The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.
∫₀¹ dr / r^0.999

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Identify the integral to be evaluated: \(\int_0^1 \frac{dr}{r^{0.999}}\).

Rewrite the integrand using exponent rules: \(\frac{1}{r^{0.999}} = r^{-0.999}\), so the integral becomes \(\int_0^1 r^{-0.999} \, dr\).

Recall the power rule for integration: For \(\int r^n \, dr\), where \(n \neq -1\), the antiderivative is \(\frac{r^{n+1}}{n+1} + C\).

Apply the power rule to the integral: \(\int_0^1 r^{-0.999} \, dr = \left[ \frac{r^{-0.999 + 1}}{-0.999 + 1} \right]_0^1 = \left[ \frac{r^{0.001}}{0.001} \right]_0^1\).

Evaluate the definite integral by substituting the limits: calculate \(\frac{1^{0.001}}{0.001} - \frac{0^{0.001}}{0.001}\), noting the behavior of \(r^{0.001}\) as \(r\) approaches 0.

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주요 개념

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Improper Integrals

Improper integrals involve integrands that are unbounded or have infinite limits. When the integrand approaches infinity at a boundary, the integral is evaluated as a limit to determine convergence or divergence.

Power Rule for Integration

The power rule states that ∫x^n dx = (x^(n+1))/(n+1) + C for n ≠ -1. This rule is essential for integrating functions with variable exponents, allowing direct evaluation of integrals involving powers of the variable.

Convergence of Integrals with Singularities

When the integrand has a singularity (like r^(-0.999) near zero), determining if the integral converges requires analyzing the behavior near the singularity. If the integral's limit exists and is finite, the integral converges.

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.∫₀¹ dr / r^0.999