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Chapter 5: Problem 60

Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{\sqrt{2}}^{2} \frac{d x}{x \sqrt{x^{2}-1}}$$

Short Answer

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Question: Evaluate the definite integral $$\int_{\sqrt{2}}^{2} \frac{d x}{x \sqrt{x^{2}-1}}$$. Answer: $$\int_{\sqrt{2}}^{2} \frac{d x}{x \sqrt{x^{2}-1}} =\ln\left|\frac{2+\sqrt{3}}{\sqrt{2}+1}\right|$$

Step by step solution

01

Recognize the Integral

The given integral is: $$\int_{\sqrt{2}}^{2} \frac{d x}{x \sqrt{x^{2}-1}}$$ We can recognize it as a standard integral. In particular, it is similar to the integral \(\int \frac{d x}{x \sqrt{x^{2}-a^{2}}}\), which has a standard result. The antiderivative for this form is given by: $$\frac{1}{a}\ln\left|\frac{x}{a}+\frac{\sqrt{x^2-a^2}}{a}\right|+C$$
02

Find the Antiderivative

Let's find the antiderivative of our given function. Using the standard result, we have: $$a = 1$$ So, the antiderivative of our given function is: $$\frac{1}{1}\ln\left|\frac{x}{1}+\frac{\sqrt{x^2-1}}{1}\right|+C=\ln\left|x+\sqrt{x^2-1}\right|+C$$
03

Apply the Fundamental Theorem of Calculus

Now, using the antiderivative, we can evaluate the definite integral using the Fundamental Theorem of Calculus: $$\int_{\sqrt{2}}^{2} \frac{d x}{x \sqrt{x^{2}-1}} =\left[\ln\left|x+\sqrt{x^2-1}\right|\right]^2_{\sqrt{2}}$$
04

Evaluate the Integral

Put the values of the bounds into the antiderivative: $$\begin{aligned} \left[\ln\left|x+\sqrt{x^2-1}\right|\right]^2_{\sqrt{2}} &= \ln\left|2+\sqrt{2^2-1}\right| - \ln\left|\sqrt{2}+\sqrt{(\sqrt{2})^2-1}\right| \\ &= \ln\left|2 + \sqrt{3}\right| - \ln\left|\sqrt{2} + 1\right| \\ \end{aligned}$$ Thus, the integral can be evaluated as: $$\int_{\sqrt{2}}^{2} \frac{d x}{x \sqrt{x^{2}-1}} =\ln\left|\frac{2+\sqrt{3}}{\sqrt{2}+1}\right|$$

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Key Concepts

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

Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a powerful tool that connects differentiation with integration, two core ideas in calculus. This theorem allows you to evaluate definite integrals using antiderivatives, which are simpler to handle compared to complex integral calculations directly.

The theorem comes in two parts:
  • First Part: It tells us that if a function is continuous on a closed interval [a, b], then the function has an antiderivative over that interval.
  • Second Part: This part is more relevant for definite integrals. It states that if you have an antiderivative \( F(x) \) of a function \( f(x) \), then the definite integral from \( a \) to \( b \) of \( f(x) \) is equal to \( F(b) - F(a) \).

This theorem makes evaluating integrals much easier, as it allows you to work with antiderivatives, as shown in the original exercise.
Antiderivatives
An antiderivative of a function is another function whose derivative is the original function. Finding antiderivatives is generally referred to as integration. It鈥檚 important in solving definite integrals.

For the given problem, we recognized the integral as a type involving \( \int \frac{dx}{x \sqrt{x^2 - 1}} \). The antiderivative for this form is known and can be written as:
\[ \ln\left|x+\sqrt{x^2-1}\right|+C \]Here, \( C \) is the constant of integration. This antiderivative is specific to this form of integral, making it easier to solve related problems.

