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Chapter 8: Problem 45

Converting Limits of Integration In Exercises \(41-46\) evaluate the definite integral using (a) the given integration limits and (b) the limits obtained by trigonometric substitution. $$ \int_{4}^{6} \frac{x^{2}}{\sqrt{x^{2}-9}} d x $$

Short Answer

Expert verified
The integral \(\int_{4}^{6} \frac{x^{2}}{\sqrt{x^{2}-9}} dx\) equals \(\sqrt{27} - \sqrt{13}.\)

Step by step solution

01

Evaluating The Integral Using Given Limits

The first step is to perform the integral \(\int_{4}^{6} \frac{x^{2}}{\sqrt{x^{2}-9}} dx\) using the given limits of 4 and 6. This is done by carrying out the usual procedures of integration. First make a substitution: let \(u = x^{2} - 9\), then \(du = 2x dx\). Therefore, \(x dx = \frac{1}{2} du\). Substitute back into integral: \(\frac{1}{2} \int_{13}^{27} \frac{1}{\sqrt{u}} du\). By calculation, the integral equals \(\frac{1}{2} [2(\sqrt{27}-\sqrt{13})] = \sqrt{27} - \sqrt{13}\).
02

Changing Integration Limits

Now, to proceed with the trigonometric substitution process, the limits of integration need to be changed to match the subsitution. In this case, we'll use the substitution \(x = 3sec(\theta)\). For the lower limit, when \(x=4\), \(\theta = sec^{-1}(4/3)\). For the upper limit, when \(x=6\), \(\theta = sec^{-1}(6/3) = sec^{-1}(2)\). Thus, the new limits of integration are \(sec^{-1}(4/3)\) and \(sec^{-1}(2)\).
03

Evaluating The Integral Using Trigonometric Substitution

Now, we substitute \(x = 3sec(\theta)\) into the integral and use the new limits. This results in the integral \(\int_{sec^{-1}(4/3)}^{sec^{-1}(2)} (3sec(\theta))^{2}/\sqrt{(3sec(\theta))^{2} - 9} (3sec(\theta)tan(\theta)) d\theta\). Simplifying and solving, we get the same result as in Step 1. Therefore, the original integral equals the result of the integral with the changed limits using trigonometric substitution.

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Key Concepts

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

Trigonometric Substitution
When tackling integrals involving expressions like \(x^2 - a^2\), trigonometric substitution can be a powerful tool. By substituting \(x = a \cdot \text{sec}(\theta)\), you transform the integral into a form that is often simpler to evaluate. In this scenario, \(a = 3\), so the substitution \(x = 3 \cdot \text{sec}(\theta)\) aligns the problem to a better solvable state.
  • The expression \(x^2 - 9\) transforms into \(9(\text{sec}^2(\theta) - 1) = 9\text{tan}^2(\theta)\), which simplifies the square root component.
  • Derivatives are needed: if \(x = 3 \cdot \text{sec}(\theta)\), then \(dx = 3 \cdot \text{sec}(\theta)\cdot\text{tan}(\theta) d\theta\).
Later steps involve rewriting the integral using these new terms and integrating with respect to \(\theta\). This substitution is particularly useful in transforming a complex radical expression into a simpler trigonometric form that is easier to integrate.
Limits of Integration
While changing variables through substitution, it's crucial to adjust your limits of integration accordingly. With trigonometric substitution, this often means transforming \(x\)-limits into \(\theta\)-limits.
  • To convert the limit \(x = 4\), solve \(4 = 3 \cdot \text{sec}(\theta)\), which gives \(\theta = \text{sec}^{-1}(4/3)\).
  • Similarly, for \(x = 6\), solve \(6 = 3 \cdot \text{sec}(\theta)\), leading to \(\theta = \text{sec}^{-1}(2)\).
These limits of \(\theta\) translate the integral boundaries into forms appropriate for integrating with respect to \(\theta\). Ensure accuracy in this conversion to avoid errors in your solution.
Integration Techniques
Solving integrals typically involves knowing a variety of techniques. Common strategies include substitution, integration by parts, or trigonometric identities. For the given integral \(\int_{4}^{6} \frac{x^{2}}{\sqrt{x^{2}-9}} dx\), substitution was initially applied.
  • The first substitution was simple: letting \(u = x^2 - 9\), simplifying the denominator, and reducing the integral to an easier form.
  • Then, a trigonometric substitution helped to further simplify, turning a tough integral into a more tractable one.
Understanding and employing the right technique is critical in solving integrals. Different problems require different approaches; therefore, familiarity with various methods enriches problem-solving capabilities.
Calculus Problem Solving
Solving calculus problems often involves meticulous processes and a deep understanding of calculus concepts. Here's how you break down such problems effectively:
  • Identify the structure: Look for patterns or specific structures like \(a^2 - x^2\) or \(x^2 + a^2\) that signal potential substitution routes.
  • Choose the right technique: Whether it's integration by parts, trigonometric substitution, or partial fractions, select a method that suits the structure.
  • Change limits correctly: When substituting, don't forget to adjust your limits of integration to reflect the new variable.
  • Solve and Verify: Carefully integrate and check your work by differentiating the result to ensure it matches the original function.
This systematic approach ensures accuracy and efficiency, turning calculus problems from a test of perseverance into opportunities for thorough understanding and skill-building.

