Evaluate the integrals in Exercises 111–114.
111. ∫₁^(ln x) (1 / t) dt,x > 1

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Identify the integral to be evaluated: \(\int_{1}^{\ln x} \frac{1}{t} \, dt\), where \(x > 1\).

Recall the Fundamental Theorem of Calculus, which states that if \(F(t)\) is an antiderivative of \(f(t)\), then \(\int_a^b f(t) \, dt = F(b) - F(a)\).

Find the antiderivative of the integrand \(\frac{1}{t}\). Since \(\frac{d}{dt}(\ln |t|) = \frac{1}{t}\), the antiderivative is \(\ln |t|\).

Apply the limits of integration by substituting the upper limit \(t = \ln x\) and the lower limit \(t = 1\) into the antiderivative: \(\ln |\ln x| - \ln |1|\).

Simplify the expression noting that \(\ln |1| = 0\), so the integral evaluates to \(\ln |\ln x|\).

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주요 개념

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Definite Integral with Variable Limits

A definite integral with variable limits involves integrating a function where the upper or lower limit is a function of a variable, not a constant. Understanding how to evaluate such integrals requires applying the Fundamental Theorem of Calculus carefully, treating the variable limit as a function.

Fundamental Theorem of Calculus

This theorem connects differentiation and integration, stating that if F is an antiderivative of f, then the definite integral of f from a to b equals F(b) - F(a). It also allows differentiation of integrals with variable limits, which is essential for evaluating integrals like ∫₁^(ln x) (1/t) dt.

Properties of the Natural Logarithm Function

The natural logarithm function, ln(x), is the inverse of the exponential function and is defined for x > 0. Recognizing that the integral of 1/t dt is ln|t| helps in evaluating the integral, especially when the upper limit is ln(x), linking the integral to logarithmic expressions.

Evaluate the integrals in Exercises 111–114.111. ∫₁^(ln x) (1 / t) dt,x > 1