mproper Integrals - Integrating over an infinite interval. n this problem our goal is to determine whether the improper integral below converges or diverges. If it converges, we will determine the value of the improper integral. sec zdz e will use the following definition: fis continuous at every point of la, b) we define f(z) dz = lim rovided that the limit exists. If the limit exists and is finite, we say that the integral converges and that the value of the improper integral is this limit. If the limit fails exist or iş infinite then, we say that the improper integral diverg art 1. e-write the improper integral as the limit. Assume -4x
mproper Integrals - Integrating over an infinite interval. n this problem our goal is to determine whether the improper integral below converges or diverges. If it converges, we will determine the value of the improper integral. sec zdz e will use the following definition: fis continuous at every point of la, b) we define f(z) dz = lim rovided that the limit exists. If the limit exists and is finite, we say that the integral converges and that the value of the improper integral is this limit. If the limit fails exist or iş infinite then, we say that the improper integral diverg art 1. e-write the improper integral as the limit. Assume -4x
Chapter3: Functions
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Transcribed Image Text:Improper Integrals - Integrating over an infinite interval.
In this problem our goal is to determine whether the improper integral below converges or diverges. If it converges, we will determine the value of the improper integral.
sec z dr
We will use the following definition:
If f is continuous at every point of [a, b) we define
f(z) de = lim
provided that the limit exists. If the limit exists and is finite, we say that the integral converges and that the value of the improper integral is this limit. If the limit fails to exist or is infinite then, we say that the improper integral diverges.
Part 1.
Re-write the improper integral as the limit. Assume -47 <t<=
sec z dz
lim
Upon evaluating the definite integral you found above, we have
lim
Part 2.
Based on your answer for Part 1. above, the improper integral either converges to a finite value or diverges. If the integral converges, state that it converges and to what value it converges. Otherwise, state that it diverges to infinity.
The integral
to
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