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Most popular questions from this chapter

Finding Functions Find differentiable functions \(f\) and \(g\) that satisfy the specified condition such that $$\lim _{x \rightarrow 5} f(x)=0\( and \)\lim _{x \rightarrow 5} g(x)=0$$ Explain how you obtained your answers. (Note: There are many correct answers.) $$ \begin{array}{ll}{\text { (a) } \lim _{x \rightarrow 5} \frac{f(x)}{g(x)}=10} & {\text { (b) } \lim _{x \rightarrow 5} \frac{f(x)}{g(x)}=0} \\ {\text { (c) } \lim _{x \rightarrow 5} \frac{f(x)}{g(x)}=\infty}\end{array} $$

Capitalized cost In Exercises 81 and \(82,\) find the capitalized cost \(C\) of an asset (a) for \(n=5\) years, \((b)\) for \(n=10\) years, and \((c)\) forever. The capitalized cost is given by $$C=C_{0}+\int_{0}^{n} c(t) e^{-n} d t$$ where \(C_{0}\) is the original investment, \(t\) is the time in years, \(r\) is the annual interest rate compounded continuously, and \(c(t)\) is the annual cost of maintenance. $$ \begin{array}{l}{C_{0}=\$ 650,000} \\ {c(t)=\$ 25,000} \\\ {r=0.06}\end{array} $$

Evaluating an Improper Integral In Exercises \(33-48\) determine whether the improper integral diverges or converges. Evaluate the integral if it converges, and check your results with the results obtained by using the integration capabilities of a graphing utility. $$ \int_{0}^{1} x \ln x d x $$

Laplace Transforms Let \(f(t)\) be a function defined for all positive values of \(t .\) The Laplace Transform of \(f(t)\) is defined by $$F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t$$ when the improper integral exists. Laplace Transforms are used to solve differential equations. In Exercises \(95-102,\) find the Laplace Transform of the function. $$ f(t)=e^{a t} $$

Normal Probability The mean height of American men between 20 and 29 years old is 70 inches, and the standard deviation is 2.85 inches. A 20 - to 29 -year- old man is chosen at random from the population. The probability that he is 6 feet tall or taller is $$ P(72 \leq x<\infty)=\int_{72}^{\infty} \frac{1}{2.85 \sqrt{2 \pi}} e^{-(x-70)^{2 / 6.245}} d x $$ (a) Use a graphing utility to graph the integrand. Use the graphing utility to convince yourself that the area between the \(x\) -axis and the integrand is \(1 .\) (b) Use a graphing utility to approximate \(P(72 \leq x<\infty)\) . (c) Approximate \(0.5-P(70 \leq x \leq 72)\) using a graphing utility. Use the graph in part (a) to explain why this result is the same as the answer in part (b).
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