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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Riemann Sum
Riemann Sums is a special type of approximation of the area under a curve by dividing it into multiple simple shapes like rectangles or trapezoids and is used in integrals when finite sums are involved. Figuring out the area of a curve is complex hence this method makes it simple. Usually, we take the help of different integration methods for this purpose. This is one of the major parts of integral calculus.
Riemann Integral
Bernhard Riemann's integral was the first systematic description of the integral of a function on an interval in the branch of mathematics known as real analysis.
Question
![**Problem: Calculus - Integration**
**Problem Statement:**
1. Evaluate the integral:
\[
\int \sin(x) e^{3x} \, dx
\]
---
**Solution Approach:**
To solve this integral, we can use the method of integration by parts. Integration by parts is given by the formula:
\[
\int u \, dv = uv - \int v \, du
\]
Let's choose \( u \) and \( dv \) such that the integration becomes simpler. In this context, we can set:
\[
u = \sin(x) \quad \text{and} \quad dv = e^{3x} \, dx
\]
Now, we need to find \( du \) and \( v \):
\[
du = \cos(x) \, dx \quad \text{and} \quad v = \int e^{3x} \, dx = \frac{1}{3} e^{3x}
\]
Plug \( u, du, v, \) and \( dv \) into the integration by parts formula:
\[
\int \sin(x) e^{3x} \, dx = \sin(x) \cdot \frac{1}{3} e^{3x} - \int \frac{1}{3} e^{3x} \cos(x) \, dx
\]
Now, we need to solve the remaining integral:
\[
\int \frac{1}{3} e^{3x} \cos(x) \, dx
\]
We repeat the integration by parts method, setting:
\[
u = \cos(x) \quad \text{and} \quad dv = \frac{1}{3} e^{3x} \, dx
\]
Then we get:
\[
du = -\sin(x) \, dx \quad \text{and} \quad v = \frac{1}{3} \cdot \frac{1}{3} e^{3x} = \frac{1}{9} e^{3x}
\]
Substitute into the integration by parts formula again:
\[
\int \frac{1}{3} e^{3x} \cos(x) \, dx = \cos(x) \cdot \frac{1}{9} e^{3x} - \int \frac{1}{9} e](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe02ff0de-5c0a-4a84-adb8-ab6e760f1e50%2F20875115-406d-419f-b5c4-5a94cc77fccc%2F8oh64ss6.png&w=3840&q=75)
Transcribed Image Text:**Problem: Calculus - Integration**
**Problem Statement:**
1. Evaluate the integral:
\[
\int \sin(x) e^{3x} \, dx
\]
---
**Solution Approach:**
To solve this integral, we can use the method of integration by parts. Integration by parts is given by the formula:
\[
\int u \, dv = uv - \int v \, du
\]
Let's choose \( u \) and \( dv \) such that the integration becomes simpler. In this context, we can set:
\[
u = \sin(x) \quad \text{and} \quad dv = e^{3x} \, dx
\]
Now, we need to find \( du \) and \( v \):
\[
du = \cos(x) \, dx \quad \text{and} \quad v = \int e^{3x} \, dx = \frac{1}{3} e^{3x}
\]
Plug \( u, du, v, \) and \( dv \) into the integration by parts formula:
\[
\int \sin(x) e^{3x} \, dx = \sin(x) \cdot \frac{1}{3} e^{3x} - \int \frac{1}{3} e^{3x} \cos(x) \, dx
\]
Now, we need to solve the remaining integral:
\[
\int \frac{1}{3} e^{3x} \cos(x) \, dx
\]
We repeat the integration by parts method, setting:
\[
u = \cos(x) \quad \text{and} \quad dv = \frac{1}{3} e^{3x} \, dx
\]
Then we get:
\[
du = -\sin(x) \, dx \quad \text{and} \quad v = \frac{1}{3} \cdot \frac{1}{3} e^{3x} = \frac{1}{9} e^{3x}
\]
Substitute into the integration by parts formula again:
\[
\int \frac{1}{3} e^{3x} \cos(x) \, dx = \cos(x) \cdot \frac{1}{9} e^{3x} - \int \frac{1}{9} e
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