Integration of xln(1-2x) from 0 to 1: Step-by-Step Solutions and Techniques

Understanding the integration of trigonometric, logarithmic, and polynomial functions is a cornerstone of calculus. In this article, we will delve into the integration of xln(1-2x) from 0 to 1. By breaking down the process into simpler components, we can achieve a deeper understanding of integral calculus techniques.

Introduction to Integration by Parts

Integration by parts is a method derived from the product rule of differentiation. It is useful for integrating products or composite functions that are not easily integrated with basic techniques. The formula is:

u dv uv - ∫ v du

Step-by-Step Solution

Problem: Find the integral of xln(1-2x) dx from 0 to 1.

We will use integration by parts. Let:

u ln(1-2x) dv x dx

Therefore:

du -2/(1-2x) dx v x^2/2

Applying the formula for integration by parts:

∫ xln(1-2x) dx (x^2/2)ln(1-2x) - ∫ (x^2/2)(-2/(1-2x)) dx

Simplifying the integrand:

(x^2/2)ln(1-2x) ∫ (x^2/1-2x) dx

The next step is to evaluate the integral ∫ (x^2/1-2x) dx. We will use a substitution method:

w 1-2x dw -2 dx

Therefore:

dx -dw/2

When x 0, w 1, and when x 1, w -1.

The integral becomes:

∫ (x^2/1-2x) dx ∫ (w^2/2w) (-dw/2) -1/2 ∫ (w 1/2) dw

This is a simpler integral:

-1/2 [w^2/2 - 2 ln|w|] C

Evaluating from w 1 to w -1 and adding the constant of integration:

-1/2 [(-1)^2/2 - 2 ln|-1|] - [-1/2 (1^2/2 - 2 ln|1|)]

Evaluating the constants:

-1/2 [1/2 - 0] - [-1/2 (1/2 - 0)] -1/4 1/4 0

Therefore, the integral simplifies to:

∫ xln(1-2x) dx (x^2/2)ln(1-2x) - 1/2 [w^2/2 - 2 ln|w|] from 0 to 1

Evaluating the integral from 0 to 1:

(1^2/2)ln(1-2(1)) - (0^2/2)ln(1-2(0)) - 1/2 [(1^2/2 - 2 ln|-1|) - (0^2/2 - 2 ln|1|)]

(1/2)ln(-1) - (0) - 1/2 (1/2 - 2 ln(-1)) 2 ln(1)

Evaluating the constants and logarithms:

(1/2)ln(-1) - 1/4 1/2 ln(-1) 2 (0)

0 - 1/4 0 0 -1/4

Conclusion

The integral of xln(1-2x) dx from 0 to 1 is -1/4. This problem showcases the applicability of integration by parts and substitution in solving complex integrals.

Key Points:

Integration by parts is a powerful technique for solving integrals of products. Substitution is used to simplify complex integrands. Constants and limits of integration are crucial in evaluating definite integrals.