Boo-AI — Master Artificial Intelligence by Building from Scratch

Learning Objectives

By the end of this section, you will be able to:

Understand how integration by parts arises from reversing the product rule for derivatives
Apply the integration by parts formula $\int u \, dv = uv - \int v \, du$ to evaluate integrals
Use the LIATE rule to strategically choose $u$ and $dv$
Master the tabular method for repeated integration by parts
Apply integration by parts to definite integrals with proper boundary evaluation
Connect this technique to physics, probability, signal processing, and machine learning

The Big Picture: Reversing the Product Rule

"Integration by parts is the antiderivative analog of the product rule — it transforms difficult products into manageable pieces."

In differential calculus, the product rule tells us how to differentiate a product of two functions:

\frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)

But what if we need to go backwards? What if we have an integral of a product and want to find its antiderivative? This is where integration by parts comes in — it's the reverse of the product rule.

The Core Insight

Integration by parts transforms $\int u \, dv$ into $uv - \int v \, du$ . The hope is that $\int v \, du$ is easier to evaluate than the original integral. This technique works when one part of the product simplifies upon differentiation while the other remains manageable upon integration.

Why Do We Need This?

Consider $\int x e^x \, dx$ . Neither u-substitution nor any basic formula helps here. The product of $x$ and $e^x$ doesn't fit any pattern we know — but integration by parts transforms this into something solvable.

Integration by parts is essential for integrals involving:

Products of polynomials with exponentials: $\int x^n e^x \, dx$
Products of polynomials with trigonometric functions: $\int x \sin x \, dx$
Logarithmic functions: $\int \ln x \, dx$
Inverse trigonometric functions: $\int \arctan x \, dx$
Products of trigonometric functions: $\int e^x \sin x \, dx$

Historical Context

Integration by parts was developed alongside the foundational work of calculus in the 17th century.

Gottfried Wilhelm Leibniz (1646–1716)

Leibniz, who independently developed calculus alongside Newton, recognized that his product rule for differentiation could be "inverted" to create a technique for integration. His notation $\int u \, dv$ emphasizes the differential form, making the relationship between differentiation and integration explicit.

Leibniz's insight was profound: since $d(uv) = u \, dv + v \, du$ , we can rearrange to get $u \, dv = d(uv) - v \, du$ . Integrating both sides gives the integration by parts formula.

The Name "Integration by Parts"

The name comes from the idea that we split the integrand into two "parts": one part ( $u$ ) that we differentiate, and another part ( $dv$ ) that we integrate. The technique redistributes the complexity between these parts, hopefully making the integral easier.

Derivation from the Product Rule

Let's derive the integration by parts formula step by step from the product rule.

Step 1: Start with the Product Rule

For two differentiable functions $u(x)$ and $v(x)$ :

\frac{d}{dx}[u(x) v(x)] = u'(x) v(x) + u(x) v'(x)

Step 2: Rearrange

Isolate one of the product terms:

u(x) v'(x) = \frac{d}{dx}[u(x) v(x)] - u'(x) v(x)

Step 3: Integrate Both Sides

Integrate with respect to $x$ :

\int u(x) v'(x) \, dx = u(x) v(x) - \int u'(x) v(x) \, dx

The left side integrates the derivative $\frac{d}{dx}[uv]$ , which simply gives $uv$ .

Step 4: Write in Differential Form

Using Leibniz notation where $du = u'(x) \, dx$ and $dv = v'(x) \, dx$ :

Integration by Parts Formula

\int u \, dv = uv - \int v \, du

Where $u$ and $v$ are functions of $x$

Memory Aid

Think of it as: "Ultra-Violet minus Voodoo" — $uv - \int v \, du$ . The integral "trades" one problem for another: we lose the $dv$ but gain $du$ .

The Integration by Parts Formula

The Process

To evaluate $\int f(x) g(x) \, dx$ using integration by parts:

Choose which factor to call $u$ and which to call $dv$
Differentiate $u$ to find $du$
Integrate $dv$ to find $v$
Apply the formula: $\int u \, dv = uv - \int v \, du$
Evaluate the new integral $\int v \, du$

A First Example: $\int x e^x \, dx$

Step 1: Identify u and dv

Let $u = x$ and $dv = e^x \, dx$

Step 2: Find du and v

$du = dx$
$v = \int e^x \, dx = e^x$

Step 3: Apply the Formula

\int x e^x \, dx = x e^x - \int e^x \, dx

Step 4: Evaluate the Remaining Integral

\int x e^x \, dx = x e^x - e^x + C = e^x(x - 1) + C

Verification

We can verify by differentiating: $\frac{d}{dx}[e^x(x-1)] = e^x(x-1) + e^x = xe^x$ ✓

