Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
![**Evaluate the Improper Integral**
\[ \int_{-\infty}^{e} 11e^{-x} \, dx \]
**Explanation:**
The task is to find the value of the improper integral given above. An improper integral is one where either the interval of integration is infinite, or the function has an infinite discontinuity.
The integral \(\int_{-\infty}^{e} 11e^{-x} \, dx\) involves an exponential decay function \(11e^{-x}\) integrated over an infinite range from \(-\infty\) to \(e\).
To evaluate this, you will typically use limits to handle the improper nature of the integral at \(-\infty\). It requires evaluating the limit:
\[ \lim_{a \to -\infty} \int_{a}^{e} 11e^{-x} \, dx. \]
This process involves finding the antiderivative of \(11e^{-x}\), which is \(-11e^{-x}\), and then applying the limits of integration to find the final value.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb7c72b9f-0930-4803-9715-bedcb9f3be97%2Fadb79640-6682-4c7e-93b1-5f2a81206328%2F0ivghrf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Evaluate the Improper Integral**
\[ \int_{-\infty}^{e} 11e^{-x} \, dx \]
**Explanation:**
The task is to find the value of the improper integral given above. An improper integral is one where either the interval of integration is infinite, or the function has an infinite discontinuity.
The integral \(\int_{-\infty}^{e} 11e^{-x} \, dx\) involves an exponential decay function \(11e^{-x}\) integrated over an infinite range from \(-\infty\) to \(e\).
To evaluate this, you will typically use limits to handle the improper nature of the integral at \(-\infty\). It requires evaluating the limit:
\[ \lim_{a \to -\infty} \int_{a}^{e} 11e^{-x} \, dx. \]
This process involves finding the antiderivative of \(11e^{-x}\), which is \(-11e^{-x}\), and then applying the limits of integration to find the final value.
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