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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 8.2.6

Evaluate the integrals in Exercises 1–24 using integration by parts.
∫(from 1 to e) x³ ln(x) dx

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1
Identify the integral to solve: \(\int_1^e x^3 \ln(x) \, dx\).
Choose functions for integration by parts: let \(u = \ln(x)\) (which simplifies upon differentiation) and \(dv = x^3 \, dx\) (which is easy to integrate).
Compute the derivatives and integrals needed: \(du = \frac{1}{x} \, dx\) and \(v = \frac{x^4}{4}\).
Apply the integration by parts formula: \(\int u \, dv = uv - \int v \, du\), so write the integral as \(\left. \frac{x^4}{4} \ln(x) \right|_1^e - \int_1^e \frac{x^4}{4} \cdot \frac{1}{x} \, dx\).
Simplify the remaining integral to \(\frac{1}{4} \int_1^e x^3 \, dx\) and prepare to evaluate both the boundary term and this integral.

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Key Concepts

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

Integration by Parts

Integration by parts is a technique derived from the product rule of 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 simplifies the problem, especially when one function becomes simpler upon differentiation.
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Logarithmic functions like ln(x) often appear in integrals where integration by parts is useful. Since the derivative of ln(x) is 1/x, selecting ln(x) as u simplifies the integral when differentiated. Understanding how to handle ln(x) helps in breaking down complex integrals involving logarithms.
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Definite integrals calculate the net area under a curve between two points, using specified limits. After integrating, the antiderivative is evaluated at the upper and lower limits, and their difference gives the integral's value. Properly applying limits is essential for accurate evaluation of definite integrals.
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