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Chapter 7: Problem 15

Evaluate the integrals. $$\int \frac{e^{\sqrt{r}}}{\sqrt{r}} d r$$

Short Answer

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

Step by step solution

01

Identify the Integration Technique

The integral \( \int \frac{e^{\sqrt{r}}}{\sqrt{r}} d r \) suggests a substitution method could simplify it. The presence of both a function and its derivative hints at using substitution.
02

Choose an Appropriate Substitution

Let \( u = \sqrt{r} \), which implies \( r = u^2 \). Then, differentiate to find \( dr \) in terms of \( du \). We get \( \frac{dr}{du} = 2u \), so \( dr = 2u \, du \).
03

Rewrite the Integral in Terms of the New Variable

Substitute \( r = u^2 \) and \( dr = 2u \, du \) into the integral: \[ \int \frac{e^{\sqrt{r}}}{\sqrt{r}} d r = \int \frac{e^u}{u} \cdot 2u \, du = \int 2e^u \, du. \] The \( u \) terms cancel out.
04

Integrate with Respect to \( u \)

The integral simplifies to \( 2 \int e^u \, du \). The integral of \( e^u \) with respect to \( u \) is \( e^u \). Thus, the integral becomes \( 2e^u + C \), where \( C \) is the constant of integration.
05

Substitute Back for Original Variable

Replace \( u \) back with \( \sqrt{r} \) to express the integral in terms of the original variable: \[ 2e^{\sqrt{r}} + C. \]

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Key Concepts

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

Substitution Method
When dealing with complex integrals, the substitution method can simplify the process. It involves replacing a part of the integral with a new variable to make calculations easier. This often occurs when you identify a term and its derivative within the integral.

  • Start by looking for a part within your integral that can be set as a new variable, say \( u \).
  • Once chosen, express the other variables in terms of this new variable \( u \) and find \( du \).
In our example, the integral \( \int \frac{e^{\sqrt{r}}}{\sqrt{r}} dr \) led to choosing \( u = \sqrt{r} \). Consequently, \( dr \) was expressed as \( 2u \, du \). Substituting these into the integral simplifies the process, showing the power of substitution in integration.
Exponential Functions
Exponential functions are critical in calculus due to their unique properties, such as their derivative and integral being exponential functions themselves. The function \( e^x \) is particularly significant, known as the exponential function.

  • An essential property is that the derivative and integral of \( e^u \) with respect to a variable are both \( e^u \). This makes calculations straightforward.
  • In substitution, such as \( \int 2e^u \, du \) in our example, this property simplifies the problem significantly.
By understanding these exponential features, solving integrals involving \( e^x \) becomes much simpler, as they maintain their form upon differentiation or integration.
Integral Calculus
Integral calculus focuses on finding antiderivatives and areas under curves. The fundamental theorem links differential and integral calculus, simplifying the process of finding integrals.

  • Integration is essentially finding a function whose derivative matches a given function.
  • Definite integrals represent the area under the curve between two points, while indefinite integrals include a constant of integration \( C \).
In our example, the indefinite integral \( \int 2e^u \, du \) was evaluated to find an antiderivative, resulting in \( 2e^{\sqrt{r}} + C \). Integral calculus techniques, such as substitution, greatly aid in tackling complex integrals efficiently.

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Most popular questions from this chapter

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