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Chapter 5: Problem 21

\(\int \frac{d y}{\sqrt{y}(1-\sqrt{y})}=\) (A) \(4 \sqrt{1-\sqrt{y}}+C\) (B) \(\frac{1}{2} \ln |1-\sqrt{y}|+C\) (C) \(2 \ln (1-\sqrt{y})+C\) (D) \(-2 \ln |1-\sqrt{y}|+C\)

Short Answer

Expert verified
The correct answer is (D) \(-2 \ln |1-\sqrt{y}| + C\).

Step by step solution

01

Substitute Variables

Let \( u = \sqrt{y} \), which implies \( du = \frac{1}{2\sqrt{y}} dy \). This substitution makes \( dy = 2u \, du \). Substitute these into the integral.
02

Rewrite the Integral

The original integral becomes \( \int \frac{2u \, du}{u(1-u)} \). Cancel the \( u \) in the numerator and denominator to get \( 2 \int \frac{du}{1-u} \).
03

Integrate

The integral \( \int \frac{du}{1-u} \) is a standard form which results in \( -\ln|1-u| + C \). Therefore, the current integral evaluates to \( -2 \ln|1-u| + C \).
04

Substitute Back

Substitute back \( u = \sqrt{y} \) into the result. The final integral becomes \( -2 \ln |1-\sqrt{y}| + C \).

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Key Concepts

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

Understanding the Substitution Method
In calculus integration, the substitution method serves as a powerful tool to simplify complex integrals. Often, integrals contain functions or expressions that can be transformed into simpler forms by changing variables. By using a new variable, we can focus on a less complicated problem, making the integration process more manageable.
  • Choose a substitution that simplifies the integral. For instance, in the example problem, by letting \( u = \sqrt{y} \), the radical component is replaced with a neat linear form.
  • Modify the differential accordingly. If \( u = \sqrt{y} \), then \( du = \frac{1}{2\sqrt{y}} dy \) allows us to express \( dy \) in terms of \( du \), specifically \( dy = 2u \, du \).
  • Rewrite your integral with the new variables. Substitute all appearances of the original variable and its differential with the new terms to achieve a simplified expression.
When executed correctly, this technique reduces the original problem into a much simpler one, often with results that are straightforward to integrate. It not only makes calculations easier but also reinforces the understanding of variable transformations in calculus.
Exploring Definite and Indefinite Integrals
Integrals in calculus come in two primary types: definite and indefinite. These integrals are fundamental in understanding both areas under curves and the accumulation of functions.
  • Indefinite integrals (or antiderivatives) are represented without upper and lower limits. Solving these integrals includes adding a constant \( C \), accounting for all possible functions with the given derivative. For example, solving \( \int \frac{1}{1-u} \, du \), where \( u = \sqrt{y} \), results in \( -\ln |1-u| + C \).
  • Definite integrals include specific limits, representing the net area under the curve. These do not include the constant \( C \) in their final solution as the limits define a specific value.
Both types serve essential functions in calculus. While indefinite integrals provide a general form, definite integrals give precise values and practical applications in the real world. Understanding when to use each is crucial for practical and theoretical calculus problems.
The Role of Natural Logarithms in Integration
Natural logarithms are a vital part of calculus, especially in integration involving rational functions or expressions with exponential components. In the context of integration, they often emerge as results of specific integrals.
  • The natural log, denoted by \( \ln \), arises frequently when integrating fractions where the denominator has a variable of degree one. For instance, in the integral \( \int \frac{du}{1-u} \), the solution leverages the natural logarithm: \( -\ln|1-u| \).
  • Natural logs help solve more complex functions by transforming logarithmic differentiation into a straightforward form. This makes solving integrals like the one given simpler and more intuitive.
  • Attention to absolute values is essential whenever \( \ln|x| \) appears, ensuring the function remains valid across all applicable input values.
Understanding the natural logarithm is indispensable for tackling integration tasks involving growth and decay, simplifying calculations, and providing insight into more complex mathematical models and phenomena.

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Most popular questions from this chapter

\(\int \frac{\ln y}{y^{2}} d y=\) (A) \(\frac{1}{y}(1-\ln y)+C\) (B) \(\frac{1}{2 y} \ln ^{2} y+C\) (C) \(-\frac{1}{y}(\ln y+1)+C\) (D) \(\frac{\ln y}{y}-\frac{1}{y}+C\)

A particle starting at rest at \(t=0\) moves along a line so that its acceleration at time \(t\) is \(12 t \mathrm{ft} / \mathrm{sec}^{2}\). How much distance does the particle cover during the first 3 seconds? (A) 32 feet (B) 48 feet (C) 54 feet (D) 108 feet

\(\int \cos ^{2} 2 x d x=\) (A) \(\frac{x}{2}+\frac{\sin 4 x}{8}+C\) (B) \(\frac{x}{2}-\frac{\sin 4 x}{8}+C\) (C) \(\frac{x}{4}+\frac{\sin 4 x}{4}+C\) (D) \(\frac{x}{4}+\frac{\sin 4 x}{16}+C\)

\(\int \sin 2 \theta d \theta=\) (A) \(\frac{1}{2} \cos 2 \theta+C\) (B) \(-2 \cos 2 \theta+C\) (C) \(\cos ^{2} \theta+C\) (D) \(-\frac{1}{2} \cos 2 \theta+C\)

Find the acceleration (in \(\mathrm{ft} / \mathrm{sec}^{2}\) ) needed to bring a particle moving with a velocity of \(75 \mathrm{ft} / \mathrm{sec}\) to a stop in 5 seconds. (A) -3 (B) -6 (C) -15 (D) -25
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