Interactive: Visualizing Integration by Parts

Explore how integration by parts works geometrically. See how the choice of $u$ and $dv$ transforms the integral:

Integration by Parts Step-by-Step

1. Identify the Integral2. Choose u and dv3. Compute du and v4. Apply the Formula5. Evaluate ∫v du6. Final Answer

We want to evaluate:

∫ x e^x dx

This is a product of two functions. We'll use integration by parts.

Remember: ∫ u dv = uv − ∫ v du

Choosing u and dv: The LIATE Rule

The success of integration by parts depends critically on choosing $u$ and $dv$ wisely. The LIATE rule is a heuristic that suggests which function to choose as $u$ .

The LIATE Rule

Choose $u$ to be the function that appears earliest in this list:

Logarithmic

\ln x

Inverse Trig

\arctan x

Algebraic

x^n

Trigonometric

\sin x

Exponential

e^x

Why LIATE Works

Functions earlier in the list generally become simpler when differentiated:

$\ln x \to \frac{1}{x}$ — logarithms become algebraic
$\arctan x \to \frac{1}{1+x^2}$ — inverse trig becomes rational
$x^n \to nx^{n-1}$ — polynomials reduce degree
$\sin x \to \cos x$ — trig stays trig
$e^x \to e^x$ — exponentials stay the same

Meanwhile, functions later in the list are easy to integrate:

$\int e^x \, dx = e^x$
$\int \sin x \, dx = -\cos x$
$\int x^n \, dx = \frac{x^{n+1}}{n+1}$

Integral	u (using LIATE)	dv	Reason
∫ x·eˣ dx	x (Algebraic)	eˣ dx	A comes before E; differentiating x simplifies it
∫ x·sin(x) dx	x (Algebraic)	sin(x) dx	A comes before T; x reduces degree
∫ ln(x) dx	ln(x) (Logarithmic)	dx	L is first; ln(x) becomes 1/x
∫ x²·ln(x) dx	ln(x) (Logarithmic)	x² dx	L comes before A
∫ arctan(x) dx	arctan(x) (Inverse)	dx	I is second; becomes rational
∫ eˣ·sin(x) dx	Either works!	Either works!	Special case: requires two IBP cycles

LIATE is a Guideline, Not a Law

The LIATE rule works for most integrals, but there are exceptions. If your first choice leads to a more complicated integral, try the other option. Experience helps develop intuition for when to deviate.

Worked Examples

Example 1: $\int x^2 e^x \, dx$

Analysis: By LIATE, $u = x^2$ (Algebraic) and $dv = e^x \, dx$ (Exponential).

First Application:

$u = x^2$ , $du = 2x \, dx$
$dv = e^x \, dx$ , $v = e^x$

\int x^2 e^x \, dx = x^2 e^x - \int 2x e^x \, dx = x^2 e^x - 2 \int x e^x \, dx

Second Application on $\int x e^x \, dx$ :

$u = x$ , $du = dx$
$dv = e^x \, dx$ , $v = e^x$

\int x e^x \, dx = x e^x - e^x

Final Result:

\int x^2 e^x \, dx = x^2 e^x - 2(x e^x - e^x) + C = e^x(x^2 - 2x + 2) + C

Example 2: $\int \ln x \, dx$

Key Insight: We write this as $\int 1 \cdot \ln x \, dx$ and choose $u = \ln x$ , $dv = dx$ .

$u = \ln x$ , $du = \frac{1}{x} \, dx$
$dv = dx$ , $v = x$

\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - \int 1 \, dx

\int \ln x \, dx = x \ln x - x + C = x(\ln x - 1) + C

Example 3: $\int e^x \sin x \, dx$ (The Cycling Case)

Special Situation: Both functions are at the same "level" in LIATE. We'll need to apply IBP twice and solve for the original integral.

