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Chapter 7: Problem 24

Sketch the region bounded by the graphs of the algebraic functions and find the area of the region. $$ y=\frac{1}{x^{2}}, \quad y=0, x=1, x=5 $$

Short Answer

Expert verified
The area of the region bounded by the given curves is \( 4/5 \) square units.

Step by step solution

01

Sketch the Region

Draw the functions \( y = 1/x^2 \), \( y = 0 \), \( x = 1 \), and \( x = 5 \) on the same graph. You'll see a region enclosed by these graphs. This region lies above the x-axis (because \( y = 0 \) means x-axis), between \( x = 1 \) and \( x = 5 \), and below the curve \( y = 1/x^2 \).
02

Set Up the Integral

The area A between the curves can be found by the definite integral from a to b, where a and b are the x-values of the boundaries (x=1 and x=5), of the difference of the functions. So, the integral should be set up as \( A = \int_{1}^{5} (1/x^2 - 0) dx \).
03

Calculate the Integral

Evaluate the integral \( A = \int_{1}^{5} 1/x^2 dx \). For that, find the antiderivative of \( 1/x^2 \) which is \( -1/x \). Then evaluate from 1 to 5.
04

Evaluate the Definite Integral

\( A = [-1/x]_{1}^{5} = -1/5 - (-1/1) = -1/5 + 1 = 4/5 \).

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Key Concepts

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

Definite Integral
The definite integral is a fundamental concept in calculus used to find the area under a curve between two points. It is represented as \( \int_{a}^{b} f(x) \, dx \), where "a" and "b" are the limits of integration, and \( f(x) \) is the function being integrated. In this exercise, the definite integral \( \int_{1}^{5} \frac{1}{x^2} \, dx \) represents the area under the curve of \( y = \frac{1}{x^2} \) from \( x = 1 \) to \( x = 5 \).

  • "a" (1) is the lower limit, and "b" (5) is the upper limit.
  • The function \( f(x) = \frac{1}{x^2} \) is continuous and positive in this interval.
The process involves finding the antiderivative of the function and then evaluating it at the upper and lower limits. This subtraction gives the total area bounded by the curve and the x-axis between the specified limits. It鈥檚 vital to correctly evaluate the antiderivative at these points to get the right area.
Area Under a Curve
The area under a curve helps us quantify the space enclosed between the curve and the x-axis over an interval. In this example, the region is bounded by \( y = \frac{1}{x^2} \), the x-axis \( y = 0 \), and vertical lines \( x = 1 \) and \( x = 5 \).

  • The graph of \( y = \frac{1}{x^2} \) decreases as x increases, creating a sloping curve.
  • Setting up a definite integral from 1 to 5 allows calculating the exact area beneath the curve.
This calculation helps visualize how the function behaves over an interval. In our case, the area represents the space enclosed above the x-axis and below the declining curve of \( y = \frac{1}{x^2} \) within the specified bounds.
Antiderivative
An antiderivative of a function \( f(x) \) is another function whose derivative is \( f(x) \). For \( y = \frac{1}{x^2} \), the antiderivative is \( -\frac{1}{x} \). This function, when differentiated, returns \( \frac{1}{x^2} \).

  • The antiderivative allows us to simplify the process of finding the area under the curve.
  • In this exercise, it was crucial to determine \( -\frac{1}{x} \) to evaluate the definite integral from 1 to 5.
By plugging in the bounds into the antiderivative function, we perform the fundamental operation of definite integration. This produces the net accumulation, known as the area, which in our case is \( \frac{4}{5} \). Identifying the correct antiderivative is key to solving these types of problems accurately.

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Most popular questions from this chapter

Let \(n \geq 1\) be constant, and consider the region bounded by \(f(x)=x^{n}\), the \(x\) -axis, and \(x=1\). Find the centroid of this region. As \(n \rightarrow \infty\), what does the region look like, and where is its centroid?

Consider the region bounded by the graphs of \(y=x^{2 n}\) and \(y=b\), where \(b>0\) and \(n\) is a positive integer. (a) Set up the integral for finding \(M_{y}\). Because of the form of the integrand, the value of the integral can be obtained without integrating. What is the form of the integrand and what is the value of the integral? Compare with the result in part (b). (b) Is \(\bar{y}>\frac{b}{2}\) or \(\bar{y}<\frac{b}{2} ?\) Explain. (c) Use integration to find \(\bar{y}\) as a function of \(n\). (d) Use the result of part (c) to complete the table. $$ \begin{array}{|l|l|l|l|l|} \hline \boldsymbol{n} & 1 & 2 & 3 & 4 \\ \hline \boldsymbol{y} & & & & \\ \hline \end{array} $$ (e) Find \(\lim _{n \rightarrow \infty} \bar{y}\). (f) Give a geometric explanation of the result in part (e).

Find \(M_{x}, M_{y}\), and \((\bar{x}, \bar{y})\) for the laminas of uniform density \(\rho\) bounded by the graphs of the equations. $$ y=x^{2 / 3}, y=0, x=8 $$

A region bounded by the parabola \(y=4 x-x^{2}\) and the \(x\) -axis is revolved about the \(x\) -axis. A second region bounded by the parabola \(y=4-x^{2}\) and the \(x\) -axis is revolved about the \(x\) -axis. Without integrating, how do the volumes of the two solids compare? Explain.

Two models \(R_{1}\) and \(R_{2}\) are given for revenue (in billions of dollars per year) for a large corporation. The model \(R_{1}\) gives projected annual revenues from 2000 to 2005, with \(t=0\) corresponding to 2000, and \(R_{2}\) gives projected revenues if there is a decrease in the rate of growth of corporate sales over the period. Approximate the total reduction in revenue if corporate sales are actually closer to the model \(\boldsymbol{R}_{\mathbf{2}}\) $$ \begin{aligned} &R_{1}=7.21+0.58 t \\ &R_{2}=7.21+0.45 t \end{aligned} $$
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