7.2 Definite Integrals via Substitution

7.2 Definite Integrals via Substitution

Core CLEP Concept: Integration by substitution ($u$-substitution) is the calculus counterpart to the Chain Rule. When evaluating definite integrals, substitution introduces a critical procedural decision: you can either transform the limits of integration into the new variable $u$ (Method 1: The Gold Standard), or find the indefinite antiderivative in terms of $x$ before evaluating the original bounds (Method 2). On the CLEP Calculus exam, transforming limits is overwhelmingly preferred because it saves time, eliminates back-substitution errors, and is explicitly tested in "integral equivalence" questions.

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1. The Substitution Theorem for Definite Integrals

The theoretical basis for changing variables in a definite integral combines the differential Chain Rule with FTC Part 2.

Theorem (Substitution for Definite Integrals): If $g'$ is continuous on the closed interval $[a, b]$ and $f$ is continuous on the range of $u = g(x)$, then: $\int_a^b f(g(x)) \cdot g'(x) \, dx = \int_{g(a)}^{g(b)} f(u) \, du$

Derivation / Proof of the Theorem

Let $F$ be an antiderivative of $f$, such that $F'(u) = f(u)$. By the Chain Rule, the derivative of the composite function $F(g(x))$ with respect to $x$ is:

$$\frac{d}{dx} [F(g(x))] = F'(g(x)) \cdot g'(x) = f(g(x)) \cdot g'(x)$$

Applying the Fundamental Theorem of Calculus (FTC Part 2) to the left-hand integral:

$$\int_a^b f(g(x)) \cdot g'(x) \, dx = [F(g(x))]_a^b = F(g(b)) - F(g(a))$$

Now apply FTC Part 2 to the right-hand integral in terms of $u$:

$$\int_{g(a)}^{g(b)} f(u) \, du = [F(u)]_{g(a)}^{g(b)} = F(g(b)) - F(g(a))$$

Because both sides evaluate to the exact same numerical difference $F(g(b)) - F(g(a))$, the equality holds:

$$\int_a^b f(g(x)) \cdot g'(x) \, dx = \int_{g(a)}^{g(b)} f(u) \, du \quad \blacksquare$$

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2. Method 1: Transforming Limits of Integration (The Gold Standard)

Method 1 changes the limits of integration from $x$-values to $u$-values at the very moment the substitution $u = g(x)$ is defined. Once transformed, the integral is evaluated entirely in terms of $u$. You never back-substitute to $x$.

   Original Problem:    / b
                        |  f( g(x) ) * g'(x) dx
                        / a
                             |
         Transform Bounds:   |  x = a  ==>  u_lower = g(a)
                             |  x = b  ==>  u_upper = g(b)
                             v
   Transformed Integral:/ g(b)
                        |      f(u) du  =  [ F(u) ]_{g(a)}^{g(b)}  =  F(g(b)) - F(g(a))
                        / g(a)

The 5-Step Execution Protocol for Method 1

1. Choose the inner substitution variable: Identify $u = g(x)$ (look inside radicals, composite parentheses, exponents, or denominators).
2. Compute the differential: Calculate $du = g'(x) \, dx$ and balance any constant numerical scalar multipliers ($dx = \frac{du}{g'(x)}$).
3. Calculate the new integration bounds:
   • New lower bound: $u_{\text{lower}} = g(a)$
   • New upper bound: $u_{\text{upper}} = g(b)$
4. Rewrite the integral completely: Express all components in terms of $u$ and $du$, bounded from $g(a)$ to $g(b)$. Ensure no $x$ variables remain.
5. Evaluate via FTC 2: Find an antiderivative $F(u)$ and compute $F(g(b)) - F(g(a))$.

What if $g(a) > g(b)$?

It is entirely normal for the new lower bound $g(a)$ to be numerically larger than the new upper bound $g(b)$ (for instance, when $g(x)$ is a decreasing function like $\cos x$, $1/x$, or $-x^2$).

Rule: Do not arbitrarily swap the positions of the bounds without introducing a negative sign. Maintain the exact order $u_{\text{lower}} = g(a)$ at the bottom and $u_{\text{upper}} = g(b)$ at the top:

$$\int_{g(a)}^{g(b)} f(u) \, du = F(g(b)) - F(g(a))$$

Alternatively, you may apply the bounds reversal property: $\int_{g(a)}^{g(b)} f(u) \, du = -\int_{g(b)}^{g(a)} f(u) \, du$.

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3. Method 2: Indefinite Antiderivative Evaluation (The Alternative)

In Method 2, you temporarily set aside the bounds, treat the integrand as an indefinite integral, find the general antiderivative in terms of $u$, back-substitute $u = g(x)$ to return to the variable $x$, and finally evaluate using the original bounds $a$ and $b$:

$$\int f(g(x)) g'(x) \, dx = \int f(u) \, du = F(u) = F(g(x)) \implies \left. F(g(x)) \right|_a^b = F(g(b)) - F(g(a))$$

Comparison of Methods & Why Method 1 Dominates

The High-Yield CLEP Question Type: Integral Equivalence

On the CLEP exam, several questions will test your conceptual understanding of substitution without asking for a numerical answer. A typical question asks:

"Which of the following integrals is equivalent to $\int_0^2 x e^{x^2} \, dx$?"

