1. Consider the region bounded by the curves y=|x^2+x-12|, x=-5, and x=5 and the x-axis.

A. Set up a sum of integrals, not containing an absolute value symbol, that can be used to find the area of this region.

B. Find the area of the region by using your answer from part A. Don’t approximate with your calculator.

Since y < 0 for -4 < x < 3,

∫[-4,-4] y dx + ∫[-4,3] -y dx + ∫[3,5] y dx

Now just plug in the polynomial for y, and evaluate each integral.

A. To set up a sum of integrals without the absolute value symbol, we need to split the region into different sections where the function inside the absolute value changes its sign.

First, let's find the points where the function inside the absolute value, y = |x^2 + x - 12|, changes its sign. We set the expression inside the absolute value equal to zero and solve for x:

x^2 + x - 12 = 0

(x + 4)(x - 3) = 0

So, the sign of the function changes at x = -4 and x = 3.

To find the area of the region, we need to split it into three parts:
1. From x = -5 to x = -4
2. From x = -4 to x = 3
3. From x = 3 to x = 5

For each section, we will integrate the positive version of the function and subtract the negative version.

1. From x = -5 to x = -4:
The positive version of the function is y = x^2 + x - 12, and the negative version is y = -(x^2 + x - 12).
So, the integral for this section is:
∫[x=-5 to x=-4] (x^2 + x - 12) - (-(x^2 + x - 12)) dx

2. From x = -4 to x = 3:
The positive version of the function is y = x^2 + x - 12, and the negative version is y = -(x^2 + x - 12).
So, the integral for this section is:
∫[x=-4 to x=3] (x^2 + x - 12) - (-(x^2 + x - 12)) dx

3. From x = 3 to x = 5:
The positive version of the function is y = x^2 + x - 12, and the negative version is y = -(x^2 + x - 12).
So, the integral for this section is:
∫[x=3 to x=5] (x^2 + x - 12) - (-(x^2 + x - 12)) dx

B. To find the area of the region, we need to evaluate the integrals we set up in part A.

1. From x = -5 to x = -4:
∫[x=-5 to x=-4] (x^2 + x - 12) - (-(x^2 + x - 12)) dx

2. From x = -4 to x = 3:
∫[x=-4 to x=3] (x^2 + x - 12) - (-(x^2 + x - 12)) dx

3. From x = 3 to x = 5:
∫[x=3 to x=5] (x^2 + x - 12) - (-(x^2 + x - 12)) dx

To find the area, you can calculate each integral separately and then add up the results. Use the fundamental theorem of calculus to evaluate each integral and solve for the area.