# average value theorem graph confusion definite integral [duplicate]

Possible Duplicate:
average value theorem and calculus integration

I have sat for 2 hours trying to understand how the area of $R_1+R_3=R_2$ can't really get this. I don't understand how $R_1+R_3=R_2$... so $R_1$ is above the average, $R_3$ is also above average and $R_2$ is below average... $R_1$ and $R_3$ don't look like they are half of the graph...

Can someone PLEASE explain how they came up with $R_1+R_3=R_2$? It just makes no sense how they came up with that.

-

## marked as duplicate by Gerry Myerson, t.b., Asaf Karagila, J. M., Zev ChonolesDec 13 '11 at 3:28

Hint (actually just telling you): "geometrically" in this context probably means "unsigned area." – Neal Dec 7 '11 at 2:37

The intermediate value of integrals says that: $$\exists c \in (a,b) \therefore f(c) = \frac{1}{b-a} \int_{a}^{b}f(x)dx$$ or $$\exists c \in (a,b) \therefore f(c) \times (b-a) = \int_{a}^{b}f(x)dx$$

it means the area calculated by integral is equal to the area of a rectangle with the length of $b-a$ and width of $f(c)$.

Suppose we want to calculate $\int_{a}^{b}(f(x)-f(c))dx$ (Actually we have shifted down the graph of function then calculate the restricted area between the graph of the function and the coordinates between $a$ and $b$). Thus for the graph of the example function we have:

$$\int_{a}^{b}(f(x)-f(c))dx = R_1+R_2+R_3$$ (as we know $R_2$ is negative)

so in this case the $f(c)$ of the The intermediate value of integrals will be equal to zero , so the area of the rectangle is zero, hence the corresponding integral is zero. i.e.:

$$\int_{a}^{b}(f(x)-f(c))dx = 0 \Longrightarrow$$ $$R_1+R_2+R_3=0 \Longrightarrow$$ $$R_1+R_3=-R_2$$

as $R_2$ is negative geometrically we have:

$$R_1+R_3=R_2$$

-