Question

Using the integral of sec \(^{3} u\) By reduction formula 4 in Section 8.3 $$ \int \sec ^{3} u d u=\frac{1}{2}(\sec u \tan u+\ln |\sec u+\tan u|)+C $$ Graph the following functions and find the area under the curve on the given interval. $$f(x)=\sqrt{x^{2}-9} / x,[3,6]$$

Short answer

Answer: The area under the curve of the function \(f(x)=\frac{\sqrt{x^{2}-9}}{x}\) in the interval [3, 6] is $$A = 3\left[\sqrt{3} - \frac{\pi}{3}\right]$$.

Step-by-step solution

Graph the function

To graph the function, plot several points by plugging in various x values in the interval [3, 6]. After plotting the points, draw the curve that represents the function: \(f(x)=\frac{\sqrt{x^{2}-9}}{x}\).

Find the definite integral

We will now find the definite integral of the function in the given interval to find the area under the curve. This can be done by summing up the product of the function and the width of each interval. $$A = \int_3^6 f(x) dx = \int_3^6 \frac{\sqrt{x^2-9}}{x} dx$$

Simplify the integral by substitution

Let \(x=3\sec{u}\), then \(dx=3\sec{u}\tan{u}du\). When x=3, u=0 and when x=6, u=\(\frac{\pi}{3}\). The integral becomes: $$A = \int_0^{\frac{\pi}{3}} \frac{\sqrt{9\sec^2{u} - 9}}{3\sec{u}} \cdot 3\sec{u}\tan{u} du$$ $$A = \int_0^{\frac{\pi}{3}} 3\tan{u} \sqrt{\sec^2{u}-1} du$$

Use the identity\(\tan^2{u}+1=\sec^2{u}\)

Using the trigonometric identity \(\tan^2{u}+1=\sec^2{u}\), we get: $$A = \int_0^{\frac{\pi}{3}} 3\tan{u} \sqrt{\tan^2{u}} du$$ $$A = \int_0^{\frac{\pi}{3}} 3\tan^2{u} du$$ Here \(\tan^2{u}\) is non-negative, since we are on the valid interval.

Integrate

To integrate \(3\tan^2{u}\), first rewrite it as \(3(\sec^2{u} - 1)\): $$A = 3\int_0^{\frac{\pi}{3}} (\sec^2{u} - 1) du$$ Now, integrate term by term: $$A = 3\left[\int_0^{\frac{\pi}{3}} \sec^2{u} du - \int_0^{\frac{\pi}{3}} du\right]$$ To calculate the integral, we directly integrate \(\sec^2{u}\) and \(1\): $$A = 3\left[\tan{u} - u \Bigg|_0^{\frac{\pi}{3}}\right]$$

Evaluate the integral at the bounds

Evaluate the integral within the bounds of 0 and \(\frac{\pi}{3}\): $$A = 3\left[\tan{\frac{\pi}{3}} - \frac{\pi}{3} - (\tan{0} - 0)\right]$$

Simplify

Simplify the expression: $$A = 3\left[\sqrt{3} - \frac{\pi}{3}\right]$$ Now we have the area under the curve of the function \(f(x)=\frac{\sqrt{x^{2}-9}}{x}\) in the given interval [3, 6]: $$A = 3\left[\sqrt{3} - \frac{\pi}{3}\right]$$

Key concepts

Definite Integrals

Definite integrals are a core concept in integral calculus, representing the summation of an infinite number of infinitesimally small quantities. Essentially, it helps in finding the area under the curve for a given interval on a graph. A definite integral is denoted with limits of integration to signify the start and end points over which the integration is performed. In mathematical terms, it is written as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) represent the bounds.

• The process involves integrating the function \( f(x) \) over the interval \( [a, b] \).
• The result of a definite integral is a number that represents the net area between the curve of the function and the x-axis within the interval.

Definite integrals are crucial in providing insights into the behavior of functions and formulating quantitative characteristics, like total distance, total population, or total economic activity within bounds.

Substitution Method

The substitution method is a technique used to simplify the process of evaluating integrals. This method involves changing the variable of integration into a new one that simplifies the integral's evaluation process.
Consider the integral \( \int f(x) \, dx \); using substitution, you set \( x = g(u) \) and \( dx = g'(u) \, du \). This makes the integral:\[ \int f(g(u)) g'(u) \, du \].

• Identifying an appropriate substitution can make complex integrals more comprehensible and solvable.
• Especially useful when dealing with trigonometric integrals where applying trigonometric identities becomes necessary.

In our exercise, we used the substitution \( x = 3\sec{u} \), facilitating easier manipulation of trigonometric terms involved in the definite integral.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for any value of the included variables. These identities are particularly helpful in simplifying integrals involving trigonometric functions.
Some important trigonometric identities include:

• \( \tan^2{u} + 1 = \sec^2{u} \)
• \( \sin^2{u} + \cos^2{u} = 1 \)

These identities allow us to convert between different trigonometric functions. In our exercise, we utilized \( \tan^2{u} + 1 = \sec^2{u} \) to simplify the integral by converting \( \sec^2{u} - 1 \) back to \( \tan^2{u} \). This simplification makes the integration process more straightforward.

Area Under Curve

Finding the area under a curve is a common application of definite integrals. This involves computing the integral over a specified interval and often represents a physical quantity, like total displacement.
Using principles of integral calculus, the area \( A \) under a curve from \( x = a \) to \( x = b \) is found using \( A = \int_{a}^{b} f(x) \, dx \).

• The curve’s function is integrated over the interval from \( a \) to \( b \).
• The definite integral yields the net area between the curve and the x-axis.

In our exercise, the integral \( \int_3^6 \frac{\sqrt{x^2-9}}{x} \, dx \) represents the area under \( f(x) \) on the interval [3, 6]. This ensures an understanding of the extent or magnitude of the variable represented by the curve in that interval.