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Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 7, Problem 7.1.40

29–62. Integrals Evaluate the following integrals. Include absolute values only when needed.
∫₋₂² (e^{z/2}) / (e^{z/2} + 1) dz

Verified step by step guidance
1
First, observe the integral \( \int_{-2}^{2} \frac{e^{z/2}}{e^{z/2} + 1} \, dz \). Notice the integrand involves an exponential function in the numerator and denominator.
To simplify the integral, consider the substitution \( u = e^{z/2} \). Then, \( du = \frac{1}{2} e^{z/2} dz = \frac{1}{2} u \, dz \), which implies \( dz = \frac{2}{u} du \). However, since the limits are in terms of \( z \), it might be easier to explore symmetry or another approach first.
Check if the integrand has any symmetry properties by evaluating \( f(-z) \) and comparing it to \( f(z) \). This can help determine if the integral can be simplified by splitting or combining parts.
Rewrite the integrand as \( \frac{e^{z/2}}{e^{z/2} + 1} = 1 - \frac{1}{e^{z/2} + 1} \). This decomposition can make the integral easier to handle by splitting it into two integrals.
Express the original integral as \( \int_{-2}^{2} 1 \, dz - \int_{-2}^{2} \frac{1}{e^{z/2} + 1} \, dz \). Then, evaluate each integral separately, using substitution or symmetry as appropriate.

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Key Concepts

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

Definite Integrals and Symmetry

Definite integrals calculate the net area under a curve between two limits. When the integrand exhibits symmetry (even, odd, or other), this property can simplify evaluation by reducing the interval or transforming the integral.
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Substitution Method

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Exponential Functions and Their Properties

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