Problem 192 - Calculus Volume 1

Question

In the following exercises, use the evaluation theorem to express the integral as a function \(F(x)\) . $$ \int_{0}^{x} \cos t d t $$

Step-by-Step Solution

Answer

\( F(x) = \sin x \).

Step 1: Understanding the Problem

The problem asks us to evaluate the integral \( \int_{0}^{x} \cos t \ dt \) and express the result as a function \( F(x) \). This means we need to find an antiderivative, \( F(t) \), of \( \cos t \) and then apply the evaluation theorem.

Step 2: Finding the Antiderivative

The first step in finding the antiderivative of \( \cos t \) is knowing that the integral of \( \cos t \) with respect to \( t \) is \( \sin t + C \), where \( C \) is the constant of integration. So, \( F(t) = \sin t \).

Step 3: Applying the Evaluation Theorem

The evaluation theorem for definite integrals states that if \( F(t) \) is an antiderivative of \( f(t) \) on an interval \([a, b]\), then \( \int_{a}^{b} f(t) \, dt = F(b) - F(a) \). Here, \( a = 0 \) and \( b = x \).

Step 4: Calculating the Definite Integral

We calculate \( F(x) = \sin x \) and \( F(0) = \sin 0 = 0 \). Hence, the integral evaluates to \( F(x) - F(0) = \sin x - 0 = \sin x \).

Step 5: Expressing the Result as a Function

Finally, express the integral as the function \( F(x) \). Therefore, \( F(x) = \sin x \).

Key Concepts

Antiderivative

When calculating integrals, finding an antiderivative is a crucial step. An antiderivative, sometimes called the indefinite integral, reverses the differentiation process. For a function like \( \cos t \), the task is to determine the function whose derivative is \( \cos t \). In this case, the antiderivative of \( \cos t \) is \( \sin t \). This is because the derivative of \( \sin t \) is \( \cos t \). It is critical to remember that when finding an antiderivative, we add a constant, denoted by \( C \), because differentiation of a constant yields zero. Thus, the complete antiderivative is represented as \( \sin t + C \). However, when working with definite integrals, like in the original exercise, the constant does not affect the result of evaluation since it cancels out.

Evaluation Theorem

The Evaluation Theorem simplifies the computation of definite integrals. This theorem is an essential link between antiderivatives and definite integrals. It states that if \( F(t) \) is an antiderivative of \( f(t) \) on an interval \([a, b]\), then the integral from \( a \) to \( b \) of \( f(t) \, dt \) is given by \( F(b) - F(a) \). This means we can evaluate the integral by simply finding values of the antiderivative at the upper and lower limits and then subtracting.

• Compute \( F(b) \): Evaluate the antiderivative at the upper limit.
• Compute \( F(a) \): Evaluate the antiderivative at the lower limit.
• Subtract: Get the value of the definite integral as \( F(b) - F(a) \).

In the exercise, applying this theorem where \( a = 0 \) and \( b = x \) gives us \( \sin x - \sin 0 \), simplifying to \( \sin x \), since \( \sin 0 = 0 \).

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus interconnects the diverse worlds of derivatives and integrals. It consists of two main parts:

• The first part guarantees the existence of antiderivatives for continuous functions and defines a function as an integral. This is what we've done in the exercise by defining \( F(x) \) based on the integral from \( 0 \) to \( x \) of \( \cos t \, dt \).
• The second part, often called the Evaluation Theorem, is what allows us to evaluate definite integrals using antiderivatives.

Together, these concepts assure that if a function is continuous, it has an antiderivative which allows us to evaluate integrals easily. This theorem provides promise that any function's behavior, represented through an integral, can be effectively deduced if we understand its antiderivative, showcasing the powerful synergy between differentiation and integration.