/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 Evaluate the integrals using app... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
The integral evaluates to \(\frac{1}{6}(2y + 1)^{3/2} - \frac{1}{2}(2y + 1)^{1/2} + C\).

Step by step solution

01

Identify the substitution

To simplify the integral, we'll perform a substitution that will make the integral easier to solve. Notice that the expression inside the square root, \(2y + 1\), can become a simpler variable, such as \(u\). Let's set \(u = 2y + 1\).
02

Derive the differential

Differentiate \(u = 2y + 1\) with respect to \(y\) to find \(du/dy\). This gives us \(du = 2 \, dy\). Solve for \(dy\) to substitute in the integral: \(dy = \frac{du}{2}\).
03

Rewrite the integral in terms of u

Substitute \(u = 2y + 1\) and \(dy = \frac{du}{2}\) into the original integral: \(\int \frac{y}{\sqrt{2y+1}} \, dy = \int \frac{y}{\sqrt{u}} \, \frac{du}{2}\). Since \(u = 2y + 1\), solve for \(y\): \(y = \frac{u-1}{2}\). Substitute \(y = \frac{u-1}{2}\) into the integral: \(\int \frac{\frac{u-1}{2}}{\sqrt{u}} \, \frac{du}{2}\).
04

Simplify the integral

Simplify the expressions: \(\int \frac{\frac{u-1}{2}}{\sqrt{u}} \, \frac{du}{2} = \frac{1}{4}\int \frac{u-1}{\sqrt{u}} \, du\). This can be split into two separate integrals: \(\frac{1}{4}\int \frac{u}{\sqrt{u}} \, du - \frac{1}{4}\int \frac{1}{\sqrt{u}} \, du\).
05

Evaluate the integrals

Evaluate \(\int \frac{u}{\sqrt{u}} \, du\) which simplifies to \(\int \sqrt{u} \, du = \int u^{1/2} \, du\). The integral of \(u^{1/2}\) is \(\frac{2}{3}u^{3/2}\). Now evaluate \(\int \frac{1}{\sqrt{u}} \, du\), which is the same as \(\int u^{-1/2} \, du\). This integral is \(2u^{1/2}\).
06

Combine and simplify results

Combine the evaluated integrals to get: \(\frac{1}{4}\left(\frac{2}{3} u^{3/2} - 2u^{1/2}\right) = \frac{1}{6} u^{3/2} - \frac{1}{2} u^{1/2}\).
07

Substitute back for y

Replace \(u\) with the original expression in terms of \(y\), since \(u = 2y + 1\). The final expression is \(\frac{1}{6}(2y + 1)^{3/2} - \frac{1}{2}(2y + 1)^{1/2} + C\), where \(+ C\) is the integration constant.

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Key Concepts

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

Definite Integrals
Definite integrals are used to calculate the area under a curve between two specific points on the x-axis. Unlike indefinite integrals, which include a constant of integration and represent a family of curves, definite integrals have limits of integration, which are the bounds within which the area is calculated. This makes them particularly useful for determining exact quantities, such as total distance, area, volume, or even economic measures between two points.

When working with definite integrals and the substitution method, it's essential to change the limits of integration according to the substitution made. For example, if the substitution technique transforms the variable from \(y\) to \(u\), the limits for \(y\) need to be converted into limits for \(u\). This ensures that the evaluated integral remains accurate to the original function, allowing one to obtain exact results.
Substitution Method
The substitution method is a fundamental strategy for solving integrals, especially when the integrand (the function inside the integral sign) is complicated to integrate directly. The idea is to make a substitution, changing the variable of integration to simplify the integral.
  • Identify part of the function that can be substituted with a new variable, say \(u\).
  • Express \(dy\) in terms of \(du\) by differentiating \(u\) with respect to \(y\). For example, if \(u = 2y + 1\), then \(du = 2 \, dy\).
  • Replace all occurrences of the original variable and its differential in the integral.

This substitution turns a previously difficult integral into one that is more manageable. Once the integral in terms of \(u\) is solved, you must remember to substitute back the original variable to achieve the final result.
Antiderivatives
An antiderivative is essentially the reverse process of finding a derivative. While the derivative of a function gives the rate of change, the antiderivative, also known as the integral, provides a function whose derivative is the original function.

Finding an antiderivative can often involve techniques like substitution, as shown in our example. Once the integral is rewritten and simplified, the next step is to find the antiderivative of each part. For instance, if you have a term like \(\int \sqrt{u} \, du\), you find an antiderivative \(\frac{2}{3} u^{3/2} + C\). Here, \(C\) is the constant of integration added when finding indefinite integrals. This reflects the fact that there are infinitely many antiderivatives, each differing by a constant.
Integration Techniques
Integration techniques are various strategies that make handling integrals more approachable. These techniques are essential because many functions do not have straightforward antiderivatives. Some of the commonly used methods include:
  • Substitution Method: Simplifying integrals by changing the variable to transform a complex function into a basic one.
  • Integration by Parts: Used when the integral is a product of two functions, leveraging the product rule for derivatives in reverse.
  • Partial Fraction Decomposition: Breaking down complex rational functions into simpler fractions that can be integrated easily.

These techniques are valuable tools that expand the range of integrals that can be solved. By mastering them, students can solve complex integrals by recognizing which method best applies to the function they are working with. In our given exercise, the substitution method is the key technique that simplifies the problem and finds the antiderivative efficiently.

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