Textbook Question
Limits of sums Use the definition of the definite integral to evaluate the following definite integrals. Use right Riemann sums and Theorem 5.1.
β«ββ΄ (πΒ²β1) dπ
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Ch. 5 - Integration
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 5, Problem 5.5.84
Variations on the substitution method Evaluate the following integrals.
β« (π΅ + 1) β(3π΅ + 2) dπ΅
Verified step by step guidance1
Identify the integral to solve: \(\int (Z + 1) \sqrt{3Z + 2} \, dZ\).
Choose a substitution to simplify the integral. Let \(u = 3Z + 2\), so that the expression under the square root becomes \(\sqrt{u}\).
Compute the differential \(du\) in terms of \(dZ\): since \(u = 3Z + 2\), then \(\frac{du}{dZ} = 3\), which implies \(dZ = \frac{du}{3}\).
Express \(Z\) in terms of \(u\) to rewrite the factor \((Z + 1)\): from \(u = 3Z + 2\), solve for \(Z\) to get \(Z = \frac{u - 2}{3}\), so \(Z + 1 = \frac{u - 2}{3} + 1 = \frac{u + 1}{3}\).
Rewrite the integral entirely in terms of \(u\) and \(du\): substitute \((Z + 1)\) and \(\sqrt{3Z + 2}\) with their expressions in \(u\), and replace \(dZ\) with \(\frac{du}{3}\). Then simplify the integrand before integrating with respect to \(u\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Substitution Method in Integration
The substitution method simplifies integrals by changing variables to transform a complicated integral into a basic form. It involves choosing a substitution u = g(z) such that the integral becomes easier to evaluate. This method is especially useful when the integral contains a composite function and its derivative.
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07:33
Euler's Method
Chain Rule and Its Role in Integration
The chain rule in differentiation helps identify the inner function and its derivative, which guides the substitution choice in integration. Recognizing the derivative of the inner function within the integral allows for an effective substitution, turning the integral into a simpler polynomial or standard form.
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05:02Intro to the Chain Rule
Integration of Polynomial and Root Functions
Integrals involving polynomials and roots often require rewriting the root as a fractional exponent. After substitution, the integral can be expressed as a sum of powers of the variable, which can be integrated using the power rule. Understanding how to manipulate and integrate these expressions is essential for solving such integrals.
Recommended video:
07:00Taylor Polynomials
Related Practice
Textbook Question
Definite integrals Use a change of variables or Table 5.6 to evaluate the following definite integrals.
β«β/ββ βββ^Β²/β΅ dπ/ xβ(25πΒ²β 1)
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Textbook Question
Integrals with sinΒ² π and cosΒ² π Evaluate the following integrals.
β« sinΒ² π dπ
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Textbook Question
Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus
β«ββ΄ (π β 2)/βπ dπ
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Textbook Question
Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating.
β« π csc πΒ² cot πΒ² dπ
Textbook Question
On which derivative rule is the Substitution Rule based?
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