$\frac{\int_{a}^{b} f(x)\,dx}{b-a}$
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The mean value (or average value) of a function $f(x)$ over a closed interval $[a, b]$ provides the average height of the function across that interval. It is formally defined using definite integration.
To find the mean value of $f(x)$ from $x = a$ to $x = b$, we calculate the definite integral of the function over the interval, which represents the total area under the curve, and then divide this area by the length of the interval, $(b-a)$.
The formula is:
$ f_{\text{avg}} = \frac{1}{b-a} \int_{a}^{b} f(x)\,dx $
We examine the provided options to identify the correct formula for the mean value:
Therefore, the correct formula for the mean value of a function $f(x)$ from $x = a$ to $x = b$ is $\frac{\int_{a}^{b} f(x)\,dx}{b-a}$.
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If f is the derivative of some function on [a, b], then there exists a number c in (a, b) such that Integral of f with respect to x =
Which condition is not required in checking for Taylor's theorem?
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 \(\mathop \smallint \limits_a^b f\left( x \right)dx =\)