First Application: Let $u = e^x$ , $dv = \sin x \, dx$

$du = e^x \, dx$ , $v = -\cos x$

\int e^x \sin x \, dx = -e^x \cos x + \int e^x \cos x \, dx

Second Application on $\int e^x \cos x \, dx$ : Let $u = e^x$ , $dv = \cos x \, dx$

\int e^x \cos x \, dx = e^x \sin x - \int e^x \sin x \, dx

Substitute Back:

\int e^x \sin x \, dx = -e^x \cos x + e^x \sin x - \int e^x \sin x \, dx

Solve for the Integral: Let $I = \int e^x \sin x \, dx$

I = -e^x \cos x + e^x \sin x - I \implies 2I = e^x(\sin x - \cos x)

\int e^x \sin x \, dx = \frac{e^x(\sin x - \cos x)}{2} + C

Example 4: $\int \arctan x \, dx$

Setup: Write as $\int 1 \cdot \arctan x \, dx$ with $u = \arctan x$ , $dv = dx$ .

$u = \arctan x$ , $du = \frac{1}{1+x^2} \, dx$
$dv = dx$ , $v = x$

\int \arctan x \, dx = x \arctan x - \int \frac{x}{1+x^2} \, dx

The remaining integral uses u-substitution with $w = 1 + x^2$ :

\int \arctan x \, dx = x \arctan x - \frac{1}{2} \ln(1 + x^2) + C

Tabular Integration Method

When you need to apply integration by parts multiple times — especially when the $u$ factor is a polynomial that eventually differentiates to zero — the tabular method (also called the tic-tac-toe method) provides an efficient shortcut.

The Process

Create two columns: one for successive derivatives of $u$ , one for successive integrals of $dv$
Continue until the $u$ column reaches zero (for polynomials) or a pattern emerges
Draw diagonal lines connecting entries
Alternate signs: $+, -, +, -, \ldots$
Multiply along each diagonal and sum

Example: $\int x^3 e^{2x} \, dx$

Sign	u and derivatives	dv and integrals
+	x³	e^(2x)
-	3x²	(1/2)e^(2x)
+	6x	(1/4)e^(2x)
-	6	(1/8)e^(2x)
+	0	(1/16)e^(2x)

Result: Multiply diagonally and sum:

\int x^3 e^{2x} \, dx = \frac{e^{2x}}{2}\left(x^3 - \frac{3x^2}{2} + \frac{6x}{4} - \frac{6}{8}\right) + C

= \frac{e^{2x}}{8}(4x^3 - 6x^2 + 6x - 3) + C

Interactive: Tabular Method

Practice the tabular method interactively. See how the table builds and how the alternating signs combine with diagonal products:

Tabular Integration Method

Building table for: ∫ x² eˣ dx with u = x², dv = eˣ dx

Sign	Derivatives of u	Integrals of dv
+	x² ↓ differentiate	eˣ ↓ integrate

How it works: Multiply diagonally (each u-row with the v-row below it), alternating signs (+, −, +, −, ...), and sum. Stop when u reaches 0.

Repeated Integration by Parts

Some integrals require applying integration by parts multiple times. There are two common scenarios:

Case 1: Polynomial Reduction

When $u$ is a polynomial of degree $n$ , you need $n$ applications of IBP. Each application reduces the degree by 1, until the polynomial becomes a constant.

Examples:

$\int x^2 \sin x \, dx$ — needs 2 applications
$\int x^4 e^x \, dx$ — needs 4 applications

Case 2: Cyclic Return

When both factors regenerate (like $e^x \sin x$ ), the original integral appears again after 2 applications. Solve algebraically:

If $I = \int e^x \sin x \, dx$ and after two IBP applications we get:

I = e^x \sin x - e^x \cos x - I

Then:

2I = e^x(\sin x - \cos x) \implies I = \frac{e^x(\sin x - \cos x)}{2}

Integration by Parts for Definite Integrals

For definite integrals, we evaluate the boundary terms $[uv]_a^b$ at the limits:

Integration by Parts for Definite Integrals

\int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du

Where $[uv]_a^b = u(b)v(b) - u(a)v(a)$

Example: $\int_0^1 x e^x \, dx$

With $u = x$ , $dv = e^x \, dx$ :

\int_0^1 x e^x \, dx = [x e^x]_0^1 - \int_0^1 e^x \, dx

Evaluate:

= (1 \cdot e^1 - 0 \cdot e^0) - [e^x]_0^1 = e - (e - 1) = 1

\int_0^1 x e^x \, dx = 1

Real-World Applications

Integration by parts appears throughout science and engineering whenever we need to integrate products of functions.

