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

The correct option is f(ξ)(b-a).

Image Analysis:
The provided image displays the mathematical expression for a definite integral:
${\int }_a^{b}f\left(x\right)dx=$
where a is the lower limit of integration, b is the upper limit of integration, and f(x) is the integrand function.

Step-by-Step Explanation:
The Mean Value Theorem for Integrals (often called the First Mean Value Theorem for Integrals) states that if a function f(x) is continuous on a closed interval [a, b], there exists at least one value ξ in the interval [a, b] such that the definite integral of the function over the interval is equal to the value of the function at ξ multiplied by the length of the interval.

Mathematically, this is expressed as:
${\int }_a^{b}f\left(x\right)dx=f\left(\xi \right)(b-a)$

Derivation and Logic:
1. Since f(x) is continuous on the closed and bounded interval [a, b], by the Extreme Value Theorem, f(x) must attain its absolute minimum value m and absolute maximum value M on this interval:
$m\le f\left(x\right)\le M$
for all x in [a, b].

2. Integrating this inequality from a to b gives:
${\int }_a^{b}mdx\le {\int }_a^{b}f\left(x\right)dx\le {\int }_a^{b}Mdx$

3. Evaluating the outer integrals:
$m(b-a)\le {\int }_a^{b}f\left(x\right)dx\le M(b-a)$

4. Dividing the entire inequality by the length of the interval $(b-a)$ (assuming b > a):
$m\le \frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx\le M$

5. By the Intermediate Value Theorem, since f(x) is continuous, it must take on every value between its minimum m and maximum M. Therefore, there must exist some value ξ in the interval [a, b] such that:
$f\left(\xi \right)=\frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx$

6. Multiplying by $(b-a)$ yields the final relationship:
${\int }_a^{b}f\left(x\right)dx=f\left(\xi \right)(b-a)$