Find the value of the given integral. o (-2e-3tu(t) + 98(t))dt The value of the given integral is
Find the value of the given integral. o (-2e-3tu(t) + 98(t))dt The value of the given integral is
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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![**Find the value of the given integral.**
\[
\int_{0^-}^{0^+} \left(-2e^{-3t}u(t) + 9\delta(t)\right) dt
\]
**The value of the given integral is** \_\_\_\_\_\_ \_\_\_.
**Explanation:**
This expression involves an integral with limits from \(0^-\) to \(0^+\). The integral includes two components:
1. \(-2e^{-3t}u(t)\): This part of the integrand includes the unit step function \(u(t)\), which is 0 for \(t < 0\) and 1 for \(t \ge 0\). However, it will not contribute to the integral within the limits \(0^-\) to \(0^+\) because the interval effectively closes at the point \(t = 0\) and \(u(t)\) is 0 just below zero.
2. \(9\delta(t)\): This part involves the Dirac delta function \(\delta(t)\), which is zero everywhere except at \(t = 0\), where it is infinite such that its integral over the entire real line is 1. The contribution of this part at \(t = 0\) is exactly given by the coefficient of the \(\delta(t)\), which is 9.
Therefore, solving the integral over the interval \(0^-\) to \(0^+\), the only contributing term is from the \(\delta(t)\), resulting in a value of 9.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1b94d5d1-8249-4b1b-8d70-6007d482e8f0%2Fcfb191bb-9eb1-4c45-953a-09471a91a8e9%2Fqwt2t9c_processed.png&w=3840&q=75)
Transcribed Image Text:**Find the value of the given integral.**
\[
\int_{0^-}^{0^+} \left(-2e^{-3t}u(t) + 9\delta(t)\right) dt
\]
**The value of the given integral is** \_\_\_\_\_\_ \_\_\_.
**Explanation:**
This expression involves an integral with limits from \(0^-\) to \(0^+\). The integral includes two components:
1. \(-2e^{-3t}u(t)\): This part of the integrand includes the unit step function \(u(t)\), which is 0 for \(t < 0\) and 1 for \(t \ge 0\). However, it will not contribute to the integral within the limits \(0^-\) to \(0^+\) because the interval effectively closes at the point \(t = 0\) and \(u(t)\) is 0 just below zero.
2. \(9\delta(t)\): This part involves the Dirac delta function \(\delta(t)\), which is zero everywhere except at \(t = 0\), where it is infinite such that its integral over the entire real line is 1. The contribution of this part at \(t = 0\) is exactly given by the coefficient of the \(\delta(t)\), which is 9.
Therefore, solving the integral over the interval \(0^-\) to \(0^+\), the only contributing term is from the \(\delta(t)\), resulting in a value of 9.
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