f <- function(x) x^2
integrate(f, lower = 1, upper = 2)2.333333 with absolute error < 2.6e-14Appendix D — Integration
D.1 Definition
If derivatives describe the rate of change of a function, integrals do the opposite—they accumulate values over an interval.
Definition D.1 The definite integral of a function \(f(x)\) from \(a\) to \(b\) is defined as:
\[ \int_a^b f(x)\, dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x \]
where the interval \([a, b]\) is divided into \(n\) subintervals of width \(\Delta x = \frac{b - a}{n}\) and \(x_i^*\) is a sample point in the \(i\)th subinterval.
This limit represents the total area under the curve \(f(x)\) from \(a\) to \(b\).
D.1.1 Indefinite Integral
An indefinite integral (or antiderivative) is a function whose derivative is the original function.
Definition D.2 If \(F'(x) = f(x)\), then \(F(x)\) is an antiderivative of \(f(x)\), and we write:
\[ \int f(x)\, dx = F(x) + C \]
where \(C\) is an arbitrary constant.
D.2 Physical Interpretation of Integrals
Suppose your car is moving at a velocity \(f(x)\), where \(x\) is time and \(f(x)\) is in meters per second. Then:
- \(\int_a^b f(x)\, dx\) is the total distance traveled between time \(x = a\) and \(x = b\).
In this way, integration undoes differentiation. If the derivative of position is velocity, then the integral of velocity is position.
| Quantity | Symbol | Interpretation | Units (if \(x\) is time) |
|---|---|---|---|
| Velocity | \(f(x)\) | Rate of change of position | Meters per second (m/s) |
| Position | \(\int f(x)\, dx\) | Total distance traveled (accumulated) | Meters (m) |
The key idea is this: integration accumulates change. It’s the natural inverse of differentiation.
D.3 Why This Matters
Many real-world questions ask about totals or areas:
- How much profit was earned over the year?
- What is the total rainfall over a week?
- What’s the area under the likelihood curve?
- How much change accumulates over time?
All of these questions require integration. Let’s build some fluency with it.
D.4 Basic Rules of Integration
D.4.1 Constant Rule
Theorem D.1 (Constant Rule) If \(f(x) = a\), then:
\[ \int a\, dx = ax + C \]
Example D.1 Compute \(\int 5\, dx\).
Solution. Use the constant rule: \(\int 5\, dx = 5x + C\).
D.4.2 Power Rule
Theorem D.2 (Power Rule) If \(f(x) = x^n\), then:
\[ \int x^n\, dx = \frac{x^{n + 1}}{n + 1} + C,\quad \text{for } n \ne -1 \]
Example D.2 Compute \(\int x^3\, dx\).
Solution. Apply the power rule:
\[ \int x^3\, dx = \frac{x^4}{4} + C \]
Example D.3 Compute \(\int x^5 - 2x^2 + 7\, dx\).
Solution. Integrate each term using the power rule:
\[ \int x^5\, dx = \frac{x^6}{6},\quad \int -2x^2\, dx = -\frac{2x^3}{3},\quad \int 7\, dx = 7x \]
So the full result is:
\[ \frac{x^6}{6} - \frac{2x^3}{3} + 7x + C \]
D.4.3 Exponential Rule
Theorem D.3 (Exponential Rule) If \(f(x) = e^x\), then:
\[ \int e^x\, dx = e^x + C \]
D.4.4 Logarithm Rule
Theorem D.4 (Logarithm Rule) If \(f(x) = \frac{1}{x}\), then:
\[ \int \frac{1}{x}\, dx = \log|x| + C,\quad x \ne 0 \]
D.4.5 Sum Rule
Theorem D.5 (Sum Rule) If \(f(x) = g(x) + h(x)\), then:
\[ \int f(x)\, dx = \int g(x)\, dx + \int h(x)\, dx \]
Example D.4 Compute \(\int (x^2 + \frac{1}{x})\, dx\).
Solution. Apply the sum and power rules:
\[ \int x^2\, dx = \frac{x^3}{3},\quad \int \frac{1}{x}\, dx = \log|x| \]
So:
\[ \int (x^2 + \frac{1}{x})\, dx = \frac{x^3}{3} + \log|x| + C \]
D.4.6 Integration by Substitution (Chain Rule Reversed)
Substitution allows us to integrate composite functions, reversing the chain rule.
Theorem D.6 (Substitution Rule) Let \(u = g(x)\). Then:
\[ \int f(g(x)) g'(x)\, dx = \int f(u)\, du \]
Example D.5 Compute \(\int 2x \cdot \exp(x^2)\, dx\).
Solution. Let \(u = x^2\), so \(du = 2x\, dx\). Then:
\[ \int 2x \cdot \exp(x^2)\, dx = \int \exp(u)\, du = \exp(u) + C = \exp(x^2) + C \]
D.4.7 Integration by Parts (Product Rule Reversed)
Sometimes, it helps to reverse the product rule.
Theorem D.7 (Integration by Parts) If \(u = u(x)\) and \(v = v(x)\), then:
\[ \int u\, dv = uv - \int v\, du \]
Example D.6 Compute \(\int x \cdot \log(x)\, dx\).
Solution. Use integration by parts:
Let \(u = \log(x)\), so \(du = \frac{1}{x} dx\)
Let \(dv = x\, dx\), so \(v = \frac{x^2}{2}\)
Then:
\[ \int x \log(x)\, dx = \frac{x^2}{2} \log(x) - \int \frac{x^2}{2} \cdot \frac{1}{x}\, dx = \frac{x^2}{2} \log(x) - \int \frac{x}{2}\, dx = \frac{x^2}{2} \log(x) - \frac{x^2}{4} + C \]
D.5 Definite Integrals
If we want to compute the total accumulated value over a specific interval \([a, b]\), we use definite integrals.
Definition D.3 The definite integral from \(a\) to \(b\) is:
\[ \int_a^b f(x)\, dx = F(b) - F(a) \]
where \(F(x)\) is any antiderivative of \(f(x)\).
Example D.7 Compute \(\int_1^2 x^2\, dx\).
Solution. Find the antiderivative: \(F(x) = \frac{x^3}{3}\)
Evaluate:
\[ \int_1^2 x^2\, dx = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8 - 1}{3} = \frac{7}{3} \]
D.6 Integration in R
R does not compute symbolic integrals with base functions, but numerical integration is straightforward using integrate().
Example D.8 Compute \(\int_1^2 x^2\, dx\) numerically.
Returns:
7/3 = 2.333...Example D.9 Compute \(\int_0^1 x \log(x)\, dx\) numerically.
f <- function(x) ifelse(x == 0, 0, x * log(x))
integrate(f, 0, 1)-0.25 with absolute error < 3e-05This returns approximately:
-0.25Numerical integration is useful for functions that have no closed-form antiderivative or are only defined computationally.
D.7 Summary: Derivatives vs Integrals
| Operation | Symbol | Meaning |
|---|---|---|
| Derivative | \(f'(x)\) or \(\frac{df}{dx}\) | Instantaneous rate of change |
| Integral | \(\int f(x)\, dx\) | Accumulated change (area under curve) |
| Definite Int. | \(\int_a^b f(x)\, dx\) | Total change from \(x = a\) to \(x = b\) |
Together, derivatives and integrals are the fundamental tools of calculus. They describe change and accumulation—central ideas in modeling, statistics, economics, physics, and beyond.