Is a Riemann integral of a real-valued function a number or a function?

Question

For example, if we define $F(x)=\int^x_a f(t)dt$, where $f$ is Riemann integrable, then $F(x)$ is a function. Or for a 2 variables real-valued integrable function $f(x, y)$, $G(x)=\int^b_a f(x,y)dy$, then $G(x)$ is a function. But for $\int_a^b f$, is it just a number rather than a function?

Answer 1

The Riemann integral is a number.

If $f : [a,b]\to\mathbb{R}$ is a Riemann integrable function then, by definition, the Riemann integral of $f$ is given by $$ \int_{a}^{b} f(x)\,\mathrm{d}x = \lim_{\delta\to 0} \sum_{n=1}^{N} f(x_n^*) (x_{n}-x_{n-1}), $$ where $a = x_0 < x_1 < \dotsb < x_N = b$ is any partition of $[a,b]$ with $x_n - x_{n-1} < \delta$ for all $n$, and $x_n^* \in [x_{n-1}, x_n]$.

For any partition of $[a,b]$, the sum on the right will be a will be a real number (since $f(x_n^*)$ and $x_n-x_{n-1}$ are real for every $n$). Thus the term on the right is the limit of a sequence of real numbers. The limit of a sequence of real numbers is, again, a real number, hence the Riemann integral is a real number. Since the variable is not really important here, the above notation will often be abbreviated to $$ \int_{a}^{b} f, $$ but this notation indicates the same number discussed above.

That being said, there is something else going on, which might explain your confusion. Roughly speaking, the Fundamental Theorem of Calculus tells us that if there is a function $F$ with derivative $f$, then $$ F(x) = \int_{a}^{x} f(t)\,\mathrm{d}t, $$ where $a$ is an arbitrarily chosen constant in the domain of $F$, and $x$ is any element in the domain of $F$. Note that, even here, the Riemann integral is a real number: first, we fix some some point $x$ in the domain of $F$, and then we determine the value of $F$ at that point (that is, we determine $F(x)$) by evaluating a Riemann integral. This process can be used to define a function—it gives us a way of evaluating $F$ at every point in its domain—but the integral itself is not a function.

Finally, there is one other notational wrinkle: if the derivative of $F$ is $f$, then we say that $F$ is an antiderivative of $f$.^[1] The Fundamental Theorem of Calculus could be restated as "If $F$ is an antiderivative of $f$, then $$ F(x) = \int_{a}^{x} f(x)\,\mathrm{d}x, $$ where $a$ and $x$ are as above." Because of this link between derivatives, antiderivatives, and (Riemann) integrals, it is common to write $$ F = \int f $$ in order to indicate that $F$ is an antiderivative of $f$. In this notation, the "integrals" have no limits, and $\int f$ doesn't really refer to an integral at all (though it is often called the "indefinite integral"). In this case, it would be fair to say that $\int f$ is a function,^[2] but $\int f$ no longer denotes a Riemann integral.

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[1] Note the use of the indefinite article: $F$ is an antiderivative, not the antiderivative. This is important, because if $F'(x) = f(x)$ for all appropriate $x$, then $$ \frac{\mathrm{d}}{\mathrm{d}x} \bigl(F'(x) + C\bigr) = f(x) $$ for any constant $C$. Hence antiderivatives are not unique; both $x \mapsto F(x)$ and $x\mapsto F(x) + C$ are antiderivatives of $f$.

[2] It may also be worth noting that it is a little bit of a lie to say that $$ \int f $$ is a function—in reality, it is probably better to think of it as an equivalence class of functions, which are equivalent up to addition of a constant. But this distinction is typically irrelevant in elementary courses, and is easily dealt with in more advanced settings.

Answer 2

That depends on how you consider the parameters $a$ and $b$, if they are fixed numbers, then $\int_a^bf$ is a number, but if you consider them as variables, you have a function of two variables

$$F(a,b)=\int\limits_a^bf(t)dt,$$

where $F(x)=\int_a^xf(t)dt$ is just the particular case obtained by fixing the value of $a$ and changing the notation of the variable $b$ by $x$.

So, it is mostly a matter of perspective.