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

Guida verificata passo dopo passo

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.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.

Durata del video:

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

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.

Video consigliato:

Percorso guidato

06:18

Integration by Parts for Definite Integrals

Logarithmic Functions in Integration

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.

Video consigliato:

5:26

Graphs of Logarithmic Functions

Definite Integrals and Limits of Integration

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.

Video consigliato:

Percorso guidato