Antiderivatives transform the problem into a more manageable form. Once you've identified the correct antiderivative, you move on to evaluate definite integrals using specific bounds.
Definite Integral Evaluation
Evaluating definite integrals involves finding the value over a specific interval. The process uses the antiderivative you've found and applies it to the upper and lower limits of the integral.

In the original exercise, after determining the antiderivative as \( \ln\left|x+\sqrt{x^2-1}\right| \), we apply the definite integral evaluation.

Steps to evaluate involve:
  • Substitute the upper limit into the antiderivative.
  • Substitute the lower limit into the antiderivative.
  • Subtract the two results.

Using the expression:
\[\left[\ln\left|x+\sqrt{x^2-1}\right|\right]^2_{\sqrt{2}} \]Substituting, we get:
\[ \ln\left|2 + \sqrt{3}\right| - \ln\left|\sqrt{2} + 1\right| \]
This results in the final evaluated integral:
\[ \ln\left|\frac{2+\sqrt{3}}{\sqrt{2}+1}\right| \]
Understanding these steps ensures you are accurately working through problems involving definite integrals.

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Most popular questions from this chapter

The population of a culture of bacteria has a growth rate given by \(p^{\prime}(t)=\frac{200}{(t+1)^{r}}\) bacteria per hour, for \(t \geq 0,\) where \(r > 1\) is a real number. In Chapter 6 it is shown that the increase in the population over the time interval \([0, t]\) is given by \(\int_{0}^{t} p^{\prime}(s) d s\). (Note that the growth rate decreases in time, reflecting competition for space and food.) a. Using the population model with \(r=2,\) what is the increase in the population over the time interval \(0 \leq t \leq 4 ?\) b. Using the population model with \(r=3,\) what is the increase in the population over the time interval \(0 \leq t \leq 6 ?\) c. Let \(\Delta P\) be the increase in the population over a fixed time interval \([0, T] .\) For fixed \(T,\) does \(\Delta P\) increase or decrease with the parameter \(r ?\) Explain. d. A lab technician measures an increase in the population of 350 bacteria over the 10 -hr period [0,10] . Estimate the value of \(r\) that best fits this data point. e. Looking ahead: Use the population model in part (b) to find the increase in population over the time interval \([0, T],\) for any \(T > 0 .\) If the culture is allowed to grow indefinitely \((T \rightarrow \infty)\) does the bacteria population increase without bound? Or does it approach a finite limit?

Continuity at the endpoints Assume \(f\) is continuous on \([a, b]\) and let \(A\) be the area function for \(f\) with left endpoint \(a\). Let \(m\) and \(M^{*}\) be the absolute minimum and maximum values of \(f\) on \([a, b],\) respectively. a. Prove that \(m^{*}(x-a) \leq A(x) \leq M^{*}(x-a),\) for all \(x\) in \([a, b] .\) Use this result and the Squeeze Theorem to show that \(A\) is continuous from the right at \(x=a\) b. Prove that \(m^{*}(b-x) \leq A(b)-A(x) \leq M^{*}(b-x),\) for all \(x\) in \([a, b] .\) Use this result to show that \(A\) is continuous from the left at \(x=b\)

An integral equation Use the Fundamental Theorem of Calculus, Part \(1,\) to find the function \(f\) that satisfies the equation $$ \int_{0}^{x} f(t) d t=2 \cos x+3 x-2 $$ Verify the result by substitution into the equation.

Definite integrals Use a change of variables or Table 5.6 to evaluate the following definite integrals. $$\int_{0}^{2} \frac{2 x}{\left(x^{2}+1\right)^{2}} d x$$

Consider the integral \(I=\int \sin ^{2} x \cos ^{2} x d x\) a. Find \(I\) using the identity \(\sin 2 x=2 \sin x \cos x\) b. Find \(I\) using the identity \(\cos ^{2} x=1-\sin ^{2} x\) c. Confirm that the results in parts (a) and (b) are consistent and compare the work involved in the two methods.
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