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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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    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).2011-12-08