Integral calculus
Integration is the inverse operation of differentiation: starting from how something changes, you reconstruct how much of it there is.
01Undoing the derivative
Differentiating is easy and mechanical: there are rules for everything. The inverse operation —given the derivative, find the function— is called finding an antiderivative, and it is not mechanical: it is a craft.
The fact that all the antiderivatives of a function differ only by a constant is a direct consequence of the mean value theorem: if two functions have the same derivative, their difference has zero derivative and is therefore constant.
Every integral is checked by differentiating the result. If it does not give the integrand, it is wrong. It is the only branch of calculus where you can check your own answer in thirty seconds: it is worth doing it every time.
02Basic integrals
The table of derivatives read backwards:
| \( \int f \, dx \) | Result | \( \int f \, dx \) | Result |
|---|---|---|---|
| \( \int x^n \, dx \quad (n \ne -1) \) | xn+1/(n + 1) + C | \( \int \operatorname{sin} x \, dx \) | −cos x + C |
| \( \int \dfrac{dx}{x} \) | ln|x| + C | \( \int \cos x \, dx \) | sin x + C |
| \( \int e^x \, dx \) | ex + C | \( \int \sec^2 x \, dx \) | tan x + C |
| \( \int a^x \, dx \) | ax/ln a + C | \( \int \dfrac{dx}{1 + x^{2}} \) | arctan x + C |
∫dx/x = ln|x| + C, with the bars. Without them, the result would not hold for negative x, where the function 1/x exists just the same.
The integral is linear: the integral of a sum is the sum of the integrals and constants come out front. On the other hand, there is no product rule or quotient rule: that is where the methods begin.
03The three methods that solve almost everything
Substitution: the chain rule in reverse
If the integrand contains a function and, as a factor, something resembling its derivative, it pays to call that function u:
By parts: the product rule in reverse
Partial fractions: for quotients of polynomials
A rational function is decomposed into fractions with first- or second-degree denominators, which are already in the table. It is the method that later solves the inverse Laplace transforms of any circuit.
Each step appears when you ask for it. At the end, the numerical check differentiates the result and compares it with the integrand.
04Integrating in a circuit
| Relation | Differentiating | Integrating |
|---|---|---|
| Capacitor | \( i = C\cdot \dfrac{dv}{dt} \) | \( v = \dfrac{1}{C}\cdot \int i \, dt + v(0) \) |
| Inductor | \( v = L\cdot \dfrac{di}{dt} \) | \( i = \dfrac{1}{L}\cdot \int v \, dt + i(0) \) |
| Charge | \( i = \dfrac{dq}{dt} \) | \( q = \int i \, dt \) |
| Energy | \( p = \dfrac{dW}{dt} \) | \( W = \int p \, dt = \int v\cdot i \, dt \) |
The constants of integration in that right-hand column are the initial conditions: the voltage the capacitor already had or the current that was already flowing through the inductor. Mathematically they are “any constant”; physically, the state of the circuit at time zero.
If a 1 µF capacitor receives a constant current of 2 mA starting from 0 V:
v(t) = (1/C)∫i dt = (1/10−6)·(2·10−3·t) = 2000·t volts, a ramp of 2 V per millisecond. This is the principle of the sawtooth generator and of the dual-slope ADC.
The op amp integrator does this computation in real time: vo = −(1/RC)∫vi dt. Its practical version appears in the operational amplifiers course.
05In the lab
Solve ∫x·cos x dx, ∫x/(x² + 1) dx and ∫dx/(x² − 4) by the appropriate method, and check all three by differentiating the result.
Build an op amp integrator (R = 10 kΩ, C = 100 nF) and drive it with a 1 kHz square wave. Check that the output is triangular and measure its slope; compare it with V/(RC). Explain why a resistor in parallel with C is needed.
Charge a capacitor with constant current and measure the voltage ramp. Compute the accumulated charge as the area under the current curve and verify q = C·V.
06Common mistakes
- Forgetting the + C. In a circuit, that means forgetting the initial condition.
- Inventing a product rule: ∫f·g dx is not ∫f dx · ∫g dx.
- Substituting without changing the dx to du: the substitution includes the differential.
- Choosing u and dv badly in integration by parts, and getting an integral harder than the original.
- Writing ln x without the absolute value.
- Applying xn+1/(n+1) with n = −1, which gives a division by zero: that case is the logarithm.
07Self-assessment
∫(3x² − 4x + 5) dx.
x³ − 2x² + 5x + C. Check: differentiating returns the integrand. ✔
∫e−2x dx.
With u = −2x: −e−2x/2 + C.
∫x·ex dx.
By parts, with u = x and dv = exdx: x·ex − ex + C.
What does the constant of integration represent in v = (1/C)∫i dt + C₁?
The initial voltage of the capacitor, v(0).
∫sin(ωt) dt.
−cos(ωt)/ω + C. The 1/ω is the factor from the chain rule in reverse.
A 10 nF capacitor receives a constant 5 µA. How fast does its voltage rise?
dv/dt = i/C = 5·10−6/10−8 = 500 V/s, or 0.5 V per millisecond.
08Further reading
- James Stewart. Single Variable Calculus: Early Transcendentals. 7th ed. (Spanish edition, “Cálculo de una variable. Trascendentes tempranas”), Cengage Learning, 2012. Chapters 5 and 7 cover antiderivatives and all the integration techniques, with a strategy for choosing among them.
- Murray R. Spiegel. Advanced Calculus (Schaum’s Outline Series). (Spanish edition, “Cálculo superior (serie Schaum)”), McGraw-Hill. Tables and hundreds of solved integrals: useful as a desk reference.
- Hebe Rabuffetti. Introducción al análisis matemático (Cálculo 1) (in Spanish). El Ateneo. Integration methods explained with the detail of each substitution.