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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
Tutti i libri di testoHass 15th EditionCh. 8 - Techniques of IntegrationProblema 8.2.63a
Capitolo 8, Problema 8.2.63a

Consider the region bounded by the graphs of
y = ln(x), y = 0, and x = e.
a. Find the area of the region.

Guida verificata passo dopo passo
1
Identify the region bounded by the curves: the graph of \(y = \ln(x)\), the line \(y = 0\), and the vertical line \(x = e\). Note that \(y = 0\) is the x-axis.
Determine the interval for \(x\) over which the region lies. Since \(y = \ln(x)\) intersects \(y = 0\) when \(\ln(x) = 0\), solve for \(x\): \(\ln(x) = 0 \implies x = 1\). So the region is bounded between \(x = 1\) and \(x = e\).
Set up the integral for the area between the curves. The area \(A\) is given by the integral of the top function minus the bottom function over the interval \([1, e]\). Here, the top function is \(y = \ln(x)\) and the bottom function is \(y = 0\), so:
\[ A = \int_{1}^{e} (\ln(x) - 0) \, dx = \int_{1}^{e} \ln(x) \, dx \]
To evaluate the integral \(\int \ln(x) \, dx\), use integration by parts. Let \(u = \ln(x)\) and $dv = dx$. Then, \(du = \frac{1}{x} dx\) and \(v = x\). Apply the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Substitute back to find the definite integral from 1 to \(e\) and express the area in terms of these evaluated expressions.

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Definite Integrals and Area Under a Curve

The definite integral of a function over an interval represents the net area between the curve and the x-axis. When the function is positive, the integral gives the exact area under the curve. Calculating the area bounded by curves often involves setting up and evaluating definite integrals.
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Definition of the Definite Integral

Natural Logarithm Function Properties

The natural logarithm function, y = ln(x), is defined for x > 0 and is the inverse of the exponential function e^x. It is continuous and increasing, with ln(1) = 0. Understanding its behavior helps in setting integration limits and interpreting the region bounded by y = ln(x), y = 0, and vertical lines.
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Properties of Functions

Setting Integration Limits from Boundary Curves

To find the area of a region bounded by curves, it is essential to identify the correct limits of integration. These limits correspond to the intersection points of the bounding curves. In this problem, the vertical boundary x = e and the x-values where y = ln(x) meets y = 0 determine the integration interval.
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Improper Integrals: Infinite Intervals
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