Mathematics · Ch 16 — Integration

Second Fundamental Theorem of Integral Calculus

16.8.3

Second Fundamental Theorem of Integral Calculus

The Second Fundamental Theorem: The Bridge from Indefinite to Definite

The Second Fundamental Theorem of Integral Calculus makes evaluating definite integrals practical. Instead of calculating limits of sums, it lets you use an antiderivative (indefinite integral) to find the exact value.

Theorem 2 (Second Fundamental Theorem of Integral Calculus)

Let $f$ be a continuous function on the closed interval $[a, b]$ , and let $F$ be an antiderivative of $f$ (meaning $F'(x) = f(x)$ for all $x$ in $[a, b]$ ). Then:

\int_{a}^{b} f(x) \, dx = \big[ F(x) \big]_{a}^{b} = F(b) - F(a)

The definite integral of $f$ from $a$ to $b$ is simply the difference between the values of any antiderivative $F$ at the upper limit $b$ and the lower limit $a$ .

Remarks on the Theorem

Practical utility: the theorem provides a straightforward method for calculating a definite integral without computing the limit of a sum — the primary reason we can solve most definite-integral problems in practice.
Core operation: the crucial step is finding a function $F$ whose derivative is the integrand $f$ . This process solidifies the deep connection between differentiation and integration as inverse processes.
A critical condition on continuity: $f$ must be well-defined and continuous on the entire closed interval $[a, b]$ . If $f$ is not continuous at some point within $[a, b]$ , the theorem cannot be applied directly. For example, consider $\int_{-2}^{3} \frac{1}{x^2 (x^2 - 1)} \, dx$ . The integrand $f(x) = \frac{1}{x^2 (x^2 - 1)}$ is not defined (hence not continuous) at $x = 0$ and $x = \pm 1$ . Since $[-2, 3]$ contains $x = -1, 0, 1$ , evaluating this integral using the Second Fundamental Theorem in its basic form would be erroneous.

Watch out

Always check that the integrand $f(x)$ is continuous on the entire interval $[a, b]$ before applying the Second Fundamental Theorem. If it is not, a different approach (like splitting the integral at the points of discontinuity) is needed.

Steps for Calculating $\int_{a}^{b} f(x) \, dx$ …

Theorem 2

The Second Fundamental Theorem of Integral Calculus

This theorem is the bridge that turns the hard work of finding an antiderivative into a direct method for evaluating a definite integral. It tells us that instead of computing limits of sums, we can simply evaluate the antiderivative at the endpoints.

Statement of the Theorem

f

The hypotheses are precise and non-negotiable:

$f$ must be continuous on the closed interval $[a, b]$ .
$F$ must be an antiderivative of $f$ , meaning $F'(x) = f(x)$ for every $x$ in $[a, b]$ .

Watch out

If $f$ is not continuous on the entire interval $[a, b]$ , the theorem does not apply. For example, the integral $\int_{-2}^{3} \frac{1}{x^2(x^2-1)} \, dx$ is meaningless because the integrand is not defined for $-1 < x < 1$ , which lies inside $[-2, 3]$ .

Complete Proof

›Proof

We begin with the definition of the definite integral as a limit of Riemann sums. Let $P = \{x_0, x_1, x_2, \dots, x_n\}$ be a partition of $[a, b]$ where $a = x_0 < x_1 < x_2 < \dots < x_n = b$ , and let $\Delta x_k = x_k - x_{k-1}$ .

Since $F$ is an antiderivative of $f$ , we have $F'(x) = f(x)$ . By the Mean Value Theorem, applied to $F$ on each subinterval $[x_{k-1}, x_k]$ , there exists some $c_k$ in $(x_{k-1}, x_k)$ such that

$F(x_k) - F(x_{k-1}) = F'(c_k)(x_k - x_{k-1}) = f(c_k) \Delta x_k.$

Now sum this telescoping expression over all $k$ from $1$ to $n$ :

$\sum_{k=1}^n \big[F(x_k) - F(x_{k-1})\big] = \sum_{k=1}^n f(c_k) \Delta x_k.$

The left-hand side telescopes:

$F(x_1) - F(x_0) + F(x_2) - F(x_1) + \dots + F(x_n) - F(x_{n-1}) = F(x_n) - F(x_0) = F(b) - F(a).$

Therefore,

$F(b) - F(a) = \sum_{k=1}^n f(c_k) \Delta x_k.$

Now take the limit as the norm of the partition (the maximum subinterval length) goes to zero. The right-hand side becomes the Riemann integral of $f$ over $[a, b]$ , provided the limit exists. Since $f$ is continuous on $[a, b]$ , it is Riemann integrable, and the limit exists and equals the definite integral. Hence,

$F(b) - F(a) = \lim_{\|P\| \to 0} \sum_{k=1}^n f(c_k) \Delta x_k = \int_a^b f(x) \, dx.$

This completes the proof. …

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Second Fundamental Theorem of Integral Calculus