Calculus I · 120 h · Topic 7 of 10

The definite integral

The definite integral is an area, and an area is a sum of infinitely many infinitely thin rectangles. From it come the mean value and the RMS value of any signal.

Riemann sums Area Mean value RMS value Properties

01An area built from rectangles

How much charge did a varying current deliver? How much energy did a resistor dissipate? The two questions are the same: the area under a curve. And the area under an arbitrary curve has no formula, so it is constructed: the interval is split into n small pieces, a rectangle is raised over each one and they are added up.

\[ \int_a^b f (x) d x = \lim_{n \to \infty} \sum_{i = 1}^n f (x_i^*) \Delta x \quad \operatorname{with} \; \Delta x = \frac{b -a}{n} \] This is the Riemann sum. The more rectangles, the thinner they are, and the error tends to zero. The symbol ∫ is a stretched S: sum.

The sign matters

The definite integral is a signed area: whatever lies below the x-axis is subtracted. That is why the integral over one full cycle of a sinusoid is zero, even though the drawn area is not. That is exactly the zero average value of alternating current.

Lab · Riemann sums

Raise the number of rectangles and watch the sum approach the exact value.

n 6

02Properties used without a second thought

Property	What it says
Linearity	\( \int (\alpha f + \beta g) = \alpha \int f + \beta \int g \). It is superposition applied to integrals
Additivity over the interval	\( \int_a^b = \int_a^c + \int_c^b \): it can be cut into pieces, useful for piecewise functions
Orientation	\( \int_a^b = -\int_b^a \), and \( \int_a^a = 0 \)
Monotonicity	If \( f \le g \) on \( [a,\, b] \), then \( \int f \le \int g \)
Boundedness	\( m(b-a) \le \int f \le M(b-a) \), with m and M the minimum and the maximum

Mean value theorem for integrals

If f is continuous on [a, b], there is a point c where f(c) is exactly the average value:

\[ f_{\operatorname{avg}} = \frac{1}{b -a} \int_a^b f (x) d x = f (c) \]

Geometrically: the rectangle of height f(c) has the same area as the region under the curve. This is what an average-responding instrument measures, and what an analog multimeter indicates on DC.

03Average value and RMS value of a signal

Here the definite integral stops being an exercise and becomes the definition of two quantities that are measured every day in the lab:

\[ V_{\operatorname{avg}} = \frac{1}{T} \int_0^T v (t) d t \quad V_{\operatorname{rms}} = \sqrt{\frac{1}{T} \int_0^T v^2 (t) d t} \] The RMS value, or root mean square, is the square root of the mean of the square. Its definition comes from requiring it to dissipate the same power as a DC value of the same size.

Where 0.707 comes from

For v(t) = V_p·sin(ωt), the square is V_p²·sin²(ωt), and using sin²θ = (1 − cos 2θ)/2, its mean value over one period is V_p²/2.

So V_rms = V_p/√2 ≈ 0.707·V_p. The 220 V of the mains is RMS: the peak is 311 V. All of that is a definite integral solved once.

Waveform	Average value (full cycle)	RMS value
Sine wave	0	V_p/√2 = 0.707 V_p
Full-wave rectified sine wave	2V_p/π = 0.637 V_p	0.707 V_p
Half-wave rectified sine wave	V_p/π = 0.318 V_p	V_p/2 = 0.5 V_p
Symmetric square wave	0	V_p
Symmetric triangle wave	0	V_p/√3 = 0.577 V_p

That table explains why an ordinary multimeter, which measures the average value of the rectified signal and multiplies by 1.11, gets it wrong with non-sinusoidal waves: for a square wave it reads 11% too high. A “true RMS” instrument really does compute the integral of the square. See digital measuring instruments.

04When there is no antiderivative: numerical integration

Many functions have no elementary antiderivative —e^−x² is the classic case— and often you have only measured points, with no formula. In those cases you integrate numerically, which means settling for a well-chosen finite Riemann sum:

Method	Idea	Error
Rectangles	Approximates f by a constant on each piece	Proportional to Δx
Midpoint	The constant is taken at the center of the piece	Proportional to Δx²
Trapezoids	Joins the endpoints with a segment	Proportional to Δx²
Simpson	Fits a parabola over every two pieces	Proportional to Δx⁴

This is exactly what a digital oscilloscope does when it displays the RMS value of a captured signal: it adds up the squares of the samples, divides by the count and takes the root. The integral became a computation over an array of numbers.

05In the lab

Exercise 1 · Riemann by hand

Compute ∫₀³ x² dx with 3, 6 and 12 rectangles using the left, right and midpoint rules. Compare with the exact value (9) and see which converges fastest. Contrast with the lab.

Exercise 2 · RMS value, measured and computed

With the generator, apply a sine, a square and a triangle wave of equal peak amplitude to a resistor. Measure with an ordinary multimeter and with a true RMS one, and compare with the table. Explain the differences.

Exercise 3 · Energy of a pulse

Capture a current pulse through a known load, export the samples and compute the energy as the sum of v·i·Δt. Compare with the measured temperature rise, if the setup allows it.

06Common mistakes

Confusing area with integral: if the function changes sign, to get the geometric area you have to integrate the absolute value or split the interval.
Computing the RMS value as the root of the mean, or averaging without squaring.
Using 0.707 for any waveform: that factor applies only to the sine wave.
Forgetting the 1/T in the average or in the RMS value.
Integrating over less than one period when computing values of a periodic signal.
Measuring a non-sinusoidal signal with an average-responding multimeter and believing the number.

07Self-assessment

What is ∫₀³ x² dx?

x³/3 evaluated between 0 and 3: 9.

Why is the integral of a sinusoid over one period zero?

Because the positive and negative half-cycles have equal areas and opposite signs. Its average value is zero.

A ±10 V square signal. Average value and RMS value?

Average value 0 and RMS value 10 V: the square is always 100.

What is the peak value of a 220 V RMS mains supply?

220·√2 ≈ 311 V. Insulation is rated for the peak, not for the RMS value.

Which numerical method converges faster, trapezoids or midpoint?

Both have error proportional to Δx², but that of the midpoint rule is usually half as large. Simpson’s rule, which combines the two, is much better.

An average-responding instrument measures a square wave of 10 V peak. What does it read?

It rectifies (average value 10 V) and multiplies by 1.11: it reads 11.1 V, when the true RMS value is 10 V. An error of 11%.

08Further reading

James Stewart. Single Variable Calculus: Early Transcendentals. 7th ed. (Spanish edition, “Cálculo de una variable. Trascendentes tempranas”), Cengage Learning, 2012. Chapter 5 builds the definite integral from Riemann sums, with the full notation.
Manuel Sadosky and Rebeca Ch. de Guber. Elementos de cálculo diferencial e integral (in Spanish). Alsina. A classic, rigorous treatment of the definite integral and its properties.
Robert L. Boylestad. Introductory Circuit Analysis. 12th ed. (Spanish edition, “Introducción al análisis de circuitos”), Pearson, 2011. Where average and RMS values are used, with the rectified waveforms worked out.

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