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Chapter 9: Problem 13

Find the area under the graph of \(y=1 / x^{2}\) for \(x \geq 2\).

Short Answer

Expert verified
The area under the graph is \(\frac{1}{2}\).

Step by step solution

01

- Set Up the Integral

The area under the graph of the function from a point to infinity is found using an improper integral. Set up the integral from 2 to infinity: \[ \text{Area} = \int_{2}^{\infty} \frac{1}{x^2} \, dx \]
02

- Solve the Integral

Solve the integral. First, rewrite the integral: \[ \int \frac{1}{x^2} \, dx = \int x^{-2} \, dx \] The integral of \(x^{-2}\) is \( -x^{-1} \). So, we have: \[ \int x^{-2} \, dx = -x^{-1} + C \]
03

- Evaluate the Improper Integral

Evaluate the limit of the improper integral: \[ \text{Area} = \int_{2}^{\infty} \frac{1}{x^2} \, dx = \lim_{{b \rightarrow \infty}} \bigg[ - \frac{1}{x} \bigg]_{2}^{b} \]This means substituting 2 and \( b \) (and then taking the limit as \( b \) approaches infinity): \[ \lim_{{b \rightarrow \infty}} \bigg[ - \frac{1}{b} + \frac{1}{2} \bigg] \]
04

- Simplify the Result

Simplify the result: The term \( -\frac{1}{b} \) approaches 0 as \( b \) approaches infinity. So, we are left with: \[ \text{Area} = 0 + \frac{1}{2} = \frac{1}{2} \]

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Key Concepts

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

area under a curve
Finding the area under a curve is a common objective in calculus. This involves integrating the function that defines the curve over a specific interval. In our exercise, we are interested in the area under the curve of the function \(y = \frac{1}{x^2}\) for \(x \geq 2\). This type of problem often appears when the interval runs to infinity requiring special techniques known as improper integrals.
integration techniques
To solve the integral \[ \text{Area} = \int_{2}^{\infty} \frac{1}{x^2} \, dx \], we first rewrite the function for ease of integration. The function can be rewritten as \( \int x^{-2} \, dx \). Integrating \(x^{-2} \) is straightforward and results in \( -x^{-1} + C \), as the integral follows the standard \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \).
limit evaluation
Performing the integral alone isn't enough if one of its bounds is infinity. We need to evaluate the improper integral using limits. Thus, we set it up as \[ \text{Area} = \lim_{b \rightarrow \infty} \left[ -\frac{1}{x} \right]_{2}^{b} \]. This means substituting and evaluating at the bounds: \[ \lim_{b \rightarrow \infty} \left[ - \frac{1}{b} + \frac{1}{2} \right] \]. The term \( -\frac{1}{b} \) approaches 0 as \( b \) approaches infinity, leaving us with \[ \frac{1}{2} \]. Thus, the area under the curve is \[ \frac{1}{2} \].

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Most popular questions from this chapter

Consider \(\int_{0}^{2} f(x) d x\), where \(f(x)=\frac{1}{12} x^{4}+3 x^{2}\). (a) Make a rough sketch of the graph of \(f^{\prime \prime}(x)\) for \(0 \leq x \leq 2 .\) (b) Find a number \(A\) such that \(\left|f^{\prime \prime}(x)\right| \leq A\) for all \(x\) satisfying \(0 \leq x \leq 2\). (c) Obtain a bound on the error of using the midpoint rule with \(n=10\) to approximate the definite integral. (d) The exact value of the definite integral (to four decimal places) is \(8.5333\), and the midpoint rule with \(n=10\) gives \(8.5089 .\) What is the error for the midpoint approximation? Does this error satisfy the bound obtained in part (c)? (e) Redo part (c) with the number of intervals doubled to \(n-20\). Is the bound on the error halved? Quartered?

If \(k>0\), show that \(\int_{e}^{\infty} \frac{k}{x(\ln x)^{k+1}} d x=1 .\)

Determine if the given expression approaches \(\mathrm{g}\) limit as \(b \rightarrow \infty\), and find that number when it does. \(5\left(b^{2}+3\right)^{-1}\)

Evaluate the following integrals: \(\int x^{2} e^{-x} d x\)

Evaluate the following integrals using techniques studied thus far. \(\int \frac{\ln x}{x^{5}} d x\)
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