Book cover for Calculus Early Transcendental Functions

Calculus Early Transcendental Functions

Ron Larson, Bruce Edwards

ISBN #9781285774770

6th Edition

8,973 Questions

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Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
Common Mistakes

Summary

Chapter 8 emphasizes the importance of mastering various integration techniques. It discusses how to match integrals with basic integration rules, and highlights the benefits of algebraic manipulation and substitution. Special focus is given to integration by parts, which is a powerful tool for evaluating products of functions. The chapter provides strategic guidelines to choose the appropriate method, ensuring that even seemingly complex integrals can be systematically reduced to simpler forms.

Learning Objectives

1

Apply and recognize appropriate basic integration rules to a variety of integrands.

2

Perform algebraic manipulations such as long division, factoring, and completing the square to rewrite integrals in a form matching a known rule.

3

Utilize substitution techniques, trigonometric identities, and the constant multiple rule to simplify integrals.

4

Solve integration by parts problems, especially for integrals involving products of algebraic and transcendental functions.

5

Analyze and reduce complex integrals into simpler components using strategic choices of functions for u and dv.

Key Concepts

CONCEPT

DEFINITION

Basic Integration Rules

A set of standard formulas used to calculate antiderivatives of common functions, including power rule, log rule, trigonometric integrals, and inverse trigonometric formulas.

Arctangent Rule

An integration rule for functions of the form ∫du/(a² + u²) which yields (1/a) arctan(u/a) + C.

Log Rule

An integration technique useful for integrals where the derivative of the denominator appears in the numerator, yielding a natural logarithm of the absolute value of the denominator.

Integration by Parts

A technique based on the product rule for differentiation, expressed as ∫u dv = uv − ∫v du, used to integrate products of functions.

Substitution

A method for simplifying integrals by changing variables so the integrand matches a standard form, often reversing the chain rule.

Example Problems

Example 1

Select the correct antiderivative. $$\frac{d y}{d x}=\frac{x}{\sqrt{x^{2}+1}}$$ (a) $2 \sqrt{x^{2}+1}+C$ (b) $\sqrt{x^{2}+1}+C$ (c) $\frac{1}{2} \sqrt{x^{2}+1}+C$ (d) $\ln \left(x^{2}+1\right)+C$

Example 2

Select the correct antiderivative. $$\frac{d y}{d x}=\frac{x}{x^{2}+1}$$ (a) $\ln \sqrt{x^{2}+1}+C$ (b) $\frac{2 x}{\left(x^{2}+1\right)^{2}}+C$ (c) arctan $x+C$ (d) $\ln \left(x^{2}+1\right)+C$

Example 3

Select the correct antiderivative. $$\frac{d y}{d x}-\frac{1}{x^{2}+1}$$ (a) $\ln \sqrt{x^{2}+1}+C$ (b) $\frac{2 x}{\left(x^{2}+1\right)^{2}}+c$ (c) arctan $x+C$ (d) $\ln \left(x^{2}+1\right)+C$

Example 4

Select the correct antiderivative. $$\frac{d y}{d x}=x \cos \left(x^{2}+1\right)$$ (a) $2 x \sin \left(x^{2}+1\right)+C$ (b) $-\frac{1}{2} \sin \left(x^{2}+1\right)+C$ (c) $\frac{1}{2} \sin \left(x^{2}+1\right)+C$ (d) $-2 x \sin \left(x^{2}+1\right)+C$

Example 5

Select the basic integration formula you can use to find the integral, and identify $u$ and $a$ when appropriate. $$\int(5 x-3)^{4} d x$$
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Step-by-Step Explanations

QUESTION

Evaluate the integral ∫ (1/(1 + x²)) dx using the Arctangent Rule.

STEP-BY-STEP ANSWER:

Step 1: Recognize that the integrand 1/(1+x²) matches the standard form for the arctan rule, where a = 1.
Step 2: According to the rule, ∫ du/(a²+u²) = (1/a) arctan(u/a) + C. Here, u = x and a = 1.
Step 3: Substitute into the formula to obtain: (1/1) arctan(x/1) + C.
Final Answer: arctan(x) + C.

Arctangent Rule Example

QUESTION

Evaluate the integral ∫ e^x sin x dx using integration by parts.

STEP-BY-STEP ANSWER:

Step 1: Choose u and dv. Let u = sin x (since its derivative, cos x, is simpler) and dv = e^x dx (since its antiderivative remains e^x).
Step 2: Compute du and v. We have du = cos x dx and v = e^x.
Step 3: Apply the integration by parts formula: ∫ e^x sin x dx = u*v − ∫ v du = e^x sin x − ∫ e^x cos x dx.
Step 4: Now evaluate ∫ e^x cos x dx by applying integration by parts again. Let u = cos x (whose derivative is −sin x) and dv = e^x dx, so that du = −sin x dx and v = e^x.
Step 5: Then ∫ e^x cos x dx = e^x cos x − ∫ e^x (−sin x) dx = e^x cos x + ∫ e^x sin x dx.
Step 6: Substitute this back into the earlier expression: ∫ e^x sin x dx = e^x sin x − [e^x cos x + ∫ e^x sin x dx].
Step 7: Combine like terms by adding ∫ e^x sin x dx to both sides: 2∫ e^x sin x dx = e^x sin x − e^x cos x.
Step 8: Solve for the original integral: ∫ e^x sin x dx = (e^x (sin x − cos x)) / 2 + C.
Final Answer: (e^x (sin x − cos x)) / 2 + C.

Integration by Parts Example

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Common Mistakes

  • Incorrectly separating the denominator when rewriting integrands—always keep the denominator intact during algebraic manipulation.
  • Choosing a poor substitution or misidentifying u and dv in integration by parts, which can lead to even more complex integrals.
  • Overlooking the need for additional algebraic steps (such as long division or adding and subtracting terms) before applying integration rules.
  • Forgetting the constant multiple rule or errors in differentiating during substitution.