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Chapter 8: Problem 18

The integrals in Exercises \(1-44\) are in no particular order. Evaluate each integral using any algebraic method or trigonometric identity you think is appropriate. When necessary, use a substitution to reduce it to a standard form. $$ \int \frac{2^{\sqrt{y}} d y}{2 \sqrt{y}} $$

Short Answer

Expert verified
\( \frac{2^{\sqrt{y}}}{\ln(2)} + C \)

Step by step solution

01

Simplify the Integral

First, observe that the integral can be simplified by noticing that the expression under the integral consists of two parts: the function inside the numerator and the function in the denominator. The integral can be rewritten as \( \int \frac{2^{\sqrt{y}}}{2 \sqrt{y}} \, dy = \int \frac{1}{2} 2^{\sqrt{y}} \frac{1}{\sqrt{y}} \, dy \). This form suggests the use of substitution.
02

Choose a Substitution

We choose a substitution to simplify the integral. Let \( u = \sqrt{y} \). Then, \( y = u^2 \) and \( dy = 2u \, du \). Substitute into the integral to get \[ \int \frac{2^{u} \, 2u \, du}{2u} = \int 2^{u} \, du. \]
03

Evaluate the Integral

Now, our integral has been transformed into \( \int 2^{u} \, du \). We apply the integration formula for exponential functions: \( \int a^{u} \, du = \frac{a^{u}}{\ln(a)} + C \). Here, \( a = 2 \), thus \( \int 2^{u} \, du = \frac{2^{u}}{\ln(2)} + C \).
04

Substitute Back

The final step is to substitute back the original variable \( y \) for \( u \). Recall that \( u = \sqrt{y} \). So, the integral becomes \[ \frac{2^{\sqrt{y}}}{\ln(2)} + C \].

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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 powerful technique in integration. It's especially useful when an integral seems difficult to solve in its original form. This method involves replacing a part of the integral with a new variable, thereby simplifying the problem. Here's a simple step-by-step for the substitution method:
  • Choose a substitution: Identify a part of the integral that can be replaced with a new variable, say \( u \). This often simplifies the expression. For example, if you have \( y \) in the integral \( \int \frac{2^{\sqrt{y}} d y}{2 \sqrt{y}} \), you might set \( u = \sqrt{y} \).
  • Find \( dy \): Express \( dy \) in terms of \( du \). In our substitution, \( dy = 2u \, du \).
  • Substitute: Replace all occurrences of the original variable in the integral with your new variable and expressions related to it. Your goal is to transform the integral into a simpler form, as in \( \int 2^{u} \, du \).
  • Integrate: Perform the integration on the simplified integral.
  • Substitute back: Once integrated, replace the substitution variable back with the original term to return to the variable the problem started with.
Using substitution efficiently can make seemingly complex integrals much more manageable. It's a key method that you'll often see applied to exponential and trigonometric functions.
Exponential Functions
Exponential functions are a staple in integration problems and mathematics in general. These functions take the form \( a^x \), where \( a \) is a constant base and \( x \) is the exponent. In our case, we dealt with an exponential function \( 2^{\sqrt{y}} \). When integrating exponential functions like this, there's a general principle:
  • Integration formula: The rule \( \int a^u \, du = \frac{a^u}{\ln(a)} + C \), states that when integrating \( a^u \), divide by the natural log of the base \( a \).
  • Handling exponential growth: Exponential functions grow rapidly. When integrated, they model real-world processes like population growth, radioactive decay, and interest calculations effectively.
Integration of exponential functions often requires using logarithms, as seen in integration formulas. This highlights the natural connection between exponential growth and logarithmic rates.
Definite and Indefinite Integrals
The integration process can lead to either a definite or an indefinite integral, each serving a different purpose in mathematics.
  • Indefinite Integrals: This is what we calculated in the solution. An indefinite integral represents a family of functions and includes a constant \( C \). For example, our result \( \frac{2^{\sqrt{y}}}{\ln(2)} + C \) is an indefinite integral.
  • Definite Integrals: Unlike indefinite integrals, definite integrals have limits of integration. They compute the exact area under a curve between two bounds. They give a specific numerical value rather than a function.
  • When to use which: Use indefinite integrals when seeking a general solution or family of solutions. Opt for definite integrals when solving for an exact value or area.
Understanding the difference between definite and indefinite integrals is crucial in tackling various mathematical problems effectively, as they have different applications and results.

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