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Calculus I · 120 h · Topic 8 of 10

The fundamental theorem of calculus

Differentiation and integration are inverse operations. That result, which seems obvious, is the theorem that unifies all of calculus and lets you compute areas without adding infinitely many terms.

Area function Newton–Leibniz rule Antiderivative Derivative and integral Applications

01The area function

Up to here the definite integral gave a number. But if you hold the lower endpoint fixed and move the upper one, a function appears: the one that accumulates area as it advances.

\[ A (x) = \int_a^x f (t) d t \] The variable of integration is called t so it is not confused with the upper limit x. A(a) = 0, and A increases where f is positive.

This function is exactly the charge accumulated by a capacitor from the current, or the energy consumed by an installation from the power. Accumulating is integrating with a variable limit.

Lab · accumulating area

Above, f and the area accumulated up to x. Below, the function A(x) as it is drawn. Look at where A increases, where it decreases and where it has its extrema.

02First theorem: differentiating the area gives back the function

\[ \frac{d}{d x} \int_a^x f (t) d t = f (x) \] If f is continuous, the area function is differentiable and its derivative is f itself. Integrating and differentiating cancel.

The idea of the proof fits in two lines: in going from x to x + h, the area grows by a narrow strip of width h and height approximately f(x), so A(x + h) − A(x) ≈ f(x)·h. Dividing by h and taking the limit, we get A′(x) = f(x). The “approximately” becomes exact by the mean value theorem for integrals.

What this theorem solves

Before it, computing an area required adding infinitely many rectangles. After it, it is enough to find a function whose derivative is the integrand. That is why the work of finding antiderivatives is worth it: it is the shortcut to all areas.

03Second theorem: the Newton–Leibniz rule (Barrow’s rule)

\[ \int_a^b f (x) d x = F (b) -F (a) \quad \operatorname{with} \; F' = f \] Any antiderivative works: the constant cancels in the subtraction. That is why, in a definite integral, the + C does not appear.
Worked example · the area that used to cost infinitely many sums

∫₀³ x² dx: an antiderivative is F(x) = x³/3.

F(3) − F(0) = 27/3 − 0 = 9. The same 9 that the Riemann sums of the previous topic were approaching, now in one line.

Two cautions with definite integrals

  • Substitution: if you change variables, you must also change the limits, or go back to the original variable before evaluating. With u = x² + 1, the integral from x = 0 to x = 2 becomes one from u = 1 to u = 5.
  • Discontinuities: the Newton–Leibniz rule requires f to be continuous on all of [a, b]. Applying it “by rote” to ∫−11 dx/x² gives −2, an impossible result for a positive function: that integral is improper and diverges.

04Differentiating integrals with variable limits

Combining the first theorem with the chain rule:

\[ \frac{d}{d x} \int_{u (x)}^{v (x)} f (t) d t = f (v (x)) v' (x) -f (u (x)) u' (x) \]

It is used, for example, to differentiate expressions for energy accumulated between two moving instants, and it is the gateway to the general Leibniz rule that appears in Calculus II.

05The two directions in a circuit

Differentiating and integrating, back and forth

The capacitor relation is written both ways: i = C·dv/dt and v(t) = v(0) + (1/C)∫₀t i dτ. They are the same law, read in the two directions that the fundamental theorem authorizes. The same goes for the inductor.

  • Energy meter. The kWh meter integrates the instantaneous power: its reading is literally the area function of p(t). Differentiating that reading gives back the power.
  • Dual-slope ADC. It integrates the input voltage for a fixed time and then integrates a reference until it returns to zero; the time it takes is proportional to the input. The whole method is the fundamental theorem built into a circuit, and that is why it rejects mains noise so well: it averages.
  • Charging a capacitor with pulsed current. The final voltage does not depend on the shape of the pulses, only on the total area: on the charge delivered.
  • Average value of a signal. A well-chosen low-pass filter is an approximate integrator: that is why the average value of the signal appears at the output of a PWM, with an RC.

06In the lab

Exercise 1 · Newton–Leibniz versus Riemann

Compute ∫₀π sin x dx with the Newton–Leibniz rule and with Riemann sums of 10, 50 and 200 terms. Check that the sum converges to 2 and estimate how many terms are needed for three correct decimal places.

Exercise 2 · PWM and its average value

Generate a PWM signal with variable duty cycle, filter it with an RC and measure the DC output voltage. Check that it is V·D, the average value, and explain the result as an integral over a period.

Exercise 3 · Measured area function

Capture the charging current of a capacitor and compute the accumulated charge point by point in a spreadsheet (sum of i·Δt). Plot q(t) next to the measured voltage and check that they are proportional, with constant C.

07Common mistakes

  • Applying the Newton–Leibniz rule across a discontinuity of the integrand inside the interval.
  • Changing variables and not changing the limits.
  • Writing the + C in a definite integral.
  • Differentiating ∫ax² f(t)dt as f(x), forgetting the chain-rule factor 2x.
  • Confusing the variable of integration with the upper limit: that is why t is written inside.
  • Believing that an op amp integrator integrates forever: without a feedback resistor, any offset accumulates and saturates the output.

08Self-assessment

If A(x) = ∫₀x (t² + 1) dt, what is A′(x)?

x² + 1, by the first fundamental theorem.

Compute ∫₁e dx/x.

ln x between 1 and e: ln e − ln 1 = 1.

Why is the constant of integration not needed in a definite integral?

Because it cancels in the subtraction: (F(b) + C) − (F(a) + C) = F(b) − F(a).

Differentiate G(x) = ∫₀x² sin t dt.

G′(x) = sin(x²)·2x: first theorem plus the chain rule.

A current of 2 mA charges a 1 µF capacitor for 5 ms from 0 V. Final voltage?

q = ∫i dt = 2 mA · 5 ms = 10 µC, and v = q/C = 10 µC/1 µF = 10 V.

What does an electric energy meter measure, mathematically?

The area function of the instantaneous power: W(t) = ∫p dt. Its derivative is the power.

09Further reading

  • James Stewart. Single Variable Calculus: Early Transcendentals. 7th ed. (Spanish edition, “Cálculo de una variable. Trascendentes tempranas”), Cengage Learning, 2012. Section 5.3 presents the two fundamental theorems with the proof and the intuition of the strip.
  • Tom M. Apostol. Calculus, Volume 1. 2nd ed. (Spanish edition, “Calculus, volumen 1”), Reverté. It builds all of calculus starting from the integral, so the fundamental theorem sits at the center.
  • Michael Spivak. Calculus. 3rd ed. (Spanish edition, “Calculus”), Reverté, 2012. The most careful proof, with all the hypotheses in plain view.
Development of the topic “The fundamental theorem of calculus” of Calculus I (Level 1), based on the curriculum of the UTN Electronic Engineering program, 2023 curriculum — Ordinance No. 1849 of the UTN Higher Council. Back to the Topic Map · catto.ar