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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 8, Problem 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.

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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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Definite Integrals and Area Under a Curve

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