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Ch. 5 - Integrals
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 5, Problem 5.PE.56

Evaluate the integrals in Exercises 47–68.


∫₀¹/² x³ (1 + 9x⁴)⁻³/² dx

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1
Identify the integral to solve: \(\int_0^{\frac{1}{2}} x^3 (1 + 9x^4)^{-\frac{3}{2}} \, dx\).
Look for a substitution that simplifies the expression inside the integral. Notice that the term \(1 + 9x^4\) is raised to a power and \(x^3\) is multiplied outside. Consider letting \(u = 1 + 9x^4\).
Compute the differential \(du\) by differentiating \(u\) with respect to \(x\): \(du = 36x^3 \, dx\). This allows you to express \(x^3 \, dx\) in terms of \(du\) as \(x^3 \, dx = \frac{du}{36}\).
Change the limits of integration from \(x\) to \(u\): when \(x=0\), \(u = 1 + 9(0)^4 = 1\); when \(x=\frac{1}{2}\), \(u = 1 + 9\left(\frac{1}{2}\right)^4 = 1 + 9 \times \frac{1}{16} = 1 + \frac{9}{16} = \frac{25}{16}\).
Rewrite the integral in terms of \(u\) using the substitution and new limits: \(\int_1^{\frac{25}{16}} u^{-\frac{3}{2}} \times \frac{1}{36} \, du\). Then, proceed to integrate \(u^{-\frac{3}{2}}\) 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.

Definite Integrals

A definite integral calculates the net area under a curve between two specific limits. It involves evaluating the antiderivative at the upper and lower bounds and subtracting these values. This process provides a numerical value representing the accumulation of quantities over the interval.
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Substitution Method

The substitution method simplifies integrals by changing variables to transform a complex integral into a more manageable form. By setting a part of the integrand as a new variable, the integral often becomes easier to evaluate, especially when dealing with composite functions.
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Power Functions and Exponents

Understanding how to manipulate power functions and exponents is essential for integrating expressions involving terms like x³ and (1 + 9x⁴)⁻³/². This includes applying exponent rules and recognizing how powers affect differentiation and integration.
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