The picture is below of how to separate this into 2 different regions, which you have to because it's not continuous over the whole function. It "breaks" at x = 2. So the way to separate this is to take the integral from x = 0 to x = 2 and then add it to the integral for x = 2 to x = 3. In order to integrate each one of those "parts" of that absolute value function we have to determine the equation for each line that makes up that part.

For the integral from [0, 2], the equation of the line is -3x + 6;

For the integral from [2, 3], the equation of the line is 3x - 6.

We integrate then:

$\int\limits^2_0 {-3x+6} \, dx+\int\limits^3_2 {3x-6} \, dx$ and

$-\frac{3x^2}{2}+6x\left \} {{2} \atop {0}} \right. +\frac{3x^2}{2}-6x\left \} {{3} \atop {2}} \right.$ sorry for the odd representation; that's as good as it gets here!

Using the First Fundamental Theorem of Calculus, we get:

(6 - 0) + (-4.5 - (-6)) = 6 + 1.5 = 7.5

5 0

3 years ago

A bakeware company is designing a prototype for a new cake pan. The pan needs to have a height of 1.5 inches so that the cake la

koban [17]

Answer:

Step-by-step explanation:

took the test hope it helps =\/

7 0

2 years ago