To solve this, perform the transformation:

• Let $u = x^2 \implies du = 2x \, dx \implies x \, dx = \frac{1}{2} \, du$.
• Lower bound: when $x = 0 \implies u = 0^2 = 0$.
• Upper bound: when $x = 2 \implies u = 2^2 = 4$.
• Transformed equivalent integral: $\frac{1}{2} \int_0^4 e^u \, du$.

Students who rely exclusively on Method 2 often struggle with these questions because they never learn to transform limits systematically.

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4. Comprehensive Worked Examples by Function Type

Example 1: Algebraic Power with Constant Adjustment

Problem: Evaluate $\int_0^1 x \left( 1 + 2x^2 \right)^3 \, dx$.

• Step 1: Let $u = 1 + 2x^2$.
• Step 2: $du = 4x \, dx \implies x \, dx = \frac{1}{4} \, du$.
• Step 3: Transform limits:
  • Lower limit ($x = 0$): $u(0) = 1 + 2(0)^2 = 1$.
  • Upper limit ($x = 1$): $u(1) = 1 + 2(1)^2 = 3$.
• Step 4: Rewrite the integral: $$\int_0^1 x \left( 1 + 2x^2 \right)^3 \, dx = \int_1^3 u^3 \left( \frac{1}{4} \, du \right) = \frac{1}{4} \int_1^3 u^3 \, du$$
• Step 5: Evaluate: $$\frac{1}{4} \left[ \frac{u^4}{4} \right]_1^3 = \frac{1}{16} \left[ 3^4 - 1^4 \right] = \frac{1}{16} [81 - 1] = \frac{80}{16} = 5$$

Example 2: Trigonometric Function with Transcendental Bounds

Problem: Evaluate $\int_0^{\pi/4} \tan x \sec^2 x \, dx$.

• Step 1: Let $u = \tan x$.
• Step 2: $du = \sec^2 x \, dx$.
• Step 3: Transform limits:
  • Lower limit ($x = 0$): $u(0) = \tan(0) = 0$.
  • Upper limit ($x = \pi/4$): $u(\pi/4) = \tan(\pi/4) = 1$.
• Step 4: Rewrite and evaluate: $$\int_0^1 u \, du = \left[ \frac{u^2}{2} \right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}$$

(Alternative Substitution: Setting $u = \sec x \implies du = \sec x \tan x \, dx$ transforms limits to $u(0) = 1$ and $u(\pi/4) = \sqrt{2}$, yielding $\int_1^{\sqrt{2}} u \, du = \left[\frac{u^2}{2}\right]_1^{\sqrt{2}} = \frac{2}{2} - \frac{1}{2} = \frac{1}{2}$.)

Example 3: Natural Logarithm with Rational Integrand

Problem: Evaluate $\int_1^e \frac{(\ln x)^2}{x} \, dx$.

• Step 1: Let $u = \ln x$.
• Step 2: $du = \frac{1}{x} \, dx$.
• Step 3: Transform limits:
  • Lower limit ($x = 1$): $u(1) = \ln(1) = 0$.
  • Upper limit ($x = e$): $u(e) = \ln(e) = 1$.
• Step 4: Rewrite and evaluate: $$\int_0^1 u^2 \, du = \left[ \frac{u^3}{3} \right]_0^1 = \frac{1}{3} - 0 = \frac{1}{3}$$

Example 4: Exponential with Radical Exponent

Problem: Evaluate $\int_1^4 \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx$.

• Step 1: Let $u = \sqrt{x} = x^{1/2}$.
• Step 2: $du = \frac{1}{2\sqrt{x}} \, dx \implies \frac{dx}{\sqrt{x}} = 2 \, du$.
• Step 3: Transform limits:
  • Lower limit ($x = 1$): $u(1) = \sqrt{1} = 1$.
  • Upper limit ($x = 4$): $u(4) = \sqrt{4} = 2$.
• Step 4: Rewrite and evaluate: $$\int_1^4 \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx = \int_1^2 e^u (2 \, du) = 2 \int_1^2 e^u \, du = 2 [e^u]_1^2 = 2(e^2 - e)$$

Example 5: Inverse Trigonometric Form

Problem: Evaluate $\int_0^{1/3} \frac{1}{1 + 9x^2} \, dx$.

