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S =/2e-3x/4 3x -3x/4 (A) 4 2x - 3x/4 (B) 3 (C) - 1/2 e X (D) 1/2 e e e ₂-3x/4 - 3x/4 dx + + 3 - & e 3 z fe е 3x/4 -3x/4 113 2 -3x/4 -3x/4 + C + C +C + C

S =/2e-3x/4 3x -3x/4 (A) 4 2x - 3x/4 (B) 3 (C) - 1/2 e X (D) 1/2 e e e ₂-3x/4 - 3x/4 dx + + 3 - & e 3 z fe е 3x/4 -3x/4 113 2 -3x/4 -3x/4 + C + C +C + C

Calculus: Early Transcendentals
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
### Evaluating the Integral Given the integral: \[ \int \frac{x}{2} e^{-3x/4} \, dx \] ### Multiple Choice Options: - **(A)** \[-\frac{3x}{4} e^{-3x/4} + \frac{3}{4} e^{-3x/4} + C\] - **(B)** \[-\frac{2x}{3} e^{-3x/4} - \frac{8}{9} e^{-3x/4} + C\] - **(C)** \[-\frac{x}{2} e^{-3x/4} + \frac{3}{8} e^{-3x/4} + C\] - **(D)** \[\frac{x}{2} e^{-3x/4} - \frac{1}{2} e^{-3x/4} + C\] ### Explanation: To solve the given integral, one can use integration by parts, where we choose: \[ u = \frac{x}{2} \quad \text{and} \quad dv = e^{-3x/4} \, dx \] Then calculate \( du \) and \( v \): \[ du = \frac{1}{2} \, dx \] \[ v = \int e^{-3x/4} \, dx = -\frac{4}{3} e^{-3x/4} \] Applying the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \] Results in: \[ \int \frac{x}{2} e^{-3x/4} \, dx = \left( \frac{x}{2} \right) \left( -\frac{4}{3} e^{-3x/4} \right) - \int \left( -\frac{4}{3} e^{-3x/4} \right) \left( \frac{1}{2} \, dx \right) \] Simplify: \[ -\frac{2x}{3} e^{-3x/4} + \frac{2}{3} \int e^{-3x/4} \, dx \] \[ \int e^{-3x/4} \, dx = -\frac{4}{3} e^{-3x/4}
Transcribed Image Text:### Evaluating the Integral Given the integral: \[ \int \frac{x}{2} e^{-3x/4} \, dx \] ### Multiple Choice Options: - **(A)** \[-\frac{3x}{4} e^{-3x/4} + \frac{3}{4} e^{-3x/4} + C\] - **(B)** \[-\frac{2x}{3} e^{-3x/4} - \frac{8}{9} e^{-3x/4} + C\] - **(C)** \[-\frac{x}{2} e^{-3x/4} + \frac{3}{8} e^{-3x/4} + C\] - **(D)** \[\frac{x}{2} e^{-3x/4} - \frac{1}{2} e^{-3x/4} + C\] ### Explanation: To solve the given integral, one can use integration by parts, where we choose: \[ u = \frac{x}{2} \quad \text{and} \quad dv = e^{-3x/4} \, dx \] Then calculate \( du \) and \( v \): \[ du = \frac{1}{2} \, dx \] \[ v = \int e^{-3x/4} \, dx = -\frac{4}{3} e^{-3x/4} \] Applying the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \] Results in: \[ \int \frac{x}{2} e^{-3x/4} \, dx = \left( \frac{x}{2} \right) \left( -\frac{4}{3} e^{-3x/4} \right) - \int \left( -\frac{4}{3} e^{-3x/4} \right) \left( \frac{1}{2} \, dx \right) \] Simplify: \[ -\frac{2x}{3} e^{-3x/4} + \frac{2}{3} \int e^{-3x/4} \, dx \] \[ \int e^{-3x/4} \, dx = -\frac{4}{3} e^{-3x/4}
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