roots of the equation. Let’s solve the quadratic equation:

\( x^2 - 18x + 72 = (x - 6)(x - 12) = 0 \)

We have two solutions: x = 6 and x = 12. Since we are already given x = 12 as the x-intercept, we only need to consider x = 6 for point B.

Now, we can set up the integral:

\( \text{Area} = \int_{6}^{12} (-6x + 72 - (x - 12)^2) \, dx \)

To evaluate the integral:

\( \int_{6}^{12} (-6x + 72 - x^2 + 24x - 144) \, dx \)
\( \int_{6}^{12} (-x^2 + 18x - 72) \, dx \)

Evaluating the antiderivative:

\( \left[ \frac{-x^3}{3} + \frac{18x^2}{2} - 72x \right]_{6}^{12} \)
\( \left[ -\frac{x^3}{3} + 9x^2 - 72x \right]_{6}^{12} \)

Substitute x = 12 and x = 6:

\( \left[ -\frac{(12)^3}{3} + 9(12)^2 - 72(12) \right] - \left[ -\frac{(6)^3}{3} + 9(6)^2 - 72(6) \right] \)
\( \left[ -\frac{1728}{3} + 9(144) - 864 \right] - \left[ -\frac{216}{3} + 9(36) - 432 \right] \)
\( \left[ -576 + 1296 - 864 \right] - \left[ -72 + 324 - 432 \right] \)
\( \left[ -144 \right] - \left[ -180 \right] \)
\( -144 + 180 \)
\( 36 \)

The exact area of the shaded region is 36 square units.

- (x - 12)2.
Find the area of the shaded region formed by the positive axes and the graphs of f(x) =
—6x + 72 and g(x) —
Give
an exact answer.
x— 12)2)
Incorrect
assessment activity
g(x)
— —6x2 + 144x — 792
