The first fundamental theorem of calculus

When you reach for it

An integral \int_a^b f(t)\,dt is a number. Let the upper end move and it becomes a function: the amount built up from a to x. Pipaluk measures sea ice in a fjord. If r(t) is the rate at which the ice thickens, in centimeters per day, then \int_0^x r(t)\,dt is how much thicker the ice is on day x than on day 0. A natural question is how fast that amount changes as x moves.

The answer, the first fundamental theorem of calculus, is that the rate of change is the integrand at the moving end. You reach for it whenever an integral has a variable limit and you need its derivative, its tangent line, or its rising and falling, and especially when the integral itself cannot be written with the usual functions. It is the first half of the fundamental theorem of calculus, the link between the two halves of the subject.

The procedure

Day 1: the accumulation function

Let f be continuous on an interval I and let a be a fixed number in I. The accumulation function of f from a is

The input of A is the upper limit x. The letter t is a dummy variable (The definite integral as a limit of Riemann sums): it runs from a to x inside the integral and is gone once the integral is a number, so \int_a^x f(s)\,ds is the same function. A different letter is used for the dummy so that x has one job, the end point. For x < a the integral runs backward and \int_a^x = -\int_x^a; also A(a) = 0 (Signed area and integrals from geometry).

When the graph of f is made of lines and circles, values of A come from geometry: A(x) is the area above the axis minus the area below it, between t = a and t = x.

Day 1: the mean value theorem for integrals

Theorem. If f is continuous on [a, b] with a < b, there is a number c in [a, b] with

In pictures: the region under the graph has the same signed area as a rectangle on [a, b] whose height is a value of f. That height, \displaystyle \frac{1}{b - a}\int_a^b f(x)\,dx, is the average value of f.

Proof. By the The extreme value theorem, f has an absolute minimum m = f(d) and an absolute maximum M = f(e) on [a, b]. Since m \le f(x) \le M there, the comparison property (Properties of definite integrals) gives

Dividing by b - a > 0, the number \displaystyle \mu = \frac{1}{b - a}\int_a^b f(x)\,dx lies between f(d) and f(e). If d = e, then f is constant and any c works. Otherwise f is continuous on the closed interval between d and e, so the intermediate value theorem gives a c between them with f(c) = \mu. Multiplying by b - a gives the theorem. \blacksquare

Day 1: the first fundamental theorem

Theorem. Let f be continuous on an interval I containing a, and let A(x) = \int_a^x f(t)\,dt. Then A is differentiable at every x in I (one-sided at an end point of I), and

Proof. Fix x in I and take h \ne 0 with x + h in I. Additivity splits off the new strip:

If h > 0, the mean value theorem for integrals on [x, x + h] gives a c_h in that interval with \int_x^{x+h} f(t)\,dt = f(c_h)\,h. If h < 0, it gives a c_h in [x + h, x] with \int_{x+h}^{x} f(t)\,dt = f(c_h)(-h), and reversing the limits gives again \int_x^{x+h} f(t)\,dt = f(c_h)\,h. Either way,

As h \to 0, x - |h| \le c_h \le x + |h| and both outer bounds tend to x, so c_h \to x by the squeeze theorem; because f is continuous at x, f(c_h) \to f(x). So the difference quotient has limit f(x), which is A'(x). \blacksquare

Two things follow at once. A differentiable function is continuous, so A is continuous on I. And A is an antiderivative of f: every continuous function has one, even when no formula for it exists.

Day 2: a function in the limit

If the upper limit is a differentiable function g(x) with values in I, then \int_a^{g(x)} f(t)\,dt = A(g(x)) is a composition, and the chain rule gives

A variable lower limit is turned into an upper one by reversing: \int_{h(x)}^{b} f(t)\,dt = -\int_b^{h(x)} f(t)\,dt, whose derivative is -f(h(x))\,h'(x). With both limits variable, split at any constant c in I, which additivity allows in any order:

None of this requires a formula for the integral. The integrands e^{-t^2} and \sin(t^2) are continuous everywhere, but no antiderivative of either can be written with polynomials, roots, exponentials, logarithms and trigonometric functions. Their accumulation functions are new functions, and the theorem still gives their derivatives exactly.

Worked examples

Boundaries

• The theorem needs f continuous on an interval containing a and x. At a jump of f the accumulation function is still continuous but has a corner, and A' does not exist there (Example 7).
• The proof uses continuity only at the point x itself, through f(c_h) \to f(x); continuity on the whole interval is what guarantees the mean value theorem for integrals on each strip.
• The interval must not reach past a point where f is undefined. For \displaystyle \int_1^x \frac{dt}{t} the interval is (0, \infty); at x = -1 the integral would cross t = 0, where \displaystyle \frac{1}{t} is unbounded, and the formula \displaystyle \frac{1}{x} says nothing.
• The chain rule version needs g differentiable and its values in I.
• The variable x may appear only in the limits. If it also appears in the integrand, as in \int_0^x x\,f(t)\,dt, take it outside first (Exercise 23).
• The mean value theorem for integrals needs continuity on the closed interval (Exercise 13), and its c need not be unique (Example 2).

