INTEGRATING ABSOLUTE VALUE FUNCTIONS

Problem 1 :

Solution :

Vertex of the absolute value function is (3/2, 0)

Area cannot be negative. So, the required area is 9/2 square units.

Problem 2 :

Solution :

x² - 4x + 3

Decomposing into linear factors, we get

(x - 3)(x - 1)

By drawing the sign diagram, we can easily figure out which part will positive and negative.

So, the required area is 4 square units.

Problem 3 :

Solution :

x³ - 5x² + 6x

Decomposing into linear factors, we get

= x(x² - 5x + 6)

= x (x - 2)(x - 3)