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Chapter 5: Problem 52

Evaluate the integrals using appropriate substitutions. $$ \int \frac{e^{\sqrt{2 y+1}}}{\sqrt{2 y+1}} d y $$

Short Answer

Expert verified
The integral evaluates to \( e^{\sqrt{2y+1}} + C \).

Step by step solution

01

Identify the Inner Function for Substitution

Recognize that the expression under the square root, \( \sqrt{2y+1} \), complicates the integration. Define a substitution to simplify the integral. Let \( u = \sqrt{2y+1} \), which implies that \( u^2 = 2y + 1 \).
02

Differentiate and Solve for dy

Take the derivative of both sides with respect to \( y \) to find \( dy \) in terms of \( du \). Differentiating, we get \( \frac{d}{dy}(u^2) = \frac{d}{dy}(2y + 1) \), which simplifies to \( 2u \frac{du}{dy} = 2 \). Solving for \( \frac{du}{dy} \) gives us \( du = \frac{1}{u} dy \), so \( dy = u \, du \).
03

Substitute in the Integral

Substitute \( u = \sqrt{2y+1} \) and \( dy = u \, du \) back into the integral. The original integral \( \int \frac{e^{\sqrt{2y+1}}}{\sqrt{2y+1}} \, dy \) becomes \( \int \frac{e^u}{u} \cdot u \, du \), which simplifies to \( \int e^u \, du \).
04

Integrate

The integral \( \int e^u \, du \) is straightforward, as the antiderivative of \( e^u \) is \( e^u \). Thus, we get \( e^u + C \), where \( C \) is the constant of integration.
05

Substitute Back to Original Variable

Express the result back in terms of \( y \). Since we let \( u = \sqrt{2y+1} \), substitute back to obtain \( e^{\sqrt{2y+1}} + C \) as the final answer for the original integral.

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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 tool in calculus, especially when dealing with complex integrals. It makes the integration process easier by transforming a complicated integral into a simpler one. In our problem, the inner function \( \sqrt{2y+1} \) is difficult to integrate directly. By letting \( u = \sqrt{2y+1} \), we simplify the expression. This method involves:
  • Identifying an "inner function" \( g(y) \) as \( \sqrt{2y+1} \)
  • Substituting \( u = g(y) \), converting \( dy \) in terms of \( du \)
  • Transforming the integral to be in terms of \( u \) instead of \( y \)
The transformation makes integration much more straightforward. Once integrated, you'll substitute back to the original variable to obtain the final result.
Antiderivative
An antiderivative is simply the reverse of differentiation. It refers to a function whose derivative is the original function we started with. When you integrate a function, you are finding its antiderivative.For example, in our integral \( \int e^u \, du \), the antiderivative of \( e^u \) is known to be \( e^u \). The simplicity of the exponential function makes finding its antiderivative a direct operation. Remember, the process of finding an antiderivative involves:
  • Recognizing the basic form of the function being integrated
  • Using known antiderivatives, such as \( \int e^x \, dx = e^x + C \)
  • Adding a constant \( C \) to express the general form of potential antiderivatives
Given these steps, integration becomes a straightforward search for the reverse operation of differentiation.
Definite Integrals
Though our example involves an indefinite integral, it's helpful to understand definite integrals as well. A definite integral calculates the net area under a curve between specific limits. However, the technique of substitution can still apply to both indefinite and definite integrals.Definite integrals involve:
  • Setting limits of integration on the variable \( y \)
  • Transforming these limits to the new variable \( u \) once a substitution is made
  • Computing the antiderivative as normal
  • Plugging in the transformed limits to find the area
The process of finding a definite integral encompasses using the Fundamental Theorem of Calculus, by evaluating at upper and lower bounds. This turns the art of integration into a practical tool for solving real-world problems.

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