Problem 182 - Calculus Volume 1

Question

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2 . $$ \int_{0}^{2 \pi} \cos \theta d \theta $$

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Identify the Integral Format

The given integral to evaluate is \( \int_{0}^{2\pi} \cos \theta \, d\theta \). This is a definite integral with limits from \(0\) to \(2\pi\). We will use the Fundamental Theorem of Calculus, Part 2 to solve this.

Step 2: Find the Antiderivative

The Fundamental Theorem of Calculus, Part 2 tells us that to evaluate a definite integral \( \int_{a}^{b} f(x) \, dx \), we first find an antiderivative \( F(x) \) of \( f(x) \). For \( f(\theta) = \cos \theta \), an antiderivative is \( F(\theta) = \sin \theta \).

Step 3: Apply the Limits to the Antiderivative

Now, using the maximum and minimum limits of integration and the antiderivative \( F(\theta) = \sin \theta \), we evaluate \[ F(b) - F(a) = \sin(2\pi) - \sin(0) \].

Step 4: Calculate the Values

Since \( \sin(2\pi) = 0 \) and \( \sin(0) = 0 \), substituting into the expression, we get \( 0 - 0 = 0 \).

Step 5: Conclusion

The evaluated definite integral \( \int_{0}^{2\pi} \cos \theta \, d\theta \) is \( 0 \). This confirms that cosine over a full period of \(0\) to \(2\pi\) cancels out symmetrically, returning zero net area under the curve.

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a key principle that links together the concepts of differentiation and integration, two core branches of calculus. It comprises two parts, but Part 2 is especially relevant when dealing with definite integrals. This part tells us that if we have a continuous function on a closed interval, then the definite integral of that function can be found by taking an antiderivative of the function and evaluating it at the limits of integration.

For any function \(f(x)\), the theorem states:

• First, find an antiderivative \(F(x)\) of \(f(x)\) such that \(F'(x) = f(x)\).
• Then evaluate the definite integral \(\int_{a}^{b} f(x) \, dx\) as \(F(b) - F(a)\).

This theorem is immensely powerful because it allows us to turn a complex process, integration, into one involving evaluation of antiderivatives at particular points.

Antiderivative

An antiderivative of a function is another function whose derivative is the original function. Essentially, it reverses the process of differentiation.

When finding the definite integral using the Fundamental Theorem of Calculus, finding the antiderivative is a crucial step. Here's how to understand what an antiderivative is:

• For any continuous function \(f(x)\), an antiderivative \(F(x)\) satisfies \( F'(x) = f(x) \).
• Many functions can have multiple antiderivatives, differing by a constant. If \(F(x)\) is an antiderivative of \(f(x)\), then \(F(x) + C\) is also an antiderivative for any constant \(C\).

For our example, given \(f(\theta) = \cos \theta\), the antiderivative is \(F(\theta) = \sin \theta\). This is because the derivative of \(\sin \theta\) with respect to \(\theta\) is \(\cos \theta\).

Evaluation of Integrals

The evaluation of integrals is the process by which the numerical value of a definite integral is calculated. The Fundamental Theorem of Calculus provides a straightforward method to perform this by evaluating the antiderivative at the bounds of integration.

Here’s how you can think about it:

• First, identify the antiderivative \(F(x)\) of the function \(f(x)\) that is being integrated.
• Then, apply the evaluation: \(F(b) - F(a)\), where \(b\) and \(a\) are the upper and lower limits, respectively.
• This approach simplifies finding the area under a curve between two points or solving other integral problems where you need a definite solution.

In our specific problem, the integral \(\int_{0}^{2\pi} \cos \theta \, d\theta \) was evaluated as \(0\), demonstrating how cosine, when integrated over a full period, results in zero net area. This insight reflects how functions can be interpreted and analyzed over intervals using calculus.