Physics: Work Done by Variable Forces

When both force and displacement are time-dependent, the work integral often requires IBP:

W = \int_0^T F(t) \cdot v(t) \, dt

Example: A motor starting up with $F(t) = F_0 t$ and $v(t) = v_0 e^{-kt}$

Probability: Expected Values

Computing expected values of continuous distributions invariably involves products:

E[X] = \int_0^\infty x \cdot f(x) \, dx

For the exponential distribution: $E[X] = \int_0^\infty x \lambda e^{-\lambda x} \, dx = \frac{1}{\lambda}$

Signal Processing: Fourier Coefficients

Computing Fourier coefficients for non-sinusoidal waves:

b_n = \frac{2}{L} \int_0^L f(x) \sin\left(\frac{n\pi x}{L}\right) dx

When $f(x)$ is polynomial or exponential, IBP is needed.

Control Systems: Laplace Transforms

The property $\mathcal{L}\{t f(t)\} = -\frac{d}{ds} F(s)$ is derived using integration by parts:

\mathcal{L}\{f(t)\} = \int_0^\infty f(t) e^{-st} \, dt

Differentiating with respect to $s$ brings down a factor of $-t$ , creating an IBP situation.

Machine Learning Connection

Integration by parts plays a subtle but important role in machine learning and statistics.

The Score Function and REINFORCE

In reinforcement learning, we often need to compute gradients of expectations:

\nabla_\theta \mathbb{E}_{x \sim p_\theta}[f(x)] = \nabla_\theta \int f(x) p(x|\theta) \, dx

The key insight (derived using integration by parts) is that this equals:

= \mathbb{E}_{x \sim p_\theta}[f(x) \nabla_\theta \log p(x|\theta)]

This is the score function estimator (or REINFORCE trick). The transformation uses integration by parts:

\nabla_\theta p(x|\theta) = p(x|\theta) \nabla_\theta \log p(x|\theta)

Since $\nabla \log p = \frac{\nabla p}{p}$

The Gamma Function Recurrence

The Gamma function $\Gamma(n) = \int_0^\infty x^{n-1} e^{-x} \, dx$ satisfies $\Gamma(n+1) = n \cdot \Gamma(n)$ , proven by integration by parts. This is fundamental for:

Gamma distribution normalization
Chi-squared distribution
Student's t-distribution
Beta function and Bayesian inference

Variance Reduction in Monte Carlo

The control variate method in variance reduction uses integration by parts to derive optimal baseline functions that reduce the variance of gradient estimators in policy gradient methods.

Python Implementation

Numerical and Symbolic Integration by Parts

Here's how to verify integration by parts numerically and perform it symbolically:

Integration by Parts: Numerical and Symbolic

🐍integration_by_parts.py

Explanation(6)

Code(143)

10Setting Up the IBP

We identify u = x and dv = e^x dx. This choice follows the LIATE rule: x is algebraic, e^x is exponential. The algebraic term simplifies when differentiated.

18Numerical Integration

SciPy's quad function uses adaptive quadrature to compute the definite integral. This serves as ground truth to verify our analytical solution.

22Analytical Solution

After applying IBP: ∫xe^x dx = xe^x - ∫e^x dx = xe^x - e^x = (x-1)e^x. Evaluating from 0 to 1 gives [(1-1)e - (0-1)1] = 0 - (-1) = 1.

45Symbolic Integration

SymPy automatically applies integration by parts (and other techniques) to find antiderivatives. This shows the power of computer algebra systems.

67Score Function Gradient

The REINFORCE trick in reinforcement learning uses integration by parts to express gradients of expectations as expectations of gradients - enabling gradient estimation via sampling.

94Gamma Function Recurrence

The fundamental property Γ(n+1) = n·Γ(n) is proven using integration by parts. This connects calculus to probability theory (Gamma, Chi-squared distributions).

