The Derivative of a Definite Integral Function

Similar to how one can think of a derivative as a function that yields a tangent-slope for any given $x$, one can create a function using a definite integral that gives the area under the graph of some non-negative valued function from some specified value to any given $x$.

More generally, for any continuous and integrable function $f$ and some specified value $a$, one can create the function $F$ defined by

$$F(x) = \int_a^x f(t)\,dt$$

Note that we use a variable of integration different from $x$ (i.e., $t$), so that we don't confuse it with the independent variable for $F$.

Of course, we have spent a long time now developing the ability to find the derivative of any function expressible as a combination of the simple functions typically encountered in an algebra or precalculus course (e.g., root functions, trigonometric functions, exponential and logarithmic functions, etc.).

A function defined by a definite integral in the way described above, however, is potentially a different beast. One might wonder -- what does the derivative of such a function look like?

Of course, we answer that question in the usual way. We apply the definition of the derivative.

$$\begin{array}{rcl} \displaystyle{F'(x)} &=& \displaystyle{\lim_{h \rightarrow 0} \frac{F(x+h) - F(x+h)}{h}}\\\\ &=& \displaystyle{\lim_{h \rightarrow 0} \frac{\displaystyle{\int_a^{x+h} f(t)\,dt - \int_a^x f(t)\,dt}}{h}} \end{array}$$

Recalling that one of the properties of definite integrals assures us that $\int_a^x f(t)\,dt + \int_x^{x+h} f(t)\,dt = \int_a^{x+h} f(t)\,dt$, we see that the difference in the numerator above must equal $\int_x^{x+h} f(t)\,dt$.


$$F'(x) = \lim_{h \rightarrow 0} \, \frac{1}{h} \int_x^{x+h} f(t)\,dt$$

Notably, the expression on the right is, by definition, the average value of $f(t)$ on the interval $[x,x+h]$, which must (by the Mean Value Theorem for Integrals) equal $f(c_h)$ for some $c_h$ in this interval.

Of course, as $h$ changes, the value of $c_h$ may change.

To find $\lim_{h \rightarrow 0} f(c_h)$, we can use the squeeze theorem. Note that

$$x \le c_h \le x+h \quad \textrm{ and } \quad \lim_{h \rightarrow 0} x = \lim_{h \rightarrow 0} (x+h) = x, \quad \textrm{ so } \quad \lim_{h \rightarrow 0} c_h = x$$

Finally, appealing to the continuity of $f$, we have

$$F'(x) = \lim_{h \rightarrow 0} f(c_h) = f(x)$$

We have thus found the derivative we sought. The derivative of $\displaystyle{\int_a^x f(t)\,dt}$ is $f(x)$.

This result is not only wonderfully simple, but also establishes the inverse-like relationship between finding the value of a derivative and finding the value of a definite integral. That is to say, one can "undo" the effect of taking a definite integral, in a certain sense, through differentiation.

Such a relationship is of course of significant importance and consequence -- and thus forms the other half of the Fundamental Theorem of Calculus (i.e., "Part I") presented below.

The Fundamental Theorem of Calculus

Part I : If $f$ is continuous on $[a,b]$, and $\displaystyle{F(x) = \int_a^x f(t)\,dt}$, then $F'(x) = f(x)$ for all $x$ in $(a,b)$.

Part II : If $f$ is continuous on $[a,b]$ and $F$ is an antiderivative of $f$ in that same interval, then

$$\int_a^b f(x)\,dx = F(b)-F(a)$$