Absolute area of integrals and signs Hi I have a quick question about the absolute area of a graph bounded by an interval.
For example lets say I have $f(x)=3-x$ in the closed interval $(0,5)$.
If I were to find the absolute area I would want to break up the graph of this area into parts, basically where there is a discontinuity. Right?
So for this example when $x=3$ we get $0$. So we can break our interval from $(0,5)$ into $(0,3)$ and $(3,5)$. To calculate this we need to find the integral of the equation at these new intervals and add them together.
The book says the $\int_0^3 (3-x)dx + \int_3^5 (x-3) dx$. I understand how to compute this.
My question is why am I trying to find the integral for $(x-3)$ for the second part. I realize this is $-(x-3)$. But why? The book doesn't really give me any explanation. Is it a +- pattern?
 A: I think when you say "discontinuity" you're talking broadly about critical points. In this case the critical points correspond to the roots where the function $f$ changes signs. 
Anyway to answer your question: the task is to compute the absolute area, which is "roughly" given by height times width. In this case width is given by the length of the interval, which in the language of integration is encapsulated by the $dx$, when we integrate. However, "height" should refer to something positive. At least colloquially that is (what does negative height even mean, right??). So, when we want the area under a graph, we should really be computing the following integral
$$\int_I |f(x)|\,dx$$
where $I$ is the interval under consideration. Now in your case $f(x)$ is given by $3-x$. But the sign of this function depends on $x$. In particular, the sign changes at the "critical point" which is where $f(x)=0$, and this happens at $x=3$. So consider computing the integral of $|f|$
$$\int_{0}^{5}|f(x)|\,dx=\int_{0}^{5}|3-x|\,dx$$
To the left of the critical point $x=3$, $3-x$ is positive so you can simply drop the absolute value and use $3-x$. To the right of $x=3$, $3-x$ is negative, hence you need to add a negative sign to make it positive: $|3-x|=-(3-x)=x-3$. Therefore the integral you should be computing boils down to
$$\int_{0}^{5}|3-x|\,dx=\int_{0}^{3}|3-x|\,dx+\int_{3}^{5}|3-x|\,dx=\int_{0}^{3}3-x\,dx+\int_{3}^{5}-(3-x)\,dx$$
Long story short, we want height to be something positive so we look for values of $x$ where $f$ is negative and simply give them a negative sign to make it positive.
A: When integrating, area below the x-axis is negative. For example, take the function $g(x)=x-1$ and integrate from 0 to 1.
$$ \int_{0}^{1} (x-1) dx = \left[ \frac{x^2}{2} - x \right]_{0}^{1} = - \frac{1}{2} $$
This represents the green region in the image below:

We know that the area of the green region is $\frac{1}{2}$, because it's a triangle ($A=\frac{1}{2}Bh$), however the integral returns a "negative area". I put this in quotes, because negative area is really not defined. So we say the absolute area is the absolute value of all the total areas. The absolute area can be expressed by $ \int_{a}^{b} \left| f(x) \right| dx$.
For your particular example, $f(x)=3-x$, we want to find where $f(x)$ is negative and positive by finding $f(x)=0$, which in this case is $x=3$. Test the point $x=1$ to find $f(1) = 2 > 0$. Since $f(x)$ is linear, $f$ is positive on $[0,3)$ and negative on $(3,5]$. 
So to find the absolute area we need to negate all the "negative areas" or areas below the x-axis.
$$ A = \int_{0}^{3} (3-x) dx + (-1) \int_{3}^{5} (3-x) dx $$
$$ A = \left[ 3x - \frac{x^2}{2} \right]_0^3 - \left[ 3x - \frac{x^2}{2} \right]_3^5 = 4.5 - (-2) = 6.5 $$
Below is an image of the plot of $f(x)$. The green region is the "positive area" represented by the first integral, which comes out to $4.5$ and the red region is the "negative area" represented by the second integral, which comes out to $-2$.

