Fundamental theorem of calculus

The fundamental theorem of calculus (FTC) establishes the connection between derivatives and integrals, two of the main concepts in calculus. It also gives us an efficient way to evaluate definite integrals.

Suppose that f(x) is continuous on an interval [a, b]. Then F is a function that satisifies F'(x) = f(x) if and only if

for all x in [a, b].

To better understand the FTC, notice that when we have picked a particular x, the integral,

is a number, so depends only on x, not t. t in this case is a placeholder that runs through all the values between a and x when we integrate f from a to x.

Example 1

Evaluate the following integral:

From the FTC, if we can find a function F that satisfies F'(x) = x2, then:

One such F that satisfies F'(x) = x2 is:

by the power rule, so:

Corollary: Integration and differentiation are reverse processes

Consider the function F on [a, b] defined by the integral

for some continuous function f on [a, b]. Arguably the most important consequence of the FTC is that the derivative of F is:

In other words, differentiating a function F defined as an integral of f will gives us back f, implying that differentiation and integration are reverse operations. To see why this is implied by the FTC, notice that:


for all x in [a, b]. Therefore, by the FTC,

For example, given

The derivative, F'(x), is:

The FTC and chain rule

Using the corollary described in the previous section,

along with the chain rule, we can find the derivatives of functions like those below.

Example 2

Notice that the upper bound of the definite integral is x2 instead of x, so to set up the chain rule, we make substitution u = x2. The chain rule implies that:


From the corollary,


and so


Be wary of independent and placeholder variables. To avoid errors, it's important to remember exactly which variables we are integrating or differentiating with respect to.

Example 3

Find the derivative of

This problem could potentially cause confusion since we have the independent variable, x, mixed in with the placeholder variable, t, under the integral, and differentiating or integrating with respect to the wrong variable would result in the wrong answer. However, since the integral is with respect to t because of the dt term, the x2 term can be treated as constant, so we can factor it out of the integral to get

Now we can use the product rule to find :

and from the corollary, so


Proof of the fundamental theorem of calculus

Recall the original statement of FTC: Suppose f(x) is continuous on [a, b]. Then F is a function that satisfies F'(x) = f(x) if and only if

for all x in [a, b].

To prove that if , then , we first assume that

for all x in [a, b]. Adding F(a) to both sides gives us an expression for F(x):


The limit definition of F'(x) is:

Refer to the derivatives page for more information on the limit definition of a derivative.

Plugging equation (1) into the limit definition of a derivative,


As h becomes smaller, x + h approaches x, or more formally, . By continiuity of f, , so as h goes to 0, the function values f(t) for all t in the interval x ≤ t ≤ x + h also satisfy , so


Now to prove that if F'(x) = f(x) then , we assume that F'(x) = f(x) for all x in [a, b]. This also means that when we fix an x in [a, b],

for all t in the interval a ≤ t ≤ x. Plugging the above substitution into the integral gives


We can break the interval [a, x] into n intervals [a, a + h], [a + h, a + 2h],..., [a + (n - 1)h, a + nh] where . This sets us up for viewing the right side as the limit of Riemann sums:


where each ti is a random value in the ith interval [a + (i - 1)h, a + ih]. By the Mean Value Theorem, there exists a ti* in the ith interval [a + (i - 1)h, a + ih] that satisfies

Since the ti's are randomly chosen, we can replace each ti with ti*, so the limit of the Riemann sums becomes

(4)    = 

On the right side, after canceling the h's, we get the telescoping series:


From extrapolating the above, we see that all the middle terms cancel out, leaving only F(a + nh) and F(a) at the beginning and end of the telescoping sum. Also note that F(a + nh) and F(a) at the beginning and end of the telescoping sum. Also note that . Therefore,


Chaining together the series of substitutions given by (2) - (5), we get

whenever F'(x) = f(x) for all x in [a, b].