Ch. 5 - Integrals

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 5 - Integrals

Problem 5.PE.49

Chapter 5, Problem 5.PE.49

Evaluate the integrals in Exercises 47–68.

∫₁² 4 dv
v²

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Identify the integral to be evaluated: \(\int_1^2 \frac{4}{v^2} \, dv\).

Rewrite the integrand using negative exponents: \(\frac{4}{v^2} = 4v^{-2}\).

Apply the power rule for integration: For \(\int v^n \, dv = \frac{v^{n+1}}{n+1} + C\), where \(n \neq -1\).

Integrate \(4v^{-2}\) as \(4 \times \frac{v^{-2+1}}{-2+1} = 4 \times \frac{v^{-1}}{-1} = -4v^{-1}\).

Evaluate the definite integral by substituting the limits \(v=2\) and \(v=1\) into the antiderivative \(-4v^{-1}\) and find the difference.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Definite Integral

A definite integral calculates the net area under a curve between two specific limits. It is represented as ∫_a^b f(x) dx, where a and b are the lower and upper bounds. Evaluating a definite integral results in a numerical value representing the accumulated quantity over the interval.

Integrand Simplification

Before integrating, simplify the integrand if possible to make the integration process easier. In this problem, the integrand is 4/v², which can be rewritten as 4v⁻². Expressing the function in terms of powers of v allows the use of power rule integration.

Power Rule for Integration

The power rule states that ∫ x^n dx = (x^(n+1)) / (n+1) + C for any real number n ≠ -1. This rule is essential for integrating functions expressed as powers of the variable. Applying this rule to 4v⁻² helps find the antiderivative needed for evaluating the definite integral.

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