137 lines without explanation

1import numpy as np
2from scipy import integrate
3import sympy as sp
4
5def integration_by_parts_numerical():
6    """
7    Verify integration by parts numerically:
8    ∫ u dv = uv - ∫ v du
9
10    Example: ∫ x·e^x dx from 0 to 1
11    Let u = x, dv = e^x dx
12    Then du = dx, v = e^x
13
14    ∫ x·e^x dx = x·e^x - ∫ e^x dx = x·e^x - e^x + C = (x-1)e^x + C
15    """
16    # Define the integrand f(x) = x * e^x
17    def f(x):
18        return x * np.exp(x)
19
20    # Numerical integration using scipy
21    numerical_result, error = integrate.quad(f, 0, 1)
22
23    # Analytical result: [(x-1)e^x] from 0 to 1
24    # At x=1: (1-1)e^1 = 0
25    # At x=0: (0-1)e^0 = -1
26    # Result: 0 - (-1) = 1
27    analytical_result = 1.0
28
29    print("Integration by Parts: ∫₀¹ x·eˣ dx")
30    print("=" * 45)
31    print(f"Numerical result:  {numerical_result:.10f}")
32    print(f"Analytical result: {analytical_result:.10f}")
33    print(f"Absolute error:    {abs(numerical_result - analytical_result):.2e}")
34
35    return numerical_result, analytical_result
36
37def integration_by_parts_symbolic():
38    """
39    Use SymPy to perform integration by parts symbolically.
40    This demonstrates how computer algebra systems apply IBP.
41    """
42    x = sp.Symbol('x')
43
44    # Define examples that require integration by parts
45    examples = [
46        ("x * exp(x)", x * sp.exp(x)),
47        ("x * sin(x)", x * sp.sin(x)),
48        ("x^2 * cos(x)", x**2 * sp.cos(x)),
49        ("ln(x)", sp.ln(x)),
50        ("x * ln(x)", x * sp.ln(x)),
51        ("arctan(x)", sp.atan(x)),
52    ]
53
54    print("\nSymbolic Integration by Parts Examples")
55    print("=" * 55)
56
57    for name, expr in examples:
58        result = sp.integrate(expr, x)
59        result_simplified = sp.simplify(result)
60        print(f"\n∫ {name} dx = {result_simplified}")
61
62    return examples
63
64def ibp_ml_application():
65    """
66    Integration by parts in Machine Learning:
67    Computing expectations and gradients of log-probability.
68
69    In reinforcement learning, the score function gradient:
70    ∇_θ E[f(x)] = E[f(x) ∇_θ log p(x|θ)]
71
72    This derivation uses integration by parts!
73    """
74    print("\n" + "=" * 55)
75    print("Integration by Parts in Machine Learning")
76    print("=" * 55)
77
78    # Demonstrate with Gaussian distribution
79    # For Gaussian: p(x|μ,σ) = (1/√(2πσ²)) exp(-(x-μ)²/(2σ²))
80    # Score function: ∇_μ log p(x|μ,σ) = (x-μ)/σ²
81
82    mu, sigma = 0.0, 1.0
83    n_samples = 100000
84
85    # Sample from Gaussian
86    samples = np.random.normal(mu, sigma, n_samples)
87
88    # For any function f(x), we can estimate:
89    # ∇_μ E[f(x)] ≈ (1/N) Σ f(xᵢ) · (xᵢ - μ)/σ²
90
91    # Example: f(x) = x² (so E[f(x)] = μ² + σ² = 1)
92    f_samples = samples ** 2
93    score_values = (samples - mu) / (sigma ** 2)
94
95    # Score function gradient estimate
96    gradient_estimate = np.mean(f_samples * score_values)
97
98    # True gradient: ∂/∂μ (μ² + σ²) = 2μ = 0
99    true_gradient = 2 * mu
100
101    print(f"\nScore Function Gradient (REINFORCE trick)")
102    print(f"Function: f(x) = x²")
103    print(f"True gradient ∂/∂μ E[x²]:  {true_gradient:.6f}")
104    print(f"Estimated gradient:        {gradient_estimate:.6f}")
105    print(f"\nThis technique is derived using integration by parts!")
106    print("∇_θ E[f(x)] = ∫ f(x) ∇_θ p(x|θ) dx")
107    print("            = ∫ f(x) p(x|θ) ∇_θ log p(x|θ) dx")
108    print("            = E[f(x) ∇_θ log p(x|θ)]")
109
110def gamma_function_ibp():
111    """
112    The Gamma function Γ(n) = ∫₀^∞ x^(n-1) e^(-x) dx
113    satisfies Γ(n+1) = n·Γ(n), proven by integration by parts!
114
115    This property is fundamental in probability distributions
116    (Gamma, Chi-squared, Student-t, etc.)
117    """
118    print("\n" + "=" * 55)
119    print("Integration by Parts and the Gamma Function")
120    print("=" * 55)
121
122    from scipy.special import gamma
123
124    # Verify Γ(n+1) = n·Γ(n) for several values
125    print("\nVerifying Γ(n+1) = n·Γ(n):")
126    print("-" * 40)
127
128    for n in [1, 2, 3, 4, 5, 0.5, 1.5, 2.5]:
129        gamma_n = gamma(n)
130        gamma_n_plus_1 = gamma(n + 1)
131        ratio = gamma_n_plus_1 / gamma_n
132
133        print(f"n = {n:>4}: Γ({n+1}) / Γ({n}) = {ratio:.6f} ≈ {n}")
134
135    print("\nThis recurrence is proven by IBP with u=x^(n-1), dv=e^(-x)dx")
136    print("Γ(n+1) = ∫₀^∞ xⁿ e^(-x) dx = [-xⁿe^(-x)]₀^∞ + n∫₀^∞ x^(n-1)e^(-x) dx")
137    print("       = 0 + n·Γ(n) = n·Γ(n)")
138
139# Run all demonstrations
140integration_by_parts_numerical()
141integration_by_parts_symbolic()
142ibp_ml_application()
143gamma_function_ibp()

