Problem 37

Question

Evaluate the integrals. \begin{equation}\int 8 e^{(x+1)} d x\end{equation}

Step-by-Step Solution

Verified
Answer
\( 8 e^{x+1} + C \)
1Step 1: Identify the Integral
This is an integral in the form \( \int a e^{bx} dx \), where \( a = 8 \) and \( b = 1 \). We aim to apply the integration rule for exponentials.
2Step 2: Use the Exponential Integral Rule
The integral \( \int e^{ax} \, dx = \frac{1}{a} e^{ax} + C \). In our integral, \( a = 1 \), so the integral becomes: \( \int 8e^{x+1} \, dx \).
3Step 3: Simplify the Expression
Apply the rule: factor out the constant, \( 8 \), giving \( 8 \int e^{x+1} \, dx \). Now substitute \( u = x+1 \), so \( du = dx \).
4Step 4: Perform the Substitution
Now, the integral becomes \( 8 \int e^u \, du \).
5Step 5: Integrate
Integrating \( e^u \) gives \( 8 e^u + C \).
6Step 6: Substitute Back the Original Variable
Replace \( u = x+1 \) back into the expression, resulting in \( 8 e^{x+1} + C \).

Key Concepts

Integration by SubstitutionDefinite and Indefinite IntegralsExponential Function Integration
Integration by Substitution
Integration by substitution is a technique used to simplify integrals by substituting a part of the integral with a single variable. This process often makes it easier to solve the integral.
  • First, identify a part of the integrand that can be substituted. We usually choose some function within the integral whose derivative is also present.
  • Set this function equal to a new variable, typically denoted as \( u \). Compute the differential \( du \) in terms of the original variable.
  • Replace every instance of the chosen function and its differential in the integral with the new variable \( u \) and \( du \).
  • Now, the integral should look simpler and easier to evaluate. Perform the integration with respect to \( u \).
  • Finally, substitute back the original variables to express the result in terms of the initial variables.
In our example, we performed the substitution \( u = x + 1 \), transforming \( dx \) into \( du \) and simplifying the exponential function into \( e^u \). This made the integral straightforward to evaluate.
Definite and Indefinite Integrals
Integrals are a fundamental concept in calculus and they come in two types: definite and indefinite.- **Indefinite Integrals**: These involve integrating without bounds and are used to find the antiderivative of a function. They include the constant of integration, denoted by \( C \), because differentiating the antiderivative can yield infinitely many functions.- **Definite Integrals**: These have upper and lower bounds and calculate the net area under a curve within these limits. Unlike indefinite integrals, they do not include a constant of integration.When we evaluated the integral, we found an indefinite integral \( \int 8e^{x+1} \, dx \) as it was not bounded by any limits. The result was \( 8e^{x+1} + C \), incorporating the constant of integration \( C \). This highlights that without bounds, the solution remains a family of functions rather than a specific value.
Exponential Function Integration
Integrating exponential functions requires understanding their unique properties. Exponential functions take the form \( e^{ax} \), where \( e \) represents the base of the natural logarithm.
  • The integral of an exponential function \( e^{ax} \) is \( \frac{1}{a}e^{ax} + C \). This formula results from differentiating \( e^{ax} \), which always brings down the constant \( a \) from the exponent as a coefficient.
  • Sometimes, there's a coefficient outside the exponential function, like in our integral \( 8e^{x+1} \). Factor this coefficient out before integrating to simplify the process.
  • When integrating \( e^{(x+1)} \), note that the exponent components greatly influence the substitution process. We changed variables by letting \( u = x + 1 \), reducing the function to a simpler form \( e^u \).
  • Once substitution is completed, the integral of \( e^u \) is simply \( e^u + C \). Restore the original variable to get the final form.
In the given exercise, this process turns a potentially complex problem into a relatively straightforward calculation, emphasizing the elegance of exponential function integration.