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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![### Integration by Parts Example
To evaluate the integral using integration by parts:
\[
\int 2x e^{7x} \, dx
\]
First, identify the components to apply integration by parts, where typically:
\[
\int u \, dv = uv - \int v \, du
\]
Choose \( u \) and \( dv \) as follows:
- \( u = 2x \)
- \( dv = e^{7x} \, dx \)
Now, differentiate \( u \) and integrate \( dv \):
- \( du = 2 \, dx \)
- \( v = \frac{1}{7} e^{7x} \)
Plug these into the integration by parts formula:
\[
\int 2x e^{7x} \, dx = 2x \left( \frac{1}{7} e^{7x} \right) - \int \left( \frac{1}{7} e^{7x} \right) 2 \, dx
\]
Simplify this expression:
\[
\int 2x e^{7x} \, dx = \frac{2x e^{7x}}{7} - \frac{2}{7} \int e^{7x} \, dx
\]
Calculate the remaining integral:
\[
\int e^{7x} \, dx = \frac{e^{7x}}{7}
\]
Substitute this back into the equation:
\[
\frac{2x e^{7x}}{7} - \frac{2}{7} \cdot \frac{e^{7x}}{7}
\]
Simplify the final expression:
\[
\int 2x e^{7x} \, dx = \frac{2x e^{7x}}{7} - \frac{2 e^{7x}}{49} + C
\]
Therefore:
\[
\int 2x e^{7x} \, dx = \frac{2}{7} x e^{7x} - \frac{2}{49} e^{7x} + C
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff8558f12-eb43-4f23-8df4-76c708dcf941%2F0b3f14d9-aa5e-48c6-bb9e-aa098a934eec%2Fgusklkg_processed.png&w=3840&q=75)
Transcribed Image Text:### Integration by Parts Example
To evaluate the integral using integration by parts:
\[
\int 2x e^{7x} \, dx
\]
First, identify the components to apply integration by parts, where typically:
\[
\int u \, dv = uv - \int v \, du
\]
Choose \( u \) and \( dv \) as follows:
- \( u = 2x \)
- \( dv = e^{7x} \, dx \)
Now, differentiate \( u \) and integrate \( dv \):
- \( du = 2 \, dx \)
- \( v = \frac{1}{7} e^{7x} \)
Plug these into the integration by parts formula:
\[
\int 2x e^{7x} \, dx = 2x \left( \frac{1}{7} e^{7x} \right) - \int \left( \frac{1}{7} e^{7x} \right) 2 \, dx
\]
Simplify this expression:
\[
\int 2x e^{7x} \, dx = \frac{2x e^{7x}}{7} - \frac{2}{7} \int e^{7x} \, dx
\]
Calculate the remaining integral:
\[
\int e^{7x} \, dx = \frac{e^{7x}}{7}
\]
Substitute this back into the equation:
\[
\frac{2x e^{7x}}{7} - \frac{2}{7} \cdot \frac{e^{7x}}{7}
\]
Simplify the final expression:
\[
\int 2x e^{7x} \, dx = \frac{2x e^{7x}}{7} - \frac{2 e^{7x}}{49} + C
\]
Therefore:
\[
\int 2x e^{7x} \, dx = \frac{2}{7} x e^{7x} - \frac{2}{49} e^{7x} + C
\]
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