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i have sat for 2 hours trying to understand how the area of R1+R3=R2 cant really get this becuz R2 has a negative section and i think f(c) should be lower. Once again, how is R1 and R3= R2? R1 and R3 is all positive area whereas R2 has negative area so in my opinion the average area of R2 should be much lower because of the negative section of R2...Can someone please explain this to me? These are part of lecture notes.

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As geometric regions, the areas of $R_1$, $R_2$, and $R_3$ must be positive. Geometric area must be positive. (Have you ever met a triangle whose area was $-10$ cm$^2$?). And so it makes sense to claim that $Area(R_1) + Area(R_3) = Area(R_2)$.

Where many students get confused is in the interpretation of $\int f(x) \;dx$ as a sum and difference of certain areas. If continuous $f < 0$ on $[a,b]$, then $\int_a^b f(x)\;dx$ has a negative sign, and it represents the negative of the area of the region below the $x$-axis but above the curve $y = f(x)$. Notice the terminology: "Negative of the area of that region". The implication is that the (geometric) area of the region is positive, and taking the negative of it gives you the value of $\int_a^b f(x)\;dx$.

Hope this helps!

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i dont get how area can be negative, isnt all area suppose to be >=0? also, Area R1 and R3 arent half of the [a,b] interval so how can they say that R1+R3=R2? – Raynos Dec 6 '11 at 23:58
Yes, all area is supposed to be $\geq 0$, as I mentioned above. What do you mean by "Area R1 and R3 aren't half of the $[a, b]$ interval"? I'm only talking about area as in the amount of paint needed to paint those regions. Try this experiment: draw tiny squares in those regions $R_1, R_2$ and $R_3$, and count them. See how close the total number of squares for $R_1$ and $R_3$ match the total for just $R_2$ (you'll have to use tiny squares to get a reasonable approximation of the areas, though). – Shaun Ault Dec 8 '11 at 0:48
Also, the notation $\int_a^b f(x)\;dx$ must not be thought of as the same as a geometric area. $\int_a^b f(x) \;dx$ is often not an area calculation (for example, when finding work in physics: $W = \int_a^b F(x)\;dx$, where $F$ is the force acting at distance $x$). – Shaun Ault Dec 8 '11 at 0:51

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