Problem 39

Question

In Problems $39-48$, first make an appropriate substitution and then use integration by parts to evaluate each integral. $$ \int e^{\sqrt{x}} d x $$

Step-by-Step Solution

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Answer

$ \int e^{\sqrt{x}} dx = 2e^{\sqrt{x}}(\sqrt{x} - 1) + C $.

1Step 1: Choose an Appropriate Substitution

The composition inside the exponent suggests a substitution. Let $ u = \sqrt{x} $. This implies that $ x = u^2 $. Differentiating, we get $ dx = 2u \, du $. This substitution will simplify the integral.

2Step 2: Substitute and Simplify

Replace $ \sqrt{x} $ with $ u $ and $ dx $ with $ 2u \, du $ in the integral: \[ \int e^{\sqrt{x}} \, dx = \int e^u \cdot 2u \, du = 2 \int u e^u \, du \]. Now, the substitution has transformed the problem into an integral that can be solved by integration by parts.

3Step 3: Apply Integration by Parts

For $ \int u e^u \, du $, use integration by parts where $ v = e^u $ and $ dv = e^u \, du $, and set $ w' = u $ and $ dw = du $. The integration by parts formula is $ \int w' \, dv = wv - \int v \, dw $.

4Step 4: Differentiate and Integrate Parts

Differentiate and integrate to apply the parts: $ w = u $, $ dv = e^u \, du $, which means $ w = u $ and $ v = e^u $. Substitute back into the integration by parts formula: \[ \int u e^u \, du = ue^u - \int e^u \, du \].

5Step 5: Solve the Resulting Integral

The integral becomes: \[ \int u e^u \, du = ue^u - e^u = e^u(u - 1) \]. Substitute $ u = \sqrt{x} $ back into this expression.

6Step 6: Back-Substitute and Simplify

Replace $ u $ with $ \sqrt{x} $ to get $ e^{\sqrt{x}} (\sqrt{x} - 1) $. Therefore, the solution to the original integral is: \[ 2 \cdot e^{\sqrt{x}} (\sqrt{x} - 1) + C \], where $ C $ is the constant of integration.

Key Concepts

Understanding the Substitution MethodExploring Integration TechniquesGrasping Exponential Functions in Integration

Understanding the Substitution Method

The substitution method is a powerful tool for simplifying integrals, especially when dealing with composite functions. In essence, substitution involves changing variables to transform an integral into a more manageable form. Here’s how it works:

You start by identifying a part of the integrand (the function being integrated) that can be re-expressed in terms of a new variable.
This part is usually the inner function when dealing with composite functions.
For instance, in the problem $ \int e^{\sqrt{x}} \, dx $, identifying that $ u = \sqrt{x} $ is key. This substitution simplifies the integration process.
Next, complete the derivative transformation: $ dx $ changes as well. Here, $ dx = 2u \, du $.
This substitution converts the original integral into one where basic integration techniques, such as integration by parts, can be easily applied.

This technique not only simplifies the mathematics but also demonstrates the dynamic nature of integrals, showing how seemingly complex problems can be broken down with the right view.

Exploring Integration Techniques

Integration techniques are strategies that mathematicians and students use to simplify and solve integrals. Different techniques are suited for different types of integrands. Some of the primary techniques include:

Substitution: Useful for integrals involving composite functions, like $ e^{\sqrt{x}} $.
Integration by Parts: This technique, derived from the product rule for differentiation, is beneficial when the integrand is a product of two functions, such as $ u e^u $ in our case. The formula $ \int w' \, dv = wv - \int v \, dw $ helps transform the integral into a solvable form.
Partial Fractions: Applied when dealing with rational functions.
Trigonometric Integrals and Substitutions: Used when integrals include trigonometric functions, which is common in calculus.

In our example, once substitution rewrites the integral, integration by parts is applied. By letting $ v = e^u $ and $ w' = u $, we simplify the integration.

Grasping Exponential Functions in Integration

Exponential functions, such as $ e^{x} $, have distinct properties that make them fascinating and sometimes challenging to integrate. They appear often in calculus due to their ubiquitous nature in describing growth and decay processes. Important features of exponential functions:

The base of the natural exponential function is $ e $, approximately 2.71828, which is a mathematical constant.
Unlike regular polynomials, exponential functions grow by constant factors rather than constant differences, leading to functions that change continuously at varying rates.
In calculus, integrating expressions involving $ e^x $ often results in terms that involve the function itself, making them relatively straightforward under simple contexts.

In complex cases, such as $ \int e^{\sqrt{x}} \, dx $, combining substitution with techniques like integration by parts is prevalent. Overall, understanding exponential functions aids in navigating through calculus problems efficiently and with more confidence.

Problem 39

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