Question

If $f$ is continuous on $[a, b]$, then there is at least one number $c$ between $a$ and $b$ such that

• A. $\int_a^b f(x) \, dx = f(c)(b-a)$
• B. $\int_a^b f(x) \, dx = f(a)(c-b)$
• C. $\int_a^b f(x) \, dx = f(b)(c-a)$
• D. $\int_a^b f(x) \, dx = f'(c)(b-a)$

Hint:

The area under a continuous curve over an interval is always equal to the width of the interval multiplied by the function's value at some intermediate point $c$ in that interval.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This theorem is the Mean Value Theorem for Integrals. It guarantees that a continuous function on a closed interval attains its average value at some point in that interval.

Step 2: Detailed Explanation:
The average value of a continuous function $f(x)$ over the interval $[a, b]$ is defined as: \[ f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx \]
The Mean Value Theorem for Integrals states that if $f(x)$ is continuous on $[a, b]$, there must exist at least one number $c$ in the interval $[a, b]$ such that the function's value at $c$ equals this average value: \[ f(c) = \frac{1}{b-a} \int_a^b f(x) \, dx \]
Rearranging this equation by multiplying both sides by $(b-a)$: \[ \int_a^b f(x) \, dx = f(c)(b-a) \]
Geometrically, this means that the area under the curve $y = f(x)$ from $a$ to $b$ is equal to the area of a rectangle with width $(b-a)$ and height $f(c)$.

Step 3: Final Answer:
There exists a $c$ such that $\int_a^b f(x) \, dx = f(c)(b-a)$.