Detailed Notes on Integrals and Substitution Rule

5 Integrals

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5.5 The Substitution Rule

Copyright Cengage Learning. All rights reserved.

3 The Substitution Rule

• The Fundamental Theorem emphasizes the importance of finding antiderivatives.

• Antidifferentiation formulas alone do not solve certain integral evaluations.

• Introduce a new variable to assist in integral evaluation (a problem-solving strategy).

  • Example: Change from variable x to a new variable u.

Substitution: Indefinite Integrals

4 The Process of Substitution

• Let u be the quantity under the root sign in the integral.

• The differential of u is defined as:

  • $du = 2xdx$

• If dx is interpreted as a differential in this context, then:

  • The differential $2xdx$ will occur in our original integral notation.

• This results in the ability to write the integral in terms of u.

6 Verification of the Substitution Method

• To ensure correctness, check the result using the Chain Rule to differentiate the final function obtained through substitution.

• This method works for integrals expressible in the following form:

  • $ext{Integral form: } f(g(x))g'(x)dx$

7 The Substitution Rule Derived

• By applying the Chain Rule for differentiation, the following rule can be established:

  • If $u = g(x)$ is a differentiable function whose range is an interval I, and if f is continuous on I, then:

  • The Substitution Rule formulation can be expressed in terms of dx and du:

  • $ext{Integral: } ext{Integral } f(u)du = ext{Integral } f(g(x))g'(x)dx$

• Reflect upon thinking of dx and du as differentials to aid in remembering the rule.

9 Application of the Substitution Rule

• The Substitution Rule allows for the operation with dx and du after the integral signs as if they were differentials.

Example 1: Finding an Integral Using Substitution

• Task: Find the integral of a function by substituting variables.

• Solution process: Make the substitution based on the integral’s root expression.

  • Differential is adjusted accordingly to match the integral components.

  • Using the Substitution Rule leads to the solution.

• Note: Always revert to the original variable (x) in the final answer when required.

Substitution: Definite Integrals

12 Understanding Definite Integrals with Substitution

• Method 1: Evaluate the indefinite integral first, then apply the Fundamental Theorem of Calculus.

• Method 2: Change the limits of integration when changing variables, which is usually the preferable method.

14 The Substitution Rule for Definite Integrals

• If $f$ is continuous on the closed interval [a, b] and f is continuous in the range of the substitution $u = g(x)$, the following holds:

  • $ext{Integral from } a ext{ to } b f(g(x))g'(x)dx = ext{Integral from } u(a) ext{ to } u(b) f(u)du$

Example 7: Evaluating a Definite Integral

• Task: Evaluate the definite integral with a substitution.

• Solution Steps:

  • Let $u = 2x + 1$ and determine new limits of integration:

  • When $x = 0$, then $u = 2(0) + 1 = 1$.

  • When $x = 4$, then $u = 2(4) + 1 = 9$.

16 Final Steps in the Evaluation

• Upon applying the Substitution Rule, directly evaluate the expression in u without reverting to x after integrating.

Symmetry

18 Understanding Symmetry in Functions

• The Substitution Rule for Definite Integrals aids in simplifying the calculations for integrals of functions exhibiting symmetry properties.

• Integrals of Symmetric Functions Theorem:

  • If $f$ is continuous on [-a, a]:

  • (a) If $f$ is even, $ext{Integral from } -a ext{ to } a f(x)dx = 2 ext{Integral from } 0 ext{ to } a f(x)dx.$

  • (b) If $f$ is odd, $ext{Integral from } -a ext{ to } a f(x)dx = 0$ due to cancellation of areas.

20 Visual Representation

• Case of Positive and Even Function:

  • The area under the curve between $[-a, a]$ is twice that from $[0, a]$ due to symmetry.

• Case of Odd Function:

  • The integral evaluates to 0 since the positive area cancels the negative area.

Example 10: Evaluating an Even Function

• Function: $f (x) = x^{2} + 1$

• Properties: Satisfies $f (-x) = f (x)$ indicating it is an even function.

• Evaluation result:

  • $ext{Integral from } -b ext{ to } b (x^{2}+1)dx = 2 ext{Integral from } 0 ext{ to } b (x^{2}+1)dx$ yielding $2 imes ext{result}$.

Example 11: Evaluating an Odd Function

• Function: $f (x) = an(x)$

• Properties: Satisfies $f(-x) = -f(x)$ indicating it is an odd function.

• Integral evaluation leads to 0 due to symmetry:

  • $ext{Integral from } -a ext{ to } a an(x)dx = 0.$