Ely)-[i -2x -e-3x -3 15e-Qx -Le-3x 9 ㅇ @ CyJ= 15e-ax-e-3x dx 9 -3 € cy)= [15e -e-3x -3 15e-2x + -1 dx ECy)= -5 -2x -ax C-e-3x) %3D FIVE STAR. ***** FIVE STAR. *****
Ely)-[i -2x -e-3x -3 15e-Qx -Le-3x 9 ㅇ @ CyJ= 15e-ax-e-3x dx 9 -3 € cy)= [15e -e-3x -3 15e-2x + -1 dx ECy)= -5 -2x -ax C-e-3x) %3D FIVE STAR. ***** FIVE STAR. *****
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I need some help completing this intergal please
![The following is a transcription of a mathematical derivation involving integrals. It appears to calculate the expected value, \( E(y) \), for a given function using integration techniques.
1. **Expression 1:**
\[
E(y) = \int_{0}^{\infty} 15 e^{-2x} \left( \frac{0 - e^{-3x}}{-3} \right) + \left[ \frac{-1}{9} e^{-3x} \right] \, dx
\]
2. **Expression 2:**
\[
= \int_{0}^{\infty} 15 e^{-2x} \left( \frac{-e^{-3x}}{-3} \right) + \left( 0 - \frac{1}{9} \right) \, dx
\]
3. **Expression 3:**
\[
= \int_{0}^{\infty} 15 e^{-2x} \left( \frac{-e^{-3x}}{-3} \right) + \frac{-1}{9} \, dx
\]
4. **Expression 4:**
\[
E(y) = -5 \int_{0}^{\infty} e^{-2x} (e^{-3x}) - \frac{1}{9} \, dx
\]
This series of expressions shows the step-by-step transformation and simplification of an integral used to find the expected value, \( E(y) \). The process involves substitution and algebraic manipulation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F308db737-7862-446c-973d-78bf4e1d9584%2F7033815b-fb4c-48ce-a59d-275abcd742c5%2Ft0frfd.jpeg&w=3840&q=75)
Transcribed Image Text:The following is a transcription of a mathematical derivation involving integrals. It appears to calculate the expected value, \( E(y) \), for a given function using integration techniques.
1. **Expression 1:**
\[
E(y) = \int_{0}^{\infty} 15 e^{-2x} \left( \frac{0 - e^{-3x}}{-3} \right) + \left[ \frac{-1}{9} e^{-3x} \right] \, dx
\]
2. **Expression 2:**
\[
= \int_{0}^{\infty} 15 e^{-2x} \left( \frac{-e^{-3x}}{-3} \right) + \left( 0 - \frac{1}{9} \right) \, dx
\]
3. **Expression 3:**
\[
= \int_{0}^{\infty} 15 e^{-2x} \left( \frac{-e^{-3x}}{-3} \right) + \frac{-1}{9} \, dx
\]
4. **Expression 4:**
\[
E(y) = -5 \int_{0}^{\infty} e^{-2x} (e^{-3x}) - \frac{1}{9} \, dx
\]
This series of expressions shows the step-by-step transformation and simplification of an integral used to find the expected value, \( E(y) \). The process involves substitution and algebraic manipulation.
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