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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 6, Problema 6.2.39

Find the area of the region described in the following exercises.


The region bounded by y=e^x, y=e^−2x, and x=ln 4

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1
Identify the curves and boundaries that enclose the region: the curves are \( y = e^{x} \), \( y = e^{-2x} \), and the vertical line \( x = \ln 4 \).
Find the points of intersection between the two curves \( y = e^{x} \) and \( y = e^{-2x} \) by setting them equal: \( e^{x} = e^{-2x} \). Solve for \( x \) to determine the left boundary of the region.
Determine which curve is on top and which is on the bottom between the intersection point and \( x = \ln 4 \) by comparing \( e^{x} \) and \( e^{-2x} \) in that interval.
Set up the integral for the area of the region as the integral from the left intersection point to \( x = \ln 4 \) of the difference between the top curve and the bottom curve: \[ \text{Area} = \int_{x=a}^{\ln 4} \left( e^{x} - e^{-2x} \right) \, dx \], where \( a \) is the intersection point found earlier.
Evaluate the integral by integrating each term separately: \( \int e^{x} \, dx \) and \( \int e^{-2x} \, dx \), then apply the limits of integration to find the area.

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