If $f$ is continuous on $[a, b]$ then there exists a number $c$ in $(a, b)$ such that:

\[ \int_a^b f(x) \, dx = f(c)(b – a) \]

### Understanding the Mean Value Theorem for Integrals

Hello students! Today, we will learn about the Mean Value Theorem for Integrals. This theorem helps us find the average value of a function over an interval.

#### What Does the Theorem Say?

If a function $f(x)$ is continuous on the interval $[a, b]$, then there is at least one number $c$ between $a$ and $b$ such that:

\[\begin{align*} \int_a^b f(x) \, dx = f(c) \times (b – a) \end{align*}\]

This means the total area under the curve from $a$ to $b$ equals the area of a rectangle with width $(b – a)$ and height $f(c)$.

#### Why Is This Important?

It tells us that even if the function changes values, there is some point where the function’s value represents the average over the interval.

#### Simple Example

Let’s consider $f(x) = x$ on the interval $[2, 5]$.

**Step 1: **Compute the integral:

\[\begin{align*} \int_2^5 x \, dx = \left[ \dfrac{x^2}{2} \right]_2^5 = \dfrac{25}{2} – \dfrac{4}{2} = \dfrac{21}{2} \end{align*}\]

**Step 2: **Calculate $(b – a)$:

\[\begin{align*} b – a = 5 – 2 = 3 \end{align*}\]

**Step 3: **Set up the equation:

\[\begin{align*} \int_2^5 f(x) \, dx = f(c) \times (b – a) \implies \dfrac{21}{2} = f(c) \times 3 \end{align*}\]

**Step 4: **Solve for $f(c)$:

\[\begin{align*} f(c) = \dfrac{21}{2} \div 3 = \dfrac{7}{2} \end{align*}\]

**Step 5: **Find $c$ such that $f(c) = \dfrac{7}{2}$:

\[\begin{align*} f(c) = c = \dfrac{7}{2} \end{align*}\]

Since $\dfrac{7}{2} = 3.5$, and $3.5$ is between $2$ and $5$, this satisfies the theorem.

#### Key Points to Remember

– The function must be continuous on $[a, b]$.

– There exists a $c$ in $(a, b)$ where the theorem holds.

– The theorem connects the integral of a function to its average value.

#### Conclusion

The Mean Value Theorem for Integrals helps us find where a function’s value equals its average over an interval.

\[ \begin{array}{|c|} \hline \text{\Large Mean Value Theorem for Integrals:} \\ \text{If } f \text{ is continuous on } [a, b], \text{ then there exists a number } c \text{ in } (a, b) \text{ such that:} \\ \int_a^b f(x) \, dx = f(c)(b – a) \\ \hline \end{array} \]