Question

theorem: the mean - value theorem if the function f is continuous on the closed interval a,b, then there exists a number c in a,b s.t. $int_{a}^{b}f(x)dx = f(c)(b - a)$ proof:

Explanation:

Step1: Define the average value of a function

The average value of a function $y = f(x)$ on the interval $[a,b]$ is given by $\bar{y}=\frac{1}{b - a}\int_{a}^{b}f(x)dx$.

Step2: Use the extreme - value and intermediate - value theorems

Since $f(x)$ is continuous on $[a,b]$, by the extreme - value theorem, $f(x)$ attains a minimum value $m$ and a maximum value $M$ on $[a,b]$. We know that $m\leqslant\frac{1}{b - a}\int_{a}^{b}f(x)dx\leqslant M$.

Step3: Apply the intermediate - value theorem

By the intermediate - value theorem, since $f(x)$ is continuous on $[a,b]$ and $m\leqslant\frac{1}{b - a}\int_{a}^{b}f(x)dx\leqslant M$, there exists a number $c\in[a,b]$ such that $f(c)=\frac{1}{b - a}\int_{a}^{b}f(x)dx$.

Step4: Rearrange the equation

Multiply both sides of $f(c)=\frac{1}{b - a}\int_{a}^{b}f(x)dx$ by $(b - a)$ to get $\int_{a}^{b}f(x)dx=f(c)(b - a)$.

Answer:

The proof is completed as shown above.