Problem 32
Question
Find antiderinatives for the functions $$ \left(1 / e^{x}\right)+\left(1 / x^{6}\right) $$
Step-by-Step Solution
Verified Answer
The antiderivative is \(-e^{-x} - \frac{1}{5} x^{-5} + C\).
1Step 1: Identify the Function Components
The given function is \( f(x) = \frac{1}{e^x} + \frac{1}{x^6} \). This is a sum of two separate functions, \( \frac{1}{e^x} \) and \( \frac{1}{x^6} \). We need to find the antiderivatives for each of these components separately.
2Step 2: Find the Antiderivative of \( \frac{1}{e^x} \)
The term \( \frac{1}{e^x} \) can be rewritten as \( e^{-x} \). The antiderivative of \( e^{-x} \) is \( -e^{-x} + C_1 \), where \( C_1 \) is a constant of integration.
3Step 3: Find the Antiderivative of \( \frac{1}{x^6} \)
The term \( \frac{1}{x^6} \) can be rewritten as \( x^{-6} \). Using the power rule for integration, the antiderivative is \( -\frac{1}{5}x^{-5} + C_2 \), where \( C_2 \) is another constant of integration.
4Step 4: Combine the Antiderivatives
Now, combine the antiderivatives found in the previous steps. The antiderivative of the entire function is:\[ F(x) = (-e^{-x} + C_1) + \left(-\frac{1}{5}x^{-5} + C_2\right) \]Simplifying, we get:\[ F(x) = -e^{-x} - \frac{1}{5} x^{-5} + C \]where \( C = C_1 + C_2 \) is the combined constant of integration.
Key Concepts
IntegrationPower RuleFunction ComponentsConstant of Integration
Integration
Integration is a fundamental concept in calculus that allows us to find the antiderivatives of functions, essentially reversing the process of differentiation. When we integrate a function, we are finding a new function whose derivative is the original function. Think of it like finding the original path from knowing the speed per time unit that a car was traveling.
It's important to remember that integration can be applied to both simple functions and more complex expressions, like sums of multiple terms. By breaking down a complex function into simpler components, we can integrate each part individually and then combine the results.
It's important to remember that integration can be applied to both simple functions and more complex expressions, like sums of multiple terms. By breaking down a complex function into simpler components, we can integrate each part individually and then combine the results.
Power Rule
The power rule is a simple but powerful tool for finding the antiderivatives of polynomial functions. It states that the antiderivative of a function of the form \( x^n \) is \( \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \) and \( C \) represents any constant of integration.
In the process of integration, the power rule is applied to each term in a polynomial separately. If our function has terms with fractions or negative exponents, like the term \( x^{-6} \) in our exercise, the power rule still applies. By integrating \( x^{-6} \), for instance, the result is \( -\frac{1}{5}x^{-5} + C \).
The power rule makes the integration of polynomial components fast and reliable, breaking down to each power term.
In the process of integration, the power rule is applied to each term in a polynomial separately. If our function has terms with fractions or negative exponents, like the term \( x^{-6} \) in our exercise, the power rule still applies. By integrating \( x^{-6} \), for instance, the result is \( -\frac{1}{5}x^{-5} + C \).
The power rule makes the integration of polynomial components fast and reliable, breaking down to each power term.
Function Components
Functions are often composed of simpler parts or terms. In the given exercise, the function is composed of two terms: \( \frac{1}{e^x} \) and \( \frac{1}{x^6} \). To find the overall antiderivative of the entire function, we must evaluate each component separately.
The first step in solving an integration problem is identifying these components. Once identified, we can focus on finding the antiderivative of each part. This component-wise approach simplifies the problem, making it easier to manage, especially for complex functions.
Once all components are integrated separately, they are combined together, ensuring that each part remains intact within the larger function framework.
The first step in solving an integration problem is identifying these components. Once identified, we can focus on finding the antiderivative of each part. This component-wise approach simplifies the problem, making it easier to manage, especially for complex functions.
Once all components are integrated separately, they are combined together, ensuring that each part remains intact within the larger function framework.
Constant of Integration
The constant of integration, denoted as \( C \), plays a crucial role in indefinite integration. Since the process of differentiation eliminates constants, when we reverse this process through integration, a constant could have been there unseen. This is why we always add an arbitrary constant \( C \) to any antiderivative that we find.
Each component that we integrate separately may come with its own constant of integration, like \( C_1 \) or \( C_2 \). But ultimately, when these components are added together, we can combine their constants into a single constant, resulting in a simpler final expression.
Ignoring the constant of integration can lead to significant errors in understanding the full family of possible solutions that the antiderivative represents.
Each component that we integrate separately may come with its own constant of integration, like \( C_1 \) or \( C_2 \). But ultimately, when these components are added together, we can combine their constants into a single constant, resulting in a simpler final expression.
Ignoring the constant of integration can lead to significant errors in understanding the full family of possible solutions that the antiderivative represents.
Other exercises in this chapter
Problem 32
Solve with \(y_{0}=0\) and graph the solution. $$ \frac{d y}{d t}=y-1 $$
View solution Problem 32
Compute \(d y / d x\) by differentiating \(\ln y .\) This is LD: $$ y=x^{2} $$
View solution Problem 33
Assume \(10 \%\) interest (so \(a=1+i=1.1)\). Find the future value in 20 years of \(\$ 100\) deposited now.
View solution Problem 33
Solve with \(y_{0}=0\) and graph the solution. $$ \frac{d y}{d t}=-y+1 $$
View solution