20.2 Antiderivatives, Indefinite Integrals & Integration by U-Substitution

20.2 Antiderivatives, Indefinite Integrals & Integration by U-Substitution

The Concept of Antiderivatives & The Indefinite Integral

Integral calculus begins by reversing the operation of differentiation. Given a function $f$ defined on an interval $I$, a differentiable function $F$ is called an antiderivative of $f$ on $I$ if: $F'(x) = f(x) \quad \text{for all } x \in I$

Antiderivatives are not unique. If $F(x)$ is an antiderivative of $f(x)$, then for any real constant $C$, the function $G(x) = F(x) + C$ is also an antiderivative because the derivative of a constant is zero: $\frac{d}{dx}[F(x) + C] = F'(x) + 0 = f(x)$.

A fundamental theorem grounded in the Mean Value Theorem guarantees that all antiderivatives of a given function differ by at most an additive constant: if $F_1'(x) = F_2'(x)$ for all $x$ in an open interval $I$, then there exists a constant $C$ such that $F_1(x) - F_2(x) = C$. The collection of all antiderivatives of $f(x)$ is designated the indefinite integral of $f$ with respect to $x$, denoted: $\int f(x)\,dx = F(x) + C$ where the elongated symbol $\int$ is the integral sign, $f(x)$ is the integrand, $x$ is the variable of integration, $dx$ is the differential indicating the variable, and $C$ is the arbitrary constant of integration. Geometrically, $\int f(x),dx$ represents an infinite family of parallel curves that are vertical translations of one another.

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Linearity Properties & Basic Integration Rules

Like differentiation, indefinite integration is a linear operator:

1. Constant Multiple Rule: $\int c f(x),dx = c \int f(x),dx$ for any constant $c$.
2. Sum and Difference Rule: $\int [f(x) \pm g(x)],dx = \int f(x),dx \pm \int g(x),dx$.

The inverse of the power rule for differentiation yields the Power Rule for Integration: $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad \text{for any real } n \neq -1$ When $n = -1$, the power rule denominator becomes zero, reflecting a mathematical singularity. The antiderivative of $x^{-1} = \frac{1}{x}$ is the natural logarithm function: $\int \frac{1}{x}\,dx = \ln|x| + C$ The absolute value bars are mathematically required because the domain of $f(x) = \frac{1}{x}$ is all non-zero real numbers $(-\infty, 0) \cup (0, \infty)$, while $\ln(x)$ is defined only for positive reals. For $x < 0$, $\frac{d}{dx}[\ln(-x)] = \frac{1}{-x}(-1) = \frac{1}{x}$, confirming validity for all $x \neq 0$.

Exponential and trigonometric integration formulas derive directly from reversing standard differentiation rules:

• Exponential Functions: $\int e^x,dx = e^x + C$, and for general base $a > 0, a \neq 1$: $\int a^x,dx = \frac{a^x}{\ln a} + C$.
• Trigonometric Functions: $\int \cos x\,dx = \sin x + C, \quad \int \sin x\,dx = -\cos x + C$ $\int \sec^2 x\,dx = \tan x + C, \quad \int \csc^2 x\,dx = -\cot x + C$ $\int \sec x \tan x\,dx = \sec x + C, \quad \int \csc x \cot x\,dx = -\csc x + C$

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Comprehensive Integral Calculus Reference

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Method of Integration by U-Substitution

The method of $u$-substitution is the integral calculus counterpart to the Chain Rule of differential calculus. By the Chain Rule, if $F$ is an antiderivative of $f$, then: $\frac{d}{dx}[F(g(x))] = F'(g(x)) g'(x) = f(g(x)) g'(x)$ Integrating both sides with respect to $x$ yields the substitution identity: $\int f(g(x)) g'(x)\,dx = F(g(x)) + C$

By introducing an intermediate variable $u = g(x)$, the differential is $du = g'(x),dx$. Substituting transforms the complicated integral into a standard elementary integral: $\int f(g(x)) g'(x)\,dx = \int f(u)\,du = F(u) + C = F(g(x)) + C$

Systematic U-Substitution Protocol

1. Identify the Inner Function $g(x)$: Look for a composite expression where the integrand contains both an inner function $u = g(x)$ and, up to a constant multiplicative factor, its derivative $g'(x)$.
2. Compute the Differential: Differentiate to write $du = g'(x),dx$, and solve for the differential factor: $dx = \frac{du}{g'(x)}$.
3. Substitute and Balance Constants: Rewrite the entire integrand in terms of $u$ and $du$. Any constant multiplier may be factored outside the integral. Every trace of the original variable $x$ must be eliminated.
4. Integrate in Terms of $u$: Apply standard antiderivative formulas.
5. Back-Substitute: Replace $u$ with the original function $g(x)$ so that the final antiderivative is expressed in terms of the original variable $x$.

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Advanced Algebraic Substitution Techniques

In certain integrals, introducing $u = g(x)$ leaves an uncancelled polynomial factor of $x$. When this occurs, one must invert the substitution relation to express $x$ directly in terms of $u$: $x = g^{-1}(u)$.

