For example in statistics we learn that mean = E(x) of a function which is defined as
$$\mu = \int_a^b xf(x) \,dx$$
however in calculus we learn that
$$\mu = \frac {1}{b-a}\int_a^b f(x) \,dx $$
What is the difference between the means in statistics and calculus and why don't they give the same answer?
thank you
This seems to be based on confusion resulting from resemblance between the notations used in the two situations.
In probability and statistics, one learns that $\displaystyle\int_{-\infty}^\infty x f(x)\,dx$ is the mean, NOT of the function $f$, but of a random variable denoted (capital) $X$ (whereas lower-case $x$ is used in the integral) whose probability density function is $f.$ This is the same as $\displaystyle \int_a^b xf(x)\,dx$ in cases where the probability is $1$ that the random variable $X$ is between $a$ and $b.$ (The failure, in the posted question, to distinguish betweeen the lower-case $x$ used in the integral and the capital $X$ used in the expression $\operatorname E(X)$ is an error that can make it impossible to understand expressions like $\Pr(X\le x)$ and some other things.)
In calculus, the expression $\displaystyle \frac 1 {b-a} \int_a^b f(x)\,dx$ is the mean, NOT of any random variable $X,$ but of the function $f$ itself, on the interval $[a,b].\vphantom{\dfrac11}$
Notice that in probability, you necessarily have $\displaystyle \int_a^b f(x)\,dx=1$ and $f(x)\ge 0,$ and the mean $\displaystyle \int_a^b xf(x)\,dx$ is necessarily between $a$ and $b.$ But none of that applies to the calculus problem, since the quantity whose mean is found is on the $f(x)$-axis, not on the $x$-axis. $$\S \qquad\qquad \S \qquad\qquad \S$$ Postscript: Nine people including me have up-voted "Jack M"'s comment, so just to satisfy that point of view I will add some things.
If $f$ is the density function of the probability distribution of the random variable (capital) $X,$ then the mean of $g(X)$ (where $g$ is some other function) is $$ \int_{-\infty}^\infty g(x) f(x)\,dx. $$ Applying that to the situation in calculus, one can say that the density function of the uniform distribution on the interval $[a,b]$ is $1/(b-a),$ so if $X$ is a random variable with that distribution, then $$ \operatorname E(f(X)) = \int_a^b f(x) \frac 1 {b-a} \, dx. $$ And a random variable $X$ itself can be regarded as a function whose domain is a sample space $\Omega,$ with the probability measure $P$ assigning probabilities to subsets of $\Omega,$ and then you have $$ \operatorname E(X) = \int_\Omega X(\omega)\, P(d\omega). $$
Maybe you can consider the second form of mean as a sample mean, analogue to $\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$ in the discrete case.
That is: the function $f(x)$ in this case would be the realization of a random variable.
Please consider the notion of a stochastic process, where a collection of random variables $X$ is indexed, or associated with a continuous deterministic variable $t$. For instance, $X(t)$ could model the weather temperature in a given moment of time $t$, during the interval $[ a, b]$. So, $X(a)$ would not be a value, but a whole random variable, with a given mean, variance, etc. And when you write $X(t)$ you have a collection of random variables, one for each $t \in [a, b]$.
If all these variables have the same properties, one say $X(t)$ is stationary, and therefore, $\mu$, the mean of $X(t)$, can be estimated using the series of values of a realization of $X(t)$ -- in our case, some actual measurements of the weather temperature in a given range of time. In that case, your function $f(x)$ would be renamed $x(t)$ (with a lowercase $x$) to indicate one realization of the stochastic process $X(t)$. And then
$$\bar{x} = \frac{1}{b - a} \int_a^b x(t)\,dt$$
would be an estimator of $\mu$, the actual mean of the stationary process $X(t)$.
