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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Concept explainers
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
![**Evaluate the integral:**
\[ \int 8e^{(x-1)} dx \]
To solve this integral, we will use the properties of exponential functions and basic integration techniques.
### Solution:
First, observe that we can simplify the integrand by considering a substitution.
Let \(u = x - 1\). Then, \(du = dx\).
Rewriting the integral in terms of \(u\):
\[ \int 8e^{(x-1)} dx = \int 8e^u du \]
Now, we can integrate \(8e^u\):
\[ \int 8e^u du = 8 \int e^u du \]
The integral of \(e^u\) with respect to \(u\) is simply \(e^u\):
\[ 8 \int e^u du = 8e^u + C \]
Finally, substitute back \(u = x - 1\):
\[ 8e^u + C = 8e^{(x-1)} + C \]
Therefore, the evaluated integral is:
\[ \int 8e^{(x-1)} dx = 8e^{(x-1)} + C \]
Where \(C\) is the constant of integration.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1e2c0fb4-fe7c-46b9-9b1b-ce7222ffa53b%2Ff5d41b13-7c7a-48aa-aa15-be7e77f1d6f3%2Fa5jvh4_processed.png&w=3840&q=75)
Transcribed Image Text:**Evaluate the integral:**
\[ \int 8e^{(x-1)} dx \]
To solve this integral, we will use the properties of exponential functions and basic integration techniques.
### Solution:
First, observe that we can simplify the integrand by considering a substitution.
Let \(u = x - 1\). Then, \(du = dx\).
Rewriting the integral in terms of \(u\):
\[ \int 8e^{(x-1)} dx = \int 8e^u du \]
Now, we can integrate \(8e^u\):
\[ \int 8e^u du = 8 \int e^u du \]
The integral of \(e^u\) with respect to \(u\) is simply \(e^u\):
\[ 8 \int e^u du = 8e^u + C \]
Finally, substitute back \(u = x - 1\):
\[ 8e^u + C = 8e^{(x-1)} + C \]
Therefore, the evaluated integral is:
\[ \int 8e^{(x-1)} dx = 8e^{(x-1)} + C \]
Where \(C\) is the constant of integration.
![**Calculus: Differentiation of Exponential Trigonometric Functions**
**Problem:**
Find the first derivative of \( y = e^x \cos x \).
**Solution:**
To find the first derivative of \( y \) with respect to \( x \), we will use the product rule of differentiation, which is stated as follows:
If \( y = uv \), where both \( u \) and \( v \) are functions of \( x \), then the derivative of \( y \) with respect to \( x \) is given by:
\[ \frac{dy}{dx} = u'v + uv' \]
In this problem:
- Let \( u = e^x \)
- Let \( v = \cos x \)
First, we find the derivatives of \( u \) and \( v \) with respect to \( x \):
- The derivative of \( u = e^x \) with respect to \( x \) is \( u' = e^x \)
- The derivative of \( v = \cos x \) with respect to \( x \) is \( v' = -\sin x \)
Using the product rule, we combine these results:
\[ \frac{dy}{dx} = u'v + uv' \]
\[ \frac{dy}{dx} = (e^x)(\cos x) + (e^x)(-\sin x) \]
\[ \frac{dy}{dx} = e^x \cos x - e^x \sin x \]
Therefore, the first derivative of \( y = e^x \cos x \) is:
\[ \frac{dy}{dx} = e^x (\cos x - \sin x) \]
By positioning this content on an educational website, students can learn about the product rule and how it applies to differentiating exponential trigonometric functions. This step-by-step process helps in understanding these fundamental calculus principles effectively.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1e2c0fb4-fe7c-46b9-9b1b-ce7222ffa53b%2Ff5d41b13-7c7a-48aa-aa15-be7e77f1d6f3%2F9gf7clc_processed.png&w=3840&q=75)
Transcribed Image Text:**Calculus: Differentiation of Exponential Trigonometric Functions**
**Problem:**
Find the first derivative of \( y = e^x \cos x \).
**Solution:**
To find the first derivative of \( y \) with respect to \( x \), we will use the product rule of differentiation, which is stated as follows:
If \( y = uv \), where both \( u \) and \( v \) are functions of \( x \), then the derivative of \( y \) with respect to \( x \) is given by:
\[ \frac{dy}{dx} = u'v + uv' \]
In this problem:
- Let \( u = e^x \)
- Let \( v = \cos x \)
First, we find the derivatives of \( u \) and \( v \) with respect to \( x \):
- The derivative of \( u = e^x \) with respect to \( x \) is \( u' = e^x \)
- The derivative of \( v = \cos x \) with respect to \( x \) is \( v' = -\sin x \)
Using the product rule, we combine these results:
\[ \frac{dy}{dx} = u'v + uv' \]
\[ \frac{dy}{dx} = (e^x)(\cos x) + (e^x)(-\sin x) \]
\[ \frac{dy}{dx} = e^x \cos x - e^x \sin x \]
Therefore, the first derivative of \( y = e^x \cos x \) is:
\[ \frac{dy}{dx} = e^x (\cos x - \sin x) \]
By positioning this content on an educational website, students can learn about the product rule and how it applies to differentiating exponential trigonometric functions. This step-by-step process helps in understanding these fundamental calculus principles effectively.
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