For instance, consider $\int x \sqrt{x + 2},dx$. Let $u = x + 2$, which gives $du = dx$ and $x = u - 2$. Substituting yields: $\int (u - 2)\sqrt{u}\,du = \int (u - 2) u^{1/2}\,du = \int (u^{3/2} - 2u^{1/2})\,du$ Applying the power rule to each term converts a non-distributable radical into an easily integrable sum of rational powers: $\frac{2}{5}u^{5/2} - 2\left(\frac{2}{3}u^{3/2}\right) + C = \frac{2}{5}(x + 2)^{5/2} - \frac{4}{3}(x + 2)^{3/2} + C$

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Initial Value Problems & Physical Kinematics

An initial value problem (IVP) consists of a differential equation $\frac{dy}{dx} = f(x)$ paired with an auxiliary boundary condition $y(x_0) = y_0$. While indefinite integration produces a general solution containing the arbitrary constant $C$, substituting the coordinate pair $(x_0, y_0)$ isolates the unique value of $C$ defining the particular solution curve passing through that point.

In physics and engineering kinematics, rectilinear particle motion is modeled by hierarchical derivatives:

• Position: $s(t)$
• Velocity: $v(t) = s'(t) = \frac{ds}{dt} \implies s(t) = \int v(t),dt + s_0$
• Acceleration: $a(t) = v'(t) = s''(t) = \frac{dv}{dt} \implies v(t) = \int a(t),dt + v_0$

Given the acceleration function $a(t)$ and initial conditions for velocity $v(0) = v_0$ and position $s(0) = s_0$, two successive stages of indefinite integration completely reconstruct the particle's trajectory.

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Worked Exemplar: U-Substitution in a Multi-Stage Kinematic Initial Value Problem

Problem: A particle moves along a horizontal track with acceleration given by $a(t) = 12t (t^2 + 1)^2$ for $t \ge 0$. At time $t = 0$, its initial velocity is $v(0) = 5\text{ m/s}$ and its initial position is $s(0) = 2\text{ m}$. Derive the exact position function $s(t)$ and calculate the particle's position at $t = 1\text{ s}$.

Step 1: Determine Velocity via U-Substitution. Velocity is the indefinite integral of acceleration: $v(t) = \int a(t)\,dt = \int 12t (t^2 + 1)^2\,dt$ Let $u = t^2 + 1$. The differential is $du = 2t,dt$, which implies $12t,dt = 6(2t,dt) = 6,du$. Substitute into the integral: $v(t) = \int 6 u^2\,du = 6 \left(\frac{u^3}{3}\right) + C_1 = 2u^3 + C_1$ Back-substitute $u = t^2 + 1$: $v(t) = 2(t^2 + 1)^3 + C_1$

Step 2: Apply Initial Velocity to Solve for $C_1$. Use the initial condition $v(0) = 5$: $v(0) = 2(0^2 + 1)^3 + C_1 = 5 \implies 2(1) + C_1 = 5 \implies C_1 = 3$ Thus, the particular velocity function is: $v(t) = 2(t^2 + 1)^3 + 3$

Step 3: Determine Position via Integration. Position is the indefinite integral of velocity: $s(t) = \int v(t)\,dt = \int \left[ 2(t^2 + 1)^3 + 3 \right]\,dt$ Expand the cubic binomial $(t^2 + 1)^3 = (t^2)^3 + 3(t^2)^2(1) + 3(t^2)(1)^2 + 1^3 = t^6 + 3t^4 + 3t^2 + 1$: $v(t) = 2(t^6 + 3t^4 + 3t^2 + 1) + 3 = 2t^6 + 6t^4 + 6t^2 + 5$ Integrate term-by-term using the Power Rule: $s(t) = \int (2t^6 + 6t^4 + 6t^2 + 5)\,dt = \frac{2}{7}t^7 + \frac{6}{5}t^5 + 2t^3 + 5t + C_2$

Step 4: Apply Initial Position to Solve for $C_2$. Use the initial condition $s(0) = 2$: $s(0) = \frac{2}{7}(0) + \frac{6}{5}(0) + 2(0) + 5(0) + C_2 = 2 \implies C_2 = 2$ The exact closed-form position function is: $s(t) = \frac{2}{7}t^7 + \frac{6}{5}t^5 + 2t^3 + 5t + 2$

Step 5: Evaluate Position at $t = 1\text{ s}$. Substitute $t = 1$ into $s(t)$: $s(1) = \frac{2}{7}(1) + \frac{6}{5}(1) + 2(1) + 5(1) + 2 = \frac{2}{7} + \frac{6}{5} + 9$ Find a common denominator of 35: $\frac{2}{7} = \frac{10}{35}, \quad \frac{6}{5} = \frac{42}{35}, \quad 9 = \frac{315}{35}$ $s(1) = \frac{10 + 42 + 315}{35} = \frac{367}{35}\text{ meters}$