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Chapter 8: Problem 44

Evaluate the integrals in Exercises \(31-52 .\) Some integrals do not require integration by parts. $$ \int \frac{e^{\sqrt{x}}}{\sqrt{x}} d x $$

Short Answer

Expert verified
The integral evaluates to \(2e^{\sqrt{x}} + C.\)

Step by step solution

01

Identify the substitution component

Observing the integral \( \int \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx \), we note that \( \sqrt{x} \) is the expression that seems to complicate the integration. We will use substitution to simplify this integral.
02

Set the substitution

Let's set \( u = \sqrt{x} \). Then, differentiating both sides with respect to \( x \), we have \( \frac{du}{dx} = \frac{1}{2\sqrt{x}} \), which can be rewritten as \( du = \frac{1}{2\sqrt{x}} \, dx \). Thus, we find that \( dx = 2u \, du \).
03

Rewrite the integral using substitution

Substitute \( u = \sqrt{x} \) and \( dx = 2u \, du \) into the integral. The integral becomes:\[ \int \frac{e^{u}}{u} 2u \, du = 2 \int e^{u} \, du. \]
04

Simplify and integrate

Since the expression simplifies to \( 2 \int e^{u} \, du \), we can integrate easily:\[ 2 \int e^{u} \, du = 2e^{u} + C, \]where \( C \) is the constant of integration.
05

Back-substitute the original variable

Revert back to the original variable by substituting \( u = \sqrt{x} \) back into the expression:\[ 2e^{u} + C = 2e^{\sqrt{x}} + C. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The substitution method is a fundamental technique in calculus used to simplify complex integrals. Essentially, it transforms a complicated integral into a more manageable form. In our given exercise, we have the integral \( \int \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx \). The function \( \sqrt{x} \) is tricky to deal with directly, so we employ substitution to make integration straightforward.

  • First, we select the substitution \( u = \sqrt{x} \). This choice simplifies the integral because the derivative of \( \sqrt{x} \) naturally appears in the integration process.
  • Finding \( dx \) in terms of \( u \), we differentiate and rearrange, obtaining \( dx = 2u \, du \).
  • By making these substitutions, the original integral transforms into a simpler form \( 2 \int e^u \, du \).
Once substitution is complete, the integral becomes easier to evaluate. After integrating with respect to \( u \), we substitute back the original variable to get the final result.
Exponential Functions
Exponential functions, such as \( e^x \), are a vital part of calculus and appear frequently in integration problems. The base of natural logarithms, \( e \), approximately 2.718, serves as a foundation for these functions. In the context of our integral \( \int \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx \), the expression \( e^{\sqrt{x}} \) is key.

  • The exponential function naturally arises in various mathematical scenarios, including growth models and decay processes.
  • One of the strengths of exponential functions is their simple differentiation and integration rules. For instance, the derivative and integral of \( e^x \) are both \( e^x \), making calculations straightforward.
In our case, once substitution is done, integrating \( e^u \) is simple as the integral becomes \( 2e^u + C \). The elegance of the exponential function properties allows us to handle such integrals efficiently.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a central theorem linking the concept of differentiation with integration. It provides a way to evaluate definite integrals and explains the accumulation of quantities. In the given problem, where we work with the integral \( \int 2e^{u} \, du \), the theorem aids in transitioning from the antiderivative to the final solution.

  • Part of the theorem allows us to find antiderivatives, enabling the evaluation of definite integrals as differences of function values.
  • In indefinite integrals, like our problem, finding an antiderivative becomes the main goal, representing it as \( 2e^u + C \), where \( C \) is the integration constant.
Through this theorem, we can revert back to the original variable, ensuring the solution represents the original problem's structure. By back-substituting \( u = \sqrt{x} \), we directly apply these principles to yield the result \( 2e^{\sqrt{x}} + C \).

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