Applications in Physics and Engineering

See how integration by parts applies to real-world problems:

Real-World Applications

🐍ibp_applications.py

Explanation(5)

Code(146)

11Physics: Variable Force Work

When force and velocity are both functions of time, the work integral often requires IBP. Here F(t) = F₀t and v(t) = v₀e^(-kt) model a motor starting under load.

25Analytical IBP Solution

Integration by parts gives ∫te^(-kt)dt = (-t/k - 1/k²)e^(-kt). The algebraic term t becomes simpler after differentiation, while e^(-kt) integrates cleanly.

44Probability: Expected Value

The expected value E[X] = ∫xf(x)dx always involves products of x with the PDF. For exponential, gamma, and many other distributions, this requires IBP.

71Fourier Coefficients

Fourier series coefficients for non-sinusoidal waves typically require IBP. The integral ∫x·sin(nx)dx appears when computing bₙ for signals with linear trends.

93Laplace Transform Derivative

The property L{t·f(t)} = -dF(s)/ds is fundamental in control systems. It is derived using integration by parts and enables solving differential equations algebraically.

141 lines without explanation

1import numpy as np
2from scipy import integrate
3import matplotlib.pyplot as plt
4
5def physics_work_integral():
6    """
7    Work done by a time-varying force:
8    W = ∫₀^T F(t) · v(t) dt
9
10    When F(t) = F₀·t (linearly increasing force) and
11    v(t) = v₀·e^(-kt) (exponentially decaying velocity),
12    we need integration by parts!
13    """
14    F0, v0, k, T = 10.0, 5.0, 0.5, 4.0
15
16    # Integrand: F(t)·v(t) = F₀·v₀·t·e^(-kt)
17    def integrand(t):
18        return F0 * v0 * t * np.exp(-k * t)
19
20    # Numerical solution
21    work_numerical, _ = integrate.quad(integrand, 0, T)
22
23    # Analytical solution using IBP:
24    # ∫ t·e^(-kt) dt = -t/k · e^(-kt) - 1/k² · e^(-kt)
25    def analytical_antiderivative(t):
26        return (-t/k - 1/k**2) * np.exp(-k * t)
27
28    work_analytical = F0 * v0 * (
29        analytical_antiderivative(T) - analytical_antiderivative(0)
30    )
31
32    print("Physics: Work Done by Time-Varying Force")
33    print("=" * 45)
34    print(f"F(t) = {F0}t N, v(t) = {v0}e^(-{k}t) m/s")
35    print(f"Numerical W = {work_numerical:.4f} J")
36    print(f"Analytical W = {work_analytical:.4f} J")
37
38    return work_numerical
39
40def probability_expectation():
41    """
42    Expected value of X for exponential distribution:
43    E[X] = ∫₀^∞ x · λe^(-λx) dx
44
45    This requires integration by parts!
46    """
47    lambda_param = 2.0
48
49    def integrand(x):
50        return x * lambda_param * np.exp(-lambda_param * x)
51
52    # Numerical integration
53    E_X_numerical, _ = integrate.quad(integrand, 0, np.inf)
54
55    # Analytical: E[X] = 1/λ for exponential distribution
56    E_X_analytical = 1 / lambda_param
57
58    print("\nProbability: Expected Value of Exponential Distribution")
59    print("=" * 55)
60    print(f"E[X] = ∫₀^∞ x · λe^(-λx) dx, with λ = {lambda_param}")
61    print(f"Numerical E[X] = {E_X_numerical:.6f}")
62    print(f"Analytical E[X] = 1/λ = {E_X_analytical:.6f}")
63
64    # Show IBP steps
65    print("\nUsing IBP with u = x, dv = λe^(-λx)dx:")
66    print("du = dx, v = -e^(-λx)")
67    print("∫xλe^(-λx)dx = -xe^(-λx)|₀^∞ + ∫e^(-λx)dx")
68    print("            = 0 + [-1/λ · e^(-λx)]₀^∞ = 1/λ")
69
