Problem 42
Question
In Problems \(37-42, a, b\), and \(c\) are constants and \(g(x)\) is a continuous function whose derivative \(g^{\prime}(x)\) is also continuous. Use substitution to evaluate each indefinite integral. $$ \int g^{\prime}(x) e^{-g(x)} d x $$
Step-by-Step Solution
Verified Answer
The integral evaluates to \(-e^{-g(x)} + C\).
1Step 1: Identify the Substitution Variables
In this problem, we need to recognize a substitution that will simplify the integration process. We can choose a substitution function to simplify the expression inside the integral.
2Step 2: Use Substitution Method
Let \[ u = g(x) \]Then, the derivative of \( u \) with respect to \( x \) is:\[ \frac{du}{dx} = g'(x) \]Thus, \( du = g'(x) \, dx \). This matches the expression in the integral perfectly, which simplifies our integral.
3Step 3: Substitute and Simplify the Integral
With the substitution variables, substitute \( g'(x) \, dx = du \) and \( g(x) = u \) into the integral:\[ \int g'(x) e^{-g(x)} \, dx = \int e^{-u} \, du \]
4Step 4: Integrate Using Basic Integration Formula
The integral \( \int e^{-u} \, du \) is a standard integral. Its result is:\[ \int e^{-u} \, du = -e^{-u} + C \]where \( C \) is the constant of integration.
5Step 5: Substitute Back to Original Variables
Now, we substitute back \( u = g(x) \) to express the final answer in terms of \( x \):\[ -e^{-g(x)} + C \]
Key Concepts
Substitution MethodIntegration by SubstitutionContinuous Functions
Substitution Method
The substitution method is a handy tool in calculus, specifically for evaluating integrals. It involves replacing complex parts of an integral with simpler expressions.
At its core, the substitution method focuses on changing variables to make integrals easier to solve. Imagine you have a puzzle, and substitution helps by fitting the pieces correctly. For example, if you see a composite function within your integral, like the one in our problem with \(g(x)\), substitution allows us to rewrite that function. We say \(u = g(x)\), transforming the integral into a more straightforward form.
At its core, the substitution method focuses on changing variables to make integrals easier to solve. Imagine you have a puzzle, and substitution helps by fitting the pieces correctly. For example, if you see a composite function within your integral, like the one in our problem with \(g(x)\), substitution allows us to rewrite that function. We say \(u = g(x)\), transforming the integral into a more straightforward form.
- Recognize repetitive or complicated function parts in your integral.
- Determine a new variable that simplifies the function, often represented by \(u\).
- Express the differential \(du\) in terms of the original variable.
Integration by Substitution
Integration by substitution is a vital technique in calculus, where we use substitution to integrate functions more efficiently.
It's akin to reverse chain rule differentiation. By choosing a part of the integrand to substitute, you're updating the variable and differential, making the integral simpler.
For instance, in our example, the integral \(\int g'(x) e^{-g(x)} \, dx\) is transformed by letting \(u = g(x)\). The derivative \(g'(x) \, dx = du\) aligns with our goal to integrate \(e^{-u}\) instead of \(e^{-g(x)}\).
It's akin to reverse chain rule differentiation. By choosing a part of the integrand to substitute, you're updating the variable and differential, making the integral simpler.
For instance, in our example, the integral \(\int g'(x) e^{-g(x)} \, dx\) is transformed by letting \(u = g(x)\). The derivative \(g'(x) \, dx = du\) aligns with our goal to integrate \(e^{-u}\) instead of \(e^{-g(x)}\).
- Identify a general form \(\int f(g(x))g'(x) \, dx\) where substitution applies.
- Use \(u = g(x)\) to rearrange the integrand, making integration straightforward.
- Integrate using known integrals: here \(\int e^{-u} \, du = -e^{-u} + C\).
Continuous Functions
Continuous functions are integral to calculus and play a crucial role when applying integration techniques like substitution.
A function \(g(x)\) is continuous when there are no breaks, jumps, or holes in its graph, meaning it can be drawn without lifting your pencil off the paper. Continuity guarantees that the function behaves predictably over its domain.
In our problem, \(g(x)\) and its derivative \(g'(x)\) are both continuous. This is pivotal because it assures that the substitution method works smoothly. Without continuity, our integral might not cover the entire domain accurately, or it could have undefined points where substitution falls apart.
A function \(g(x)\) is continuous when there are no breaks, jumps, or holes in its graph, meaning it can be drawn without lifting your pencil off the paper. Continuity guarantees that the function behaves predictably over its domain.
In our problem, \(g(x)\) and its derivative \(g'(x)\) are both continuous. This is pivotal because it assures that the substitution method works smoothly. Without continuity, our integral might not cover the entire domain accurately, or it could have undefined points where substitution falls apart.
- Continuous functions allow evaluations over entire intervals without disruption.
- They ensure substitution and integration yield correct, meaningful results.
- Predictable behavior makes continuous functions suitable for various calculus applications.
Other exercises in this chapter
Problem 42
In Problems \(39-48\), first make an appropriate substitution and then use integration by parts to evaluate each integral. $$ \int x^{5} e^{x^{2}} d x $$
View solution Problem 42
Use partial fraction decompositions to evaluate each integral. $$ \int \frac{4 x^{2}-x-1}{(x+1)^{2}(x-3)} d x $$
View solution Problem 43
Find a comparison function for each integrand and determine whether the integral is convergent. $$ \int_{1}^{\infty} \frac{1}{\sqrt{1+x}} d x $$
View solution Problem 43
In Problems \(39-48\), first make an appropriate substitution and then use integration by parts to evaluate each integral. $$ \int x^{3} \sin \left(x^{2}\right)
View solution