According to the Mean Value Theorem, for a continuous function f (x) in the interval [a,b], there exists a value , ξ in this interval such that

Correct Answer: f(ξ)(b-a)

Step-by-step Explanation:

1. Analyze the Given Image:
The image displays the left-hand side of a definite integral equation:

${\int }_a^{b}f\left(x\right)dx=$
This represents the definite integral of a function $f\left(x\right)$ over the interval $[a,b]$.

2. Apply the Mean Value Theorem for Integrals:
The Mean Value Theorem for Integrals (also known as the First Mean Value Theorem for Integrals) states that if a function $f\left(x\right)$ is continuous on the closed interval $[a,b]$, there exists at least one value $\xi$ in the interval $[a,b]$ such that the function value at $\xi$ is equal to the average value of the function over the interval.

The average value of the function $f\left(x\right)$ on $[a,b]$ is given by:

$f_{\text{avg}}=\frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx$

3. Formulate the Equation:
Setting the function value $f\left(\xi \right)$ equal to this average value gives:

$f\left(\xi \right)=\frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx$

To find the value of the integral itself, we multiply both sides of the equation by $(b-a)$:

${\int }_a^{b}f\left(x\right)dx=f\left(\xi \right)(b-a)$

Therefore, the expression that completes the equation shown in the image is $f\left(\xi \right)(b-a)$.