/*! 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 50 $$ y=\frac{\ln x}{x^{2}}, y=0,... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 50

$$ y=\frac{\ln x}{x^{2}}, y=0, x=1, x=e $$

Short Answer

Expert verified
The area under the curve is \( \frac{1}{e} \)

Step by step solution

01

- Setting up the integral

We will use integral calculus to find the area. The integral of a function gives the area under the curve of that function. We need to set the integral for the given equation. The integral should be set up as follows: \( \int_{1}^{e} \frac{\ln x}{x^{2}} dx \).
02

- Solving the integral

We are going to use the method of integration by parts using the formula \( \int u dv = uv - \int v du \), where u and v are parts of the original function. In this case, we set \( u = \ln x \), \( dv = \frac{1}{x^{2}} dx \). Differentiating and integrating, we get \( du = \frac{1}{x} dx \) and \( v = -\frac{1}{x} \). Substituting back into the integration by parts formula, we get \( -\ln x \cdot \frac{1}{x} + \int \frac{1}{x} \cdot \frac{1}{x} dx = -\frac{\ln x}{x} - \int \frac{1}{x^{2}} dx \). Evaluating the integral of \( \frac{1}{x^{2}} \) we get \( \int \frac{1}{x^{2}} dx = -\frac{1}{x} \). Now substitute this into the equation and simplify to \( -\frac{\ln x}{x} + \frac{1}{x} \).
03

- Evaluating the integral

The next step is to evaluate the integral at x=1 and x=e. The area is: \( \left[-\frac{\ln x}{x} + \frac{1}{x} \right]_{1}^{e} = -\frac{\ln e}{e} + \frac{1}{e} - \left(-\frac{\ln 1}{1} + \frac{1}{1} \right) = -1 + \frac{1}{e} - (-1) = \frac{1}{e} \).

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Key Concepts

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

Integral Calculus
Integral calculus is a branch of mathematics focused on finding the total accumulation of quantities, which can be viewed as the area under a curve when graphed. It is the counterpart to differential calculus, which deals with the rate of change of quantities. Integral calculus is vital for many applications in science, engineering, and economics.

For example, when faced with a function like the one in the exercise provided, integral calculus allows us to determine the area bounded by the function, the x-axis, and the vertical lines at x=1 and x=e. In this case, we are interested in understanding how much 'space' the function occupies on a graph between those points.

Integral calculus is also used for more abstract problems, such as calculating probabilities, center of mass, and other measurements that depend on accumulation rather than instant variation.
Area Under the Curve
When visualizing integration in a geometrical sense, one common interpretation is finding the 'area under the curve' of a graphed function over a given interval. This idea is pivotal to applications in physics for finding quantities like displacement and work, and in probability theory for defining distributions.

For the given exercise, the area under the curve between x=1 and x=e represents the integral of the function \( \frac{\text{ln} x}{x^{2}} \) with respect to x over this interval. The integral gives us a numerical value that represents this area. However, since the function dips beneath the x-axis, we are computing a

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Most popular questions from this chapter

Approximate the integral using (a) the Trapezoidal Rule and (b) Simpson's Rule for the indicated value of \(n\). (Round your answers to three significant digits.) $$ \int_{0}^{1} \sqrt{1-x} d x, n=4 $$

Decide whether the integral is improper. Explain your reasoning. $$ \int_{0}^{1} \frac{d x}{3 x-2} $$

Probability The probability of finding between \(a\) and \(b\) percent iron in ore samples is modeled by \(P(a \leq x \leq b)=\int_{a}^{b} 2 x^{3} e^{x^{2}} d x, \quad 0 \leq a \leq b \leq 1\) (see figure). Find the probabilities that a sample will contain between (a) \(0 \%\) and \(25 \%\) and (b) \(50 \%\) and \(100 \%\) iron.

Consumer and Producer Surpluses Find the consumer surplus and the producer surplus for a product with the given demand and supply functions. Demand: \(p=\frac{60}{\sqrt{x^{2}+81}}\), Supply: \(p=\frac{x}{3}\)

Profit The net profits \(P\) (in billions of dollars per year) for The Hershey Company from 2002 through 2005 can be modeled by \(P=\sqrt{0.00645 t^{2}+0.1673}, \quad 2 \leq t \leq 5\) where \(t\) is time in years, with \(t=2\) corresponding to 2002 . Find the average net profit over that time period. (Source: The Hershey Co.)
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