Question

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 \(\mathop \smallint \limits_a^b f\left( x \right)dx =\)
1. f (ξ) (b - a)
2. f (b) (ξ - a)
3. f (a) (b - ξ)
4. 0

Answer 1

Correct Answer - Option 1 : f (ξ) (b - a)

Concept:

Mean value theorem for integrals:

Let f be continuous on [a, b]. Then there is a point x_o in (a, b) such that

\(f\left( {{x_o}} \right) = \frac{1}{{b - a}}{\rm{\;}}\mathop \smallint \limits_a^b f\left( x \right)dx\)

Calculation:

\(f\left( \xi \right) = \frac{1}{{b - a}}{\rm{\;}}\mathop \smallint \limits_a^b f\left( x \right)dx\)

\(\mathop \smallint \limits_a^b f\left( x \right)dx = f\left( \xi \right)\left( {b - a} \right)\)