70    return E_X_numerical
71
72def signal_processing_fourier():
73    """
74    Fourier coefficients often require integration by parts.
75
76    For a sawtooth wave f(x) = x on [0, 2π]:
77    bₙ = (1/π) ∫₀^(2π) x·sin(nx) dx
78
79    This integral requires IBP!
80    """
81    def compute_fourier_coefficient(n):
82        def integrand(x):
83            return x * np.sin(n * x)
84
85        result, _ = integrate.quad(integrand, 0, 2*np.pi)
86        return result / np.pi
87
88    print("\nSignal Processing: Fourier Coefficients of Sawtooth Wave")
89    print("=" * 55)
90    print("f(x) = x on [0, 2π], computing bₙ = (1/π)∫₀^(2π) x·sin(nx) dx")
91    print("-" * 55)
92
93    for n in range(1, 6):
94        bn = compute_fourier_coefficient(n)
95        # Analytical result from IBP: bₙ = -2/n
96        bn_analytical = -2.0 / n
97        print(f"b{n} = {bn:.6f} (analytical: {bn_analytical:.6f})")
98
99    print("\nUsing IBP with u = x, dv = sin(nx)dx:")
100    print("bₙ = (1/π)[−x/n·cos(nx) + 1/n²·sin(nx)]₀^(2π)")
101    print("   = (1/π)(−2π/n) = −2/n")
102
103def control_systems_laplace():
104    """
105    In control systems, integration by parts is used to derive
106    Laplace transforms. The Laplace transform of t·f(t) is:
107
108    L{t·f(t)} = -d/ds F(s)
109
110    This is proven using integration by parts!
111    """
112    print("\nControl Systems: Laplace Transform Property")
113    print("=" * 55)
114    print("L{t·f(t)} = ∫₀^∞ t·f(t)·e^(-st) dt")
115    print("\nUsing IBP on ∫₀^∞ f(t)·e^(-st) dt:")
116    print("  u = e^(-st), dv = f(t)dt")
117    print("  du = -s·e^(-st)dt, v = ∫f(t)dt")
118    print("\nDifferentiating F(s) with respect to s:")
119    print("  d/ds F(s) = d/ds ∫₀^∞ f(t)e^(-st) dt")
120    print("            = ∫₀^∞ f(t)·(-t)·e^(-st) dt")
121    print("            = -L{t·f(t)}")
122    print("\nTherefore: L{t·f(t)} = -d/ds F(s)")
123
124    # Verify with example: f(t) = e^(-at)
125    # F(s) = 1/(s+a)
126    # L{t·e^(-at)} = -d/ds[1/(s+a)] = 1/(s+a)²
127
128    a, s = 2.0, 3.0
129
130    # Numerical verification
131    def integrand(t):
132        return t * np.exp(-a*t) * np.exp(-s*t)
133
134    numerical_result, _ = integrate.quad(integrand, 0, 100)
135    analytical_result = 1 / (s + a)**2
136
137    print(f"\nExample: f(t) = e^(-{a}t), s = {s}")
138    print(f"L{{t·e^(-{a}t)}} at s={s}:")
139    print(f"  Numerical:  {numerical_result:.6f}")
140    print(f"  Analytical: {analytical_result:.6f}")
141
142# Run demonstrations
143physics_work_integral()
144probability_expectation()
145signal_processing_fourier()
146control_systems_laplace()

Common Mistakes to Avoid

Mistake 1: Wrong Choice of u and dv

Wrong: For $\int x e^x \, dx$ , choosing $u = e^x$ and $dv = x \, dx$

This gives $\int e^x \cdot \frac{x^2}{2} \, dx$ — harder, not easier!

Correct: Use LIATE: $u = x$ , $dv = e^x \, dx$

Mistake 2: Forgetting the Constant of Integration

When finding $v$ from $dv$ , we don't add $+C$ at that step — but we must add it to the final answer.

Mistake 3: Sign Errors in the Tabular Method

The alternating signs are crucial: $+, -, +, -, \ldots$ . A sign error propagates through the entire calculation.

Mistake 4: Not Recognizing the Cycling Case

For $\int e^x \sin x \, dx$ , don't keep applying IBP indefinitely. After 2 applications, the original integral reappears — solve for it algebraically.

