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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.8.26

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

Guida verificata passo dopo passo
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.
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 \( v = x \).
Apply the integration by parts formula: \( \int u \, dv = uv - \int v \, du \). Substitute the values to get \( \int \ln(x) \, dx = x \ln(x) - \int x \cdot \frac{1}{x} \, dx = x \ln(x) - \int 1 \, dx \).
Simplify the integral to \( x \ln(x) - x + C \). Now, evaluate the definite integral \( \int_0^1 \ln(x) \, dx = [x \ln(x) - x]_0^1 \).
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.

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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.
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