Problem 57 - Calculus: Early Transcendentals

Question

What is wrong with the equation? \( \displaystyle \int^{\pi}_{\pi/3} \sec \theta \tan \theta \, d\theta = \sec \theta \Bigg]^{\pi}_{\pi/3} = -3 \)

Step-by-Step Solution

Answer

The integral evaluation is correct: \(-3\). There's no actual error.

Step 1: Understand the Integration

Recognize that the integral given is \( \int \sec \theta \tan \theta \, d\theta \). The fundamental theorem of calculus tells us that we need to find an antiderivative of \( \sec \theta \tan \theta \), which is \( \sec \theta \).

Step 2: Determine the Antiderivative

The antiderivative of \( \sec \theta \tan \theta \) is indeed \( \sec \theta \). This comes from recognizing that the derivative of \( \sec \theta \) is \( \sec \theta \tan \theta \). Thus, \( F(\theta) = \sec \theta \) is the antiderivative.

Step 3: Apply the Fundamental Theorem of Calculus

Using the antiderivative \( F(\theta) = \sec \theta \), apply the Fundamental Theorem of Calculus: \( \int_{\pi/3}^{\pi} \sec \theta \tan \theta \, d\theta = \sec \theta \bigg|_{\pi/3}^{\pi} = \sec(\pi) - \sec(\pi/3) \).

Step 4: Calculate \( \sec(\pi) \) and \( \sec(\pi/3) \)

Calculate the values: \( \sec(\pi) = -1 \) because \( \sec \theta = \frac{1}{\cos \theta} \) and \( \cos(\pi) = -1 \). Also, \( \sec(\pi/3) = \frac{1}{\cos(\pi/3)} = \frac{1}{1/2} = 2 \).

Step 5: Evaluate the Definite Integral

Now substitute the calculated values: \( \sec(\pi) - \sec(\pi/3) = -1 - 2 = -3 \).

Step 6: Identify the Mistake

There is no mistake in the evaluated definite integral calculation; the result \( \int_{\pi/3}^{\pi} \sec \theta \tan \theta \, d\theta = -3 \) is correct. The error in the original statement might imply questioning rewriting of steps or conceptual misunderstanding, which isn't present here.

Key Concepts

Definite Integrals

Definite integrals are a fundamental part of calculus, helpful in calculating the accumulation of quantities, such as area under a curve, over a specified interval. When addressing definite integrals, we typically denote them as \( \int_{a}^{b} f(x) \, dx \), with\( a \) and \( b \) representing the lower and upper limits of integration, respectively.

Important aspects of definite integrals include:

• Boundaries: Definite integrals have fixed limits. These limits specify the range over which we are calculating the accumulation.
• Net Area: The result of a definite integral often represents the net area between the curve of a function and the x-axis within the indicated limits. Areas above the x-axis are considered positive, while those below are negative.
• relationship with Antiderivatives: The definite integral of a function can be solved by finding an antiderivative of that function, as demonstrated in our example problem.

Considering our exercise, the definite integral \( \int_{\pi/3}^{\pi} \sec \theta \tan \theta \, d\theta \) evaluates the accumulation of function values from \( \pi/3 \) to \( \pi \). Calculating this involves applying the fundamental theorem of calculus as well.

Antiderivative

An antiderivative, also known as an indefinite integral, is a function whose derivative is equal to the original function given. In simple terms, it's the reverse of differentiation.

When tasked with finding an antiderivative, we remember:

• Not Unique: Antiderivatives aren't unique because of the constant of integration \( C \), reflecting that multiple functions can have the same derivative.
• Recognition: To find an antiderivative, recognize the derivative rules and apply them backward. For instance, if the derivative of \( \sec \theta \) is \( \sec \theta \tan \theta \), then \( \sec \theta \) must be an antiderivative for \( \sec \theta \tan \theta \).
• Checking: Ensure accuracy by differentiating the proposed antiderivative to see if it yields the original function.

In our exercise, the key was identifying the antiderivative of \( \sec \theta \tan \theta \) as \( \sec \theta \). This recognition allowed us to move forward with applying the fundamental theorem of calculus.

Fundamental Theorem of Calculus

The fundamental theorem of calculus links differentiation and integration, showing that they are, in a sense, inverse processes. This powerful theorem has two main parts, but for definite integrals, we focus on the second part.

Key points include:

• Connection: It establishes that if \( F \) is an antiderivative of \( f \) on an interval \([a, b]\), then the definite integral of \( f \) from \( a \) to \( b \) is given by \( F(b) - F(a) \).
• Application: To solve a definite integral, find an antiderivative \( F \) for the function \( f \), then evaluate \( F \) at the upper limit and subtract the evaluation at the lower limit.
• Computation: This theorem makes computing definite integrals straightforward and less tedious than other methods.

In our exercise, the fundamental theorem of calculus allowed us to evaluate \( \int_{\pi/3}^{\pi} \sec \theta \tan \theta \, d\theta \) by finding the antiderivative \( F(\theta) = \sec \theta \), then computing \( \sec(\pi) - \sec(\pi/3) \). This difference provides the definite integral's value.