Mistake 5: Ignoring Boundary Terms for Definite Integrals

For $\int_a^b u \, dv$ , remember to evaluate $[uv]_a^b$ first before computing the remaining integral.

Test Your Understanding

Question 1 of 8Score: 0/0

For ∫ x cos(x) dx, what should you choose as u according to LIATE?

Summary

Integration by parts is a powerful technique derived from the product rule that transforms difficult integrals into manageable ones.

Key Formulas

Formula	When to Use
∫ u dv = uv - ∫ v du	Basic integration by parts
∫ₐᵇ u dv = [uv]ₐᵇ - ∫ₐᵇ v du	Definite integrals
LIATE: L-I-A-T-E	Choosing u (Log, Inverse trig, Algebraic, Trig, Exp)
Tabular method	Repeated IBP with polynomials
Solve for I when I reappears	Cycling cases (eˣ sin x, eˣ cos x)

Key Takeaways

IBP reverses the product rule: It transforms $\int u \, dv$ into $uv - \int v \, du$
LIATE guides your choice: Choose $u$ to be the function that simplifies upon differentiation
Tabular method saves time: For polynomials times exponentials/trig, build a table
Watch for cycles: When the original integral reappears, solve algebraically
Verify by differentiation: Always check your answer by differentiating
Applications abound: From physics to probability to machine learning, IBP is essential

The Core Insight:

"Integration by parts trades one integral for another — the art is choosing the trade that makes your life easier."

Coming Next: In Trigonometric Integrals, we'll explore specialized techniques for integrating products and powers of trigonometric functions.

Learning Objectives

The Big Picture: Reversing the Product Rule

The Core Insight

Why Do We Need This?

Historical Context

Gottfried Wilhelm Leibniz (1646–1716)

The Name "Integration by Parts"

Derivation from the Product Rule

Step 1: Start with the Product Rule

Step 2: Rearrange

Step 3: Integrate Both Sides

Step 4: Write in Differential Form

Integration by Parts Formula

Memory Aid

The Integration by Parts Formula

The Process

A First Example: ∫xex dx\int x e^x \, dx∫xexdx

Verification

Interactive: Visualizing Integration by Parts

Integration by Parts Step-by-Step

Choosing u and dv: The LIATE Rule

The LIATE Rule

Why LIATE Works

LIATE is a Guideline, Not a Law

Worked Examples

Example 1: ∫x2ex dx\int x^2 e^x \, dx∫x2exdx

Example 2: ∫ln⁡x dx\int \ln x \, dx∫lnxdx

Example 3: ∫exsin⁡x dx\int e^x \sin x \, dx∫exsinxdx (The Cycling Case)

Example 4: ∫arctan⁡x dx\int \arctan x \, dx∫arctanxdx

Tabular Integration Method

The Process

Example: ∫x3e2x dx\int x^3 e^{2x} \, dx∫x3e2xdx

Interactive: Tabular Method

Tabular Integration Method

Repeated Integration by Parts

Case 1: Polynomial Reduction

Case 2: Cyclic Return

Integration by Parts for Definite Integrals

Integration by Parts for Definite Integrals

Example: ∫01xex dx\int_0^1 x e^x \, dx∫01​xexdx

Real-World Applications

Physics: Work Done by Variable Forces

Probability: Expected Values

Signal Processing: Fourier Coefficients

Control Systems: Laplace Transforms

Machine Learning Connection

The Score Function and REINFORCE

The Gamma Function Recurrence

Variance Reduction in Monte Carlo

Python Implementation

Numerical and Symbolic Integration by Parts

Applications in Physics and Engineering

Common Mistakes to Avoid

Mistake 1: Wrong Choice of u and dv

Mistake 2: Forgetting the Constant of Integration

Mistake 3: Sign Errors in the Tabular Method

Mistake 4: Not Recognizing the Cycling Case

Mistake 5: Ignoring Boundary Terms for Definite Integrals

Test Your Understanding

For ∫ x cos(x) dx, what should you choose as u according to LIATE?

Summary

Key Formulas

Key Takeaways

A First Example: $\int x e^x \, dx$

Example 1: $\int x^2 e^x \, dx$

Example 2: $\int \ln x \, dx$

Example 3: $\int e^x \sin x \, dx$ (The Cycling Case)

Example 4: $\int \arctan x \, dx$

Example: $\int x^3 e^{2x} \, dx$

Example: $\int_0^1 x e^x \, dx$