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Chapter 8: Problem 18

Evaluate each integral using any algebraic method or trigonometric identity you think is appropriate, and then use a substitution to reduce it to a standard form. $$\int \frac{2^{\sqrt{y}} d y}{2 \sqrt{y}}$$

Short Answer

Expert verified
The integral evaluates to \( \frac{2^{\sqrt{y}}}{\ln 2} + C \).

Step by step solution

01

Identify the Integral Form

The given integral is \( \int \frac{2^{\sqrt{y}} \, dy}{2 \sqrt{y}} \). Notice the presence of \( \sqrt{y} \) in the denominator and as an exponent. This suggests using a substitution could simplify the integral.
02

Choose a Substitution

Let \( u = \sqrt{y} \). Then, \( y = u^2 \) and \( dy = 2u \, du \). Substitute these into the integral.
03

Substitute and Simplify

Substituting \( y = u^2 \) and \( dy = 2u \, du \) into the integral gives:\[\int \frac{2^{u} \, (2u \, du)}{2u} = \int 2^{u} \, du\]The terms \( 2u \) in the numerator and denominator cancel out.
04

Integrate the Simplified Integral

The integral \( \int 2^u \, du \) is a basic exponential integral. The antiderivative of \( 2^u \) with respect to \( u \) is \( \frac{2^u}{\ln 2} + C \), where \( C \) is an integration constant.
05

Substitute Back to Original Variable

Recall \( u = \sqrt{y} \). Substitute \( u \) back to get:\[\frac{2^{\sqrt{y}}}{\ln 2} + C\]This is the result in terms of the original variable \( y \).

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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 integration technique often used to simplify complex expressions. When you encounter an integral, identifying a substitution can transform a difficult integral into an easier, more recognizable form. In the exercise above, we dealt with integrating an expression with a square root and exponential term.
  • The first step is to choose a substitution that simplifies the integral. In this context, substituting helps manage complex square roots and exponential expressions. By setting a new variable, typically denoted as \( u \), you replace a difficult part of the integral, making it easier to work with.
  • For example, let \( u = \sqrt{y} \), transforming all parts involving \( y \) into expressions involving \( u \). This substitution turns the original integral into a more straightforward exponential integral, allowing for easier integration techniques to be applied.
Once the integration is complete, it's essential to substitute back in the original variable to present the solution in terms of the initial variable given in the problem. This approach exemplifies the cleaning power of substitution in calculus.
Exponential Integrals
Exponential integrals are integrals involving exponential functions, which often have specific forms and rules. They're a staple in calculus because exponential functions like \( a^x \), where \( a \) is a constant, can model various natural phenomena.
  • To integrate \( a^u \), where \( a \) is a constant, the antiderivative is \( \frac{a^u}{\ln a} + C \). This formula is crucial when dealing with exponential integrals, as it provides a straightforward method to find the indefinite integral.
  • In the given exercise, after employing the substitution method, the remaining integral was \( \int 2^u \, du \). Using the exponential integral rule, the solution converts to \( \frac{2^u}{\ln 2} + C \), showcasing the direct application of this rule.
Understanding the rules behind exponential integrals ensures you can quickly and accurately integrate these types of expressions, making this knowledge invaluable for more complex calculus problems.
Algebraic Simplification
Algebraic simplification plays a crucial role in solving integrals efficiently. This process involves reducing an expression to its simplest form, often making difficult problems easier to handle.
  • In integrals, simplification can involve canceling terms, rearranging factors, or combining like terms. These steps reduce the integral to a form where known rules or techniques can be easily applied.
  • During the original solution, when substituting \( u = \sqrt{y} \) and \( dy = 2u \, du \) into the integral, the terms \( 2u \) in the numerator and denominator canceled each other. This cancellation is a type of algebraic simplification that turned a complex expression into a much simpler \( \, \int 2^u \, du \, \).
By employing these simple algebraic tricks, more complex integrals become much easier to solve. Understanding how to simplify correctly ensures that you don't miss out on opportunities to use important calculus techniques like substitution or exponential integration to their fullest extent.

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Most popular questions from this chapter

Use integration by parts to obtain the formula $$ \int \sqrt{1-x^{2}} d x=\frac{1}{2} x \sqrt{1-x^{2}}+\frac{1}{2} \int \frac{1}{\sqrt{1-x^{2}}} d x $$

Evaluate the integrals without using tables. $$\int_{0}^{2} \frac{d x}{\sqrt{|x-1|}}$$

Use integration, the Direct Comparison Test, or the limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{1}^{\infty} \frac{e^{x}}{x} d x$$

The functions \(y=e^{x^{3}}\) and \(y=x^{3} e^{x^{3}}\) do not have elementary antiderivatives, but \(y=\left(1+3 x^{3}\right) e^{x^{3}}\) does. EvaluateEvaluate $$ \int\left(1+3 x^{3}\right) e^{x^{3}} d x $$ $$ \int\left(1+3 x^{3}\right) e^{x^{3}} d x $$

Use any method to evaluate the integrals. $$\int x \sin ^{2} x d x$$
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