The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.
∫₀¹ (−ln(x)) dx

Verified step by step guidance

1. Recognize that the integral is \( \int_0^1 (-\ln(x)) \, dx \). Since the integrand is \( -\ln(x) \), rewrite the integral as \( - \int_0^1 \ln(x) \, dx \) to simplify the process.
2. Recall that the integral of \( \ln(x) \) can be found using integration by parts. Set \( u = \ln(x) \) and \( dv = dx \). Then, compute \( du = \frac{1}{x} dx \) and \( v = x \).
3. Apply the integration by parts formula: \( \int u \, dv = uv - \int v \, du \). Substitute the values to get \( \int \ln(x) \, dx = x \ln(x) - \int x \cdot \frac{1}{x} \, dx = x \ln(x) - \int 1 \, dx \).
4. Simplify the integral to \( x \ln(x) - x + C \). Now, evaluate the definite integral \( \int_0^1 \ln(x) \, dx = [x \ln(x) - x]_0^1 \).
5. Evaluate the limits carefully: at \( x=1 \), \( 1 \cdot \ln(1) - 1 = 0 - 1 = -1 \); at \( x=0 \), use the limit \( \lim_{x \to 0^+} x \ln(x) = 0 \), so the expression at 0 is \( 0 - 0 = 0 \). Therefore, \( \int_0^1 \ln(x) \, dx = -1 \). Finally, multiply by \( -1 \) to get the value of the original integral.

Key Concepts

Improper Integrals and Convergence

An integral is improper if the interval is infinite or the integrand is unbounded. Here, the integrand involves ln(x), which is unbounded near 0, so we consider the limit as x approaches 0 to check convergence before evaluating.

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, often used when integrating functions like ln(x) or products involving logarithms.

Properties of the Natural Logarithm Function

The natural logarithm ln(x) is defined for x > 0 and approaches negative infinity as x approaches 0 from the right. Understanding its behavior near zero is crucial for evaluating integrals involving ln(x), especially when combined with limits.