Alright, I am being asked to solve $\int_{-4}^{4}f(x)dx$,$\int_{4}^{6}f(x)dx$,$\int_{6}^{9}f(x)dx$,$\int_{-4}^{9}f(x)dx$. $$$$So for $$\int_{-4}^{4}f(x)dx$$ I thought the way you solved these would be $f(4)-f(-4)$, which would be $0-(-1)$ which would $=1$. But, that isn't correct, and I am not sure on how to go about solving these. for $\int_{4}^{6}f(x)dx$ I did the same and got $f(6)-f(4)$ which $=4$ and was correct, so I am really lost right now. How do I solve these correctly, thanks in advance!
The comments have already answered the question, but in a nutshell: The definite integral from $a$ to $b$ of $f(x)dx$ is the sum of the areas bounded by $f(x)$ above and the $x$-axis below, minus the sum of the areas bounded by $f(x)$ below and the $x$-axis above.
One other observation: The intervals in the first three parts form the interval in the fourth part: $[-4,4], [4,6], [6,9] \to [-4, 9]$. So, having found the first three parts, add those results to get the fourth:
$$\int_{-4}^{9}f(x) dx = \int_{-4}^{4}f(x) dx + \int_{4}^{6}f(x) dx + \int_{6}^{9}f(x) dx.$$