Integration of Piecewise Functions?

[Mindrust]

Not sure what I'm doing wrong, but I can't seem to get the integral for the third and fourth piece.

 

I might be using the wrong values for [a,b]. Anyone know the answers and what intervals should be used?

 

Am I supposed to add the first two parts into the third?

[timo]

What you are doing wrong (I assume the green hooks and red crosses are supposed to indicate the parts you did correct and wrong, respectively) is that you are giving an antiderivative of f(x) for the respective x. What is asked for, however, is g(x). If you look careful at the definition of g(x) you will realize that g(x) is an arbitrary antiderivative of f(x) (or in fact should in this context not be thought of as the antiderivative at all)

 

EDIT: In an earlier version of this post I said that g(x) is not the antiderivative of f(x). That was wrong (it is an antiderivative, but a very specific one) and has been corrected now.

Edited November 27, 2011 by timo

[Mindrust]

So for the third piece of g(x), I was missing the constant at the end, C.

 

To find C, I let the third equal the second, at x=1. So...

 

2x - 0.5x^2 + C = 0.5x^2 (at x=1)

 

Then solve for C, which is -1.

 

So the answer to the third is 2x - 0.5x^2 - 1

 

I know I have the right answer, but not sure if my method is correct. Can anyone confirm? Thanks!

[timo]

Your method is correct but tedious and unnecessarily complicated to justify (as demonstrated by the fact that you are not even sure if it is the correct way). A simpler explanation is to consequently use the definition of g(x): For [math]1\leq x < 2[/math], [math]g(x) = \int_0^x \! f(x) \, dx = \int_0^1 \! f(x) \, dx + \int_1^x \! f(x) \, dx = \dots [/math].