# Peicewise integral

1. Apr 26, 2010

### hks118

1. The problem statement, all variables and given/known data
Evaluate the integral

3. The attempt at a solution
I tried splitting it into two integrals and adding the results, but it says that's wrong. I think it has something to do with the fact that the first part doesn't go all the way to zero...

2. Apr 26, 2010

### Cyosis

Splitting up the integral is the correct approach. Show your work so people can see where you go wrong.

3. Apr 26, 2010

### hks118

Ok, so I started by evaluating the first function on the interval [-2,0]

F(x)=2x
F(0)=0
F(-2)=-4

0-(-4)=4

Then I evaluated the second function on [0,4]

F(x)= 5x-x3/3
F(0)=0
F(4)=-1.3333333

Adding them gets me 2.6666666, which apparently is incorrect.

4. Apr 26, 2010

### Cyosis

Why do you say it is apparently incorrect. Does the book provide an answer or do you enter it in some kind of computer program? Either way your answer is essentially correct, but since you've chosen to use decimals you get rounding errors. Use fractions instead.

5. Apr 26, 2010

### hks118

It is a computer program. I'll try with factions, it is very finicky sometimes...

6. Apr 26, 2010

### hks118

Entered 8/3 and was right. Hate computers! haha