Problem 87 - Precalculus

Question

In calculus, the definite integral \(\int_{a}^{b} f(x) d x\) is used to find the area below the graph of a continuous function \(f\), above the \(x\) -axis, and between \(x=a\) and \(x=b\). The Fundamental Theorem of Calculus cstablishes that the definite integral \(f_{a}^{b} f(x) d x\) equals \(F(b)-F(a),\) where \(F\) is any antiderivative of a continuous function \(f.\) In Exercises \(87-90\), first shade the area corresponding to the definite integral and then use the information below to find the exact value of the area. $$\begin{array}{|l|c|c|c|c|} \hline \text { Function } & \tan x & \cot x & \sec x & \csc x \\ \hline \text { Antiderivative } & -\ln |\cos x| & \ln |\sin x| & \ln |\sec x+\tan x| & -\ln |\csc x+\cot x| \\ \hline \end{array}$$ $$\int_{0}^{\pi / 4} \tan x d x$$

Step-by-Step Solution

Answer

The area is \( \frac{1}{2} \ln(2) \).

Step 1: Identify the Definite Integral

The exercise asks us to evaluate the definite integral \( \int_{0}^{\pi/4} \tan x \, dx \). This integral represents the area under the curve of the tangent function, from \( x = 0 \) to \( x = \frac{\pi}{4} \), and above the \( x\)-axis.

Step 2: Find the Antiderivative

According to the information provided, the antiderivative of \( \tan x \) is \( -\ln |\cos x| \). This means that the indefinite integral of \( \tan x \) with respect to \( x \) is \( -\ln |\cos x| + C \), where \( C \) is the constant of integration.

Step 3: Apply the Fundamental Theorem of Calculus

To find the definite integral \( \int_{0}^{\pi/4} \tan x \, dx \), we use the Fundamental Theorem of Calculus, which states that \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). Here, \( F(x) = -\ln |\cos x| \).

Step 4: Calculate \( F(b) - F(a) \)

Substitute \( x = \frac{\pi}{4} \) into the antiderivative to get \( F(\frac{\pi}{4}) = -\ln |\cos(\frac{\pi}{4})| \). Similarly, substitute \( x = 0 \) to get \( F(0) = -\ln |\cos(0)| \). Now compute the result: 1. \( F(\frac{\pi}{4}) = -\ln |\frac{1}{\sqrt{2}}| = \ln(\sqrt{2}) \) 2. \( F(0) = -\ln |1| = 0 \). Combine these results to find the value of the definite integral: \( \ln(\sqrt{2}) - 0 \).

Step 5: Simplify the Expression

The expression \( \ln(\sqrt{2}) \) can be simplified using logarithmic identities. Since \( \sqrt{2} = 2^{1/2} \), we have: \[ \ln(\sqrt{2}) = \ln(2^{1/2}) = \frac{1}{2} \ln(2) \]Thus, the definite integral evaluates to \( \frac{1}{2} \ln(2) \).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges two central concepts in calculus: differentiation and integration. It is essentially the connection between finding the rate of change (derivatives) and finding the total amount (integrals). This theorem states that if a function is continuous over an interval, then the integral over that interval of the function's derivative is equal to the change of its antiderivative.
This is expressed as:

• Let \( f \) be a continuous function on the interval \([a, b]\).
• Let \( F \) be an antiderivative of \( f \), meaning \( F'(x) = f(x) \).
• Then the definite integral from \( a \) to \( b \) of \( f(x) \) is \( F(b) - F(a) \).

In practice, the theorem allows us to compute the definite integral by evaluating the antiderivative at the endpoints of the interval.
This powerful tool simplifies the process of finding the area under a curve, making it possible to avoid calculating complex limits.

Antiderivative

The antiderivative, also known as the indefinite integral, is a crucial concept in calculus. An antiderivative of a function \( f(x) \) is another function, commonly denoted as \( F(x) \), whose derivative is equal to \( f(x) \). Put simply, if you differentiate \( F(x) \), you get \( f(x) \).
Finding the antiderivative is like reversing the process of differentiation:

• If \( F'(x) = f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \).
• A function can have multiple antiderivatives, differing by a constant \( C \), because the derivative of a constant is zero.

The role of antiderivatives in integrals is central. They enable us to evaluate definite integrals efficiently using the Fundamental Theorem of Calculus, bypassing the need for complicated summations or geometric interpretations.
In the example provided in the exercise, \( -\ln |\cos x| \) is the antiderivative of \( \tan x \). By evaluating this antiderivative at different points, we can find the area under the tangent curve.

Area Under a Curve

The area under a curve is a geometric concept applied in calculus to find the 'total' quantity represented by a variable rate, like distance traveled over time. When we are interested in finding the area between a function \( f(x) \), the x-axis, and the vertical lines \( x = a \) and \( x = b \), we use the definite integral \( \int_{a}^{b} f(x) \, dx \).
This calculation gives us the net area, considering that areas below the x-axis are counted as negative:

• The definite integral is essentially computing the accumulation of infinitesimally small rectangles under the curve.
• Each rectangle’s height is determined by \( f(x) \), and its width is an infinitesimally small change in \( x \), denoted \( dx \).
• This process sums up all the areas of such rectangles to give the total area under the curve.

Using the Fundamental Theorem of Calculus and antiderivatives as illustrated, we can compute the exact area efficiently without relying solely on numerical approximation methods.
In our specific example, using the antiderivative \( -\ln |\cos x| \), the net area between \( x=0 \) and \( x=\pi/4 \) for the tangent curve is found to be \( \frac{1}{2} \ln(2) \). This represents an accurate measurement of space enclosed by the curve above the x-axis for the specified interval.