• Step 1: Rewrite $9x^2 = (3x)^2$. Let $u = 3x \implies du = 3 \, dx \implies dx = \frac{1}{3} \, du$.
• Step 2: Transform limits: $x = 0 \implies u = 0$; $x = 1/3 \implies u = 1$.
• Step 3: Rewrite and evaluate: $$\int_0^{1/3} \frac{1}{1 + 9x^2} \, dx = \frac{1}{3} \int_0^1 \frac{1}{1 + u^2} \, du = \frac{1}{3} [\arctan(u)]_0^1 = \frac{1}{3} \left(\arctan(1) - \arctan(0)\right) = \frac{1}{3} \left(\frac{\pi}{4} - 0\right) = \frac{\pi}{12}$$

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5. Symmetry Properties in Definite Integrals

When integrating over a symmetric interval $[-a, a]$, recognizing whether the integrand is an even function or an odd function can reduce complex integrals to simple arithmetic.

       EVEN FUNCTION: f(-x) = f(x)             ODD FUNCTION: f(-x) = -f(x)
              y                                       y
              |   y = f(x)                            |   y = f(x)
           +--+--+                                    |     +--+
          /|  |  |\                                   |    /|  |\
         / |  |  | \                                  |   / |  | \
        /  |  |  |  \                                 |  /  |  |  \
   ----+---+--+--+---+----> x                    ----+--+---+--+---+----+----> x
      -a   |     |   a                          -a   |  |   0  |   a
         Area1 = Area2                             -Area|     +Area
    int_{-a}^a = 2 * int_0^a                    int_{-a}^a = (-Area) + (+Area) = 0

Theorem: Integrals of Symmetric Functions

Let $f$ be continuous on $[-a, a]$:

1. Even Function ($f(-x) = f(x)$): The graph is symmetric with respect to the $y$-axis. $$\int_{-a}^a f(x) \, dx = 2 \int_0^a f(x) \, dx$$
2. Odd Function ($f(-x) = -f(x)$): The graph is symmetric with respect to the origin. $$\int_{-a}^a f(x) \, dx = 0$$

Proof for Odd Functions Using Substitution

$$\int_{-a}^a f(x) \, dx = \int_{-a}^0 f(x) \, dx + \int_0^a f(x) \, dx$$

In the first integral, let $u = -x \implies du = -dx$. When $x = -a, u = a$; when $x = 0, u = 0$:

$$\int_{-a}^0 f(x) \, dx = \int_a^0 f(-u) (-du) = \int_0^a f(-u) \, du$$

Because $f$ is odd, $f(-u) = -f(u)$:

$$\int_0^a [-f(u)] \, du = -\int_0^a f(u) \, du = -\int_0^a f(x) \, dx$$

Substituting back:

$$\int_{-a}^a f(x) \, dx = -\int_0^a f(x) \, dx + \int_0^a f(x) \, dx = 0 \quad \blacksquare$$

Step-by-Step Worked Example: Hybrid Symmetric Integrals

Problem: Evaluate $\int_{-\pi/3}^{\pi/3} \left( x^4 \sin(x) + \cos(x) \right) dx$.

• Step 1: Split into separate integrals: $$\int_{-\pi/3}^{\pi/3} x^4 \sin(x) \, dx + \int_{-\pi/3}^{\pi/3} \cos(x) \, dx$$
• Step 2: Parity check of term 1: Let $h(x) = x^4 \sin(x)$. $$h(-x) = (-x)^4 \sin(-x) = x^4 (-\sin x) = -x^4 \sin x = -h(x) \implies \text{ODD}$$ Therefore, $\int_{-\pi/3}^{\pi/3} x^4 \sin(x) \, dx = 0$.
• Step 3: Parity check of term 2: $\cos(-x) = \cos(x) \implies \text{EVEN}$. $$\int_{-\pi/3}^{\pi/3} \cos(x) \, dx = 2 \int_0^{\pi/3} \cos(x) \, dx = 2 [\sin x]_0^{\pi/3} = 2 \left( \sin\frac{\pi}{3} - \sin 0 \right) = 2\left(\frac{\sqrt{3}}{2} - 0\right) = \sqrt{3}$$
• Step 4: Total value: $0 + \sqrt{3} = \sqrt{3}$.

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6. Common CLEP Pitfalls & Traps

1. Evaluating the $u$-Antiderivative with Original $x$-Limits: The most catastrophic error on substitution problems is evaluating $[F(u)]_a^b$ without transforming the bounds. If $u = x^2 + 1$ and the original limits are $[0, 2]$, evaluating $F(u)$ from $0$ to $2$ yields a wrong answer. You must evaluate from $u(0) = 1$ to $u(2) = 5$.
2. Double-Substituting Bounds: If you transform the limits to $u_1$ and $u_2$, do not back-substitute $u = g(x)$ and then plug in $u_1$ and $u_2$. Either use $F(u)$ with $[g(a), g(b)]$, OR use $F(g(x))$ with $[a, b]$. Never mix them!
3. Dropping Constant Differential Fractions: If $u = 3x^2 \implies du = 6x \, dx$, the factor $x \, dx = \frac{1}{6} \, du$. Forgetting the factor $\frac{1}{6}$ in front of the integral will multiply your answer by 6.
4. Incorrect Parity Assumptions on Combined Functions: A sum of an even function and an odd function is neither even nor odd. You must split the integral term by term before applying symmetry shortcuts.