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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 8, Problem 8.2.26

Evaluate the integrals in Exercises 25–30 by using a substitution prior to integration by parts.
∫ from 0 to 1 x√(1 - x) dx

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Identify the integral to solve: \(\int_0^1 x \sqrt{1 - x} \, dx\).
Use substitution to simplify the integral. Let \(u = 1 - x\), which implies \(du = -dx\) and \(x = 1 - u\).
Change the limits of integration according to the substitution: when \(x = 0\), \(u = 1\); when \(x = 1\), \(u = 0\).
Rewrite the integral in terms of \(u\): \(\int_1^0 (1 - u) \sqrt{u} (-du)\), then reverse the limits to get rid of the negative sign, resulting in \(\int_0^1 (1 - u) \sqrt{u} \, du\).
Expand the integrand to \(\int_0^1 (1 - u) u^{1/2} \, du = \int_0^1 (u^{1/2} - u^{3/2}) \, du\), and then integrate term-by-term.

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Key Concepts

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

Substitution Method

The substitution method simplifies integrals by changing variables to transform the integral into a more manageable form. It involves choosing a substitution u = g(x) that simplifies the integrand, then rewriting the integral in terms of u and du. This technique is especially useful when the integral contains a composite function.
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Integration by Parts

Integration by parts is a technique based on the product rule for differentiation. It transforms the integral of a product of functions into simpler integrals using the formula ∫u dv = uv - ∫v du. Choosing u and dv wisely is key to simplifying the integral effectively.
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Definite Integrals and Limits of Integration

Definite integrals calculate the net area under a curve between two points, using specified limits of integration. When performing substitution, the limits must be adjusted to the new variable or the integral must be converted back to the original variable before evaluating. Proper handling of limits ensures accurate evaluation of the integral.
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