Where this goes next

The theorem shows that every continuous function has an antiderivative, namely its accumulation function. Since any two antiderivatives on an interval differ by a constant, any antiderivative F of f satisfies \int_a^b f(x)\,dx = F(b) - F(a), which is the second fundamental theorem: integrals computed in a line instead of by limits of sums. Where a rate is given, accumulation and net change reads A as a total amount, and Graphing an accumulation function uses A' = f to find where the amount is largest. The functions \int_0^x e^{-t^2}\,dt and \int_0^x \sin(t^2)\,dt, defined only by integrals, become standard functions of their own in probability and in optics.

Exercises

Day 1: values from a graph

Let B(x) = \int_0^x g(t)\,dt, with g graphed below.

1. Find B(2) and B(4).
2. Find B(5) and B(7).
3. Find B'(1), B'(4.5) and B'(6).
4. Let C(x) = \int_4^x g(t)\,dt. Find C(0) and C(7).

Day 1: the mean value theorem for integrals

5. Find every c for f(x) = x^2 on [-1, 2], using \displaystyle \int_a^b x^2\,dx = \frac{b^3 - a^3}{3}.
6. Find every c for f(x) = 2x + 1 on [1, 4], computing the integral as a trapezoid.

Day 1: derivatives of accumulation functions

7. Find \displaystyle \dfrac{d}{dx}\displaystyle\int_3^x \sqrt{t^2 + 7}\,dt and its value at x = 3.
8. Find \dfrac{d}{dx}\displaystyle\int_0^x \cos(t^3)\,dt and its value at \displaystyle x = \sqrt[3]{\pi}.
9. Let \displaystyle H(x) = \int_1^x \frac{e^t}{t}\,dt for x > 0. Find the tangent line to the graph of H at x = 1.
10. Let \displaystyle F(x) = \int_0^x \frac{t - 1}{t^2 + 1}\,dt. Find F'(0) and F''(0).

Day 1: reasoning

11. Explain why \int_2^x \cos(t^3)\,dt and \int_2^x \cos(s^3)\,ds are the same function of x, and why \int_2^x \cos(x^3)\,dx is a poor way to write it.
12. Let f be continuous on \mathbb{R}, A(x) = \int_a^x f(t)\,dt and P(x) = \int_b^x f(t)\,dt. Show that A - P is a constant, and name it.
13. Let f(x) = 0 for 0 \le x < 1 and f(x) = 2 for 1 \le x \le 2. Show that no c in [0, 2] satisfies \int_0^2 f(x)\,dx = 2f(c), and say which hypothesis fails.

Day 2: chain rule and two variable limits

14. Find \dfrac{d}{dx}\displaystyle\int_0^{x^2} \sin(t^2)\,dt.
15. Find \displaystyle \dfrac{d}{dx}\displaystyle\int_{\cos x}^{1} \sqrt{1 + t^4}\,dt.
16. Find \dfrac{d}{dx}\displaystyle\int_1^{e^x} \ln t\,dt and simplify.
17. Find \displaystyle \dfrac{d}{dx}\displaystyle\int_x^{3x} \frac{1}{t}\,dt for x > 0, and say what the answer means about the integral.
18. Find \dfrac{d}{dx}\displaystyle\int_{-x}^{x} \cos(t^2)\,dt.
19. Let \displaystyle G(x) = \int_{x^2}^{x^3} \frac{1}{1 + t^4}\,dt. Find G'(1).
20. Evaluate \displaystyle \displaystyle\lim_{h \to 0} \frac{1}{h}\int_2^{2+h} \sqrt{1 + t^3}\,dt.
21. Find the tangent line to the graph of \displaystyle y = \int_1^{x^2} \frac{\cos(\pi t)}{t}\,dt at x = 1.
22. Let f(t) = 2 for t < 1 and f(t) = -1 for t \ge 1, and A(x) = \int_0^x f(t)\,dt on [0, 3]. Find A(x), A(3), and A'(x) wherever it exists.

Day 2: reasoning

23. Let F(x) = \int_0^x x\sin(t^2)\,dt. Find F'(x), and explain why it is not x\sin(x^2).
24. Prove that \dfrac{d}{dx}\displaystyle\int_{h(x)}^{b} f(t)\,dt = -f(h(x))\,h'(x) for f continuous and h differentiable, from the first fundamental theorem.
25. Let f be integrable on [a, b] with |f(t)| \le M there, and A(x) = \int_a^x f(t)\,dt. Show that |A(x + h) - A(x)| \le M|h|, and conclude that A is continuous even where f jumps.