Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions

Problema 7.1.75

Capitolo 7, Problema 7.1.75

Bounds on e Use a left Riemann sum with at least n = 2 subintervals of equal length to approximate ln 2 = ∫[1 to 2] (dt/t) and show that ln 2 < 1. Use a right Riemann sum with n = 7 subintervals of equal length to approximate ln 3 = ∫[1 to 3] (dt/t) and show that ln 3 > 1.

Guida verificata passo dopo passo

First, recall that the natural logarithm function can be expressed as an integral: \(\ln b = \int_1^b \frac{1}{t} \, dt\) for \(b > 0\).

To approximate \(\ln 2 = \int_1^2 \frac{1}{t} \, dt\) using a left Riemann sum with \(n=2\) subintervals, divide the interval \([1,2]\) into 2 equal parts. Each subinterval has length \(\Delta t = \frac{2-1}{2} = 0.5\).

Identify the left endpoints of the subintervals: \(t_0 = 1\) and \(t_1 = 1.5\). Then, the left Riemann sum approximation is \(L_2 = \Delta t \left( \frac{1}{t_0} + \frac{1}{t_1} \right)\).

Since \(\frac{1}{t}\) is a decreasing function on \([1,2]\), the left Riemann sum overestimates the integral. Therefore, \(L_2 > \int_1^2 \frac{1}{t} \, dt = \ln 2\). Use this to show that \(\ln 2 < 1\) by evaluating the sum (without calculating the exact value here).

Next, to approximate \(\ln 3 = \int_1^3 \frac{1}{t} \, dt\) using a right Riemann sum with \(n=7\) subintervals, divide \([1,3]\) into 7 equal parts with \(\Delta t = \frac{3-1}{7} = \frac{2}{7}\). The right endpoints are \(t_i = 1 + i \Delta t\) for \(i=1\) to \(7\). The right Riemann sum is \(R_7 = \Delta t \sum_{i=1}^7 \frac{1}{t_i}\).

Since \(\frac{1}{t}\) is decreasing, the right Riemann sum underestimates the integral, so \(R_7 < \ln 3\). However, the problem asks to show \(\ln 3 > 1\), so you can compare \(R_7\) to 1 and use the inequality to conclude \(\ln 3 > 1\).

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Riemann Sums

Riemann sums approximate the value of a definite integral by dividing the integration interval into subintervals and summing the areas of rectangles. The height of each rectangle is determined by the function value at specific points (left, right, or midpoint) within each subinterval. Increasing the number of subintervals generally improves the approximation.

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Introduction to Riemann Sums

Properties of the Natural Logarithm as an Integral

The natural logarithm function ln(x) can be expressed as an integral: ln(x) = ∫₁ˣ (1/t) dt. This integral representation allows us to approximate ln(x) by evaluating the integral numerically, such as with Riemann sums, linking logarithmic values to areas under the curve y = 1/t.

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Change of Base Property

Inequalities from Left and Right Riemann Sums

For a decreasing function, left Riemann sums overestimate the integral, while right Riemann sums underestimate it; for an increasing function, the opposite holds. Since 1/t is decreasing on [1, ∞), left sums provide upper bounds and right sums provide lower bounds, enabling inequalities like ln 2 < 1 and ln 3 > 1.

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Percorso guidato

07:39

Left, Right, & Midpoint Riemann Sums

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