**How To Split Absolute Value Integrals**. For the first term recall we used the following fact about exponents. What you taking when you integrate is the area of an infinite number of rectangles to approximate the area.

After testing the intervals (1 ;0); This rule just says that you can split an area into two pieces and then add up the pieces to get the area that you started with. Is true for any function that is integrable on [ 0, 4], this is also true for f ( x) = | ( x − 1) ( x − 3) |, which means that.

### To Evaluate The Integral, We First Equate The Given Function To Zero And Find X Intercept.

Based on the material in the notes it should make sense that, provided both integrals converge, we should be able to split up the integral at any point. Then adding up the results. Now, we have an absolute value equation that can be broken down into two pieces.

### When F (X) < 0 Then Area Will Be Negative As F (X)*Dx 0.

For the first term recall we used the following fact about exponents. Evaluate them individually and then remembering that absolute value makes negative values positive should get you the answer. With the given interval, divide the integral into parts, then integrate it.

### After Testing The Intervals (1 ;0);

Press question mark to learn the rest of the keyboard shortcuts The absolute value expression is not isolated yet. And (3;1) we discover x35x2+ 6x.

### Drawing A Vertical Line At.

What you taking when you integrate is the area of an infinite number of rectangles to approximate the area. The function splits himself to two functions , one when x2. ∫ 0 4 f ( x) d x = ∫ 0 1 f ( x) d x + ∫ 1 3 f ( x) d x + ∫ 3 4 f ( x) d x.

### The Equality You Wrote Is Fairly Easy To Prove:

The example discussed in class today is: Is true for any function that is integrable on [ 0, 4], this is also true for f ( x) = | ( x − 1) ( x − 3) |, which means that. With this simplification we can do the integral.