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:
$R_1+R_3=R_2$