Problem 74 - Calculus Early Transcendentals

Question

Using the integral of \(\sec ^{3} u\) By reduction formula 4 in Section 7.3 $$\int \sec ^{3} u d u=\frac{1}{2}(\sec u \tan u+\ln |\sec u+\tan u|)+C$$ Graph the following functions and find the area under the curve on the given interval. $$f(x)=\left(4+x^{2}\right)^{1 / 2},[0,2]$$

Step-by-Step Solution

Answer

Question: Find the area under the curve of the function $$f(x) = \left(4+x^{2}\right)^{1 / 2}$$ on the interval [0, 2]. Answer: The area under the curve of the function on the given interval is $$2\Bigg[\frac{1}{2}(\sec(\arctan(1)) \tan(\arctan(1)) +\ln |\sec(\arctan(1))+\tan(\arctan(1))|\Bigg]-2\Bigg[\frac{1}{2}(\sec(0) \tan(0) +\ln |\sec(0)+\tan(0)|)\Bigg]$$ after evaluating the definite integral and simplifying.

Step 1: Identify the function and interval

We are given a function f(x), and we need to find the area under the curve of this function on the interval [0, 2]: $$f(x)=\left(4+x^{2}\right)^{1 / 2}$$ Interval: [0, 2]

Step 2: Evaluate the definite integral

To find the area under the curve on the given interval, we need to evaluate the definite integral of f(x) with respect to x, over the interval [0, 2]: $$\int_{0}^{2}\left(4+x^{2}\right)^{1 / 2} dx$$ Before we proceed with calculus, we need to find a suitable substitution for this integral so we can use the integral formula of \(\sec^3 u\) provided. The hint is that the expression inside the square root is \((\sec u)^3\). We can begin by making a substitution, \(x^2 = 4\sec^2 u - 4\).

Step 3: Use substitution to change variables

With the expression \(x^2 = 4\sec^2 u - 4\), we need to find the proper substitution for 'x' and 'dx'. We have: $$x^2= 4\sec^2 u - 4 \implies x^2+4 = 4\sec^2 u$$ Therefore, $$x = 2\tan u$$ Differentiate both sides with respect to 'u': $$dx = 2\sec^2 u \, du$$

Step 4: Replace 'x' and 'dx' in the integral

Substituting 'x' and 'dx' in the original integral, we get: $$\int_{0}^{2}\left(4+x^{2}\right)^{1 / 2} dx = \int \left(4\sec^2 u\right)^{1/2}(2\sec^{2} u\, du) = \int 2\sec u \sec^{2} u\, du = \int 2\sec^{3} u\, du$$ Now, the limits of the integral have to be changed according to the substitution (u = arctan(x/2)): Lower limit: \(u_1 = \arctan(\frac{0}{2}) = 0\) Upper limit: \(u_2 = \arctan(\frac{2}{2}) = \arctan(1)\) So we have, $$\int_{0}^{\arctan(1)} 2\,\sec^{3}u\, du$$

Step 5: Solve the integral using the given formula

Use the given integral formula for the antiderivative of \(\sec^{3}u\): $$\int \sec ^{3} u d u=\frac{1}{2}(\sec u \tan u+\ln |\sec u+\tan u|)+C$$ Now, integrate: $$\int_{0}^{\arctan(1)} 2\,\sec^{3}u\, du = 2\Bigg[\frac{1}{2}(\sec u \tan u +\ln |\sec u+\tan u|)\Bigg]_{0}^{\arctan(1)}$$

Step 6: Evaluate the definite integral using the new limits and find the area

Now, evaluate the definite integral at the new limits \(u = \arctan(1)\) and \(u = 0\) to find the area under the curve: Area = $$2\Bigg[\frac{1}{2}(\sec(\arctan(1)) \tan(\arctan(1)) +\ln |\sec(\arctan(1))+\tan(\arctan(1))|\Bigg]-2\Bigg[\frac{1}{2}(\sec(0) \tan(0) +\ln |\sec(0)+\tan(0)|)\Bigg]$$ After evaluating the definite integral and simplifying, we will find the area under the curve of the function on the given interval.

Key Concepts

Definite Integral

The concept of a definite integral is crucial in calculus, as it helps in finding the exact area under a curve over a specified interval. This is done by calculating the accumulated total of small slices of area beneath the curve. The definite integral of a function \(f(x)\) from \(a\) to \(b\) is represented as \(\int_{a}^{b} f(x) \, dx\).
This symbol tells us to sum all the tiny areas of rectangles from \(a\) to \(b)\), with \(dx\) representing the infinitesimally small width of these rectangles.Definite integrals have clear practical applications:

• Calculating Areas: Used when finding the area underneath a curve between two points.
• Physics Applications: In calculating quantities like displacement, work, and electric charge.

In practice, to evaluate a definite integral, we often need to find an antiderivative of the given function and use the limits of integration to find the actual value. This evaluation gives the net area, considering any sections of the curve below the x-axis as negative area. This accounts for the true accumulation over the interval.

Reduction Formula

A reduction formula is a technique used in integral calculus to simplify the integration of complex functions. It systematically reduces an integral into a simpler form, often involving the original integral with altered or reduced parameters. This approach is particularly useful in recurring integral forms, such as those involving trigonometric and exponential functions.For example, the reduction formula for the integral of \(\sec^3 u\) in this context is listed as:\[\int \sec ^{3} u \, du = \frac{1}{2}(\sec u \tan u + \ln |\sec u + \tan u|) + C \]Reduction formulas are advantageous for:

• Simplification: Offering a more manageable form, making the integral easier to evaluate.
• Recursive Solutions: Allowing step-by-step simplification, often reducing computational overhead.

In mathematical problems, especially in calculus, employing a reduction formula can transform a challenging integral into a series of steps, each more straightforward than the last, until a solvable form is reached.

Trigonometric Substitution

Trigonometric substitution is a smart technique in integral calculus used to simplify the integration of functions involving radicals, typically those of the form \(\sqrt{a^2 + x^2}\), \(\sqrt{a^2 - x^2}\), or \(\sqrt{x^2 - a^2}\). By substituting a trigonometric function, the variable is transformed in a way that simplifies the radical expression, turning it into a trigonometric identity.Consider the original exercise where we handle the integral\(\int \sqrt{4+x^2} \, dx\). Here, trigonometric substitution helps by linking the radicals to trigonometric identities. Specifically:

• Choosing the Substitution: We set \(x = 2\tan u\) because \(\tan^2 u\) relates to the expression under the square root.
• Differential Transformation: Deriving \(dx = 2\sec^2 u \, du\) helps in changing the integral's variables.

This transforms the integral into one involving \(\sec u\), facilitating a resolution via the reduction formula. Trigonometric substitution reveals the power of relating algebraic expressions to trigonometric identities, simplifying complex integrals by utilizing known trigonometric functions and their derivatives.