Problem 82 - University Calculus: Early Transcendentals

Question

Another proof of the Evaluation Theorem a. Let \(a=x_{0}

Step-by-Step Solution

Answer

\( F(b) - F(a) = \int_a^b f(x) \, dx \) by using the Mean Value Theorem and properties of Riemann sums.

Step 1: Understand the Problem

We need to use the properties of antiderivatives and the Mean Value Theorem to show that a sum involving an antiderivative over a partition corresponds to the value of the definite integral.

Step 2: Apply Fundamental Property of Antiderivatives

Recognize, for any antiderivative \( F \) of \( f \), the expression \( F(b) - F(a) \) can be written as a sum over partitions of \([a, b]\). Specifically, start by expressing \( F(b) - F(a) \) as the sum: \[ F(b) - F(a) = F(x_n) - F(x_0) = \sum_{i=1}^{n} [F(x_i) - F(x_{i-1})]. \] This makes use of the telescopic sum, where intermediate terms cancel out.

Step 3: Apply the Mean Value Theorem

For each interval \([x_{i-1}, x_i]\), apply the Mean Value Theorem: there exists a \(c_i\) such that \[ F(x_i) - F(x_{i-1}) = f(c_i)(x_i - x_{i-1}). \] Substituting this into the sum gives: \[ F(b) - F(a) = \sum_{i=1}^{n} f(c_i) (x_i - x_{i-1}), \] which is a Riemann sum for \( f \) on \([a, b]\).

Step 4: Connect with Definite Integral

Recall that the expression \( \sum_{i=1}^{n} f(c_i)(x_i - x_{i-1}) \) is a Riemann sum for the function \( f \) on the interval \([a, b]\). As the partition gets finer, the Riemann sum converges to the definite integral, thus: \[ \lim_{\max(x_i - x_{i-1}) \to 0} \sum_{i=1}^{n} f(c_i)(x_i - x_{i-1}) = \int_a^b f(x) \, dx. \] This shows that \( F(b) - F(a) = \int_{a}^{b} f(x) \, dx \), completing the proof.

Key Concepts

Mean Value Theorem

The Mean Value Theorem (MVT) is a fundamental concept in calculus. It states that for a continuous function on a closed interval and differentiable on the open interval, there exists a point where the function's instantaneous rate of change (derivative) equates to the average rate of change over that interval. This theorem is applied in calculus to find points in intervals where these conditions hold. In the context of the problem, MVT is used to connect discrete changes in antiderivatives with continuous functions. For each subinterval \([x_{i-1}, x_i]\), the theorem guarantees a point \(c_i\) with \(F(x_i) - F(x_{i-1}) = f(c_i)(x_i - x_{i-1})\). This step is crucial for transforming sum expressions into a Riemann sum, bridging discrete and continuous analysis.

Antiderivatives

Antiderivatives are functions which reverse the process of differentiation. If you differentiate an antiderivative of a function \(f\), you get back \(f\). This property is integral to solving problems involving integrals, like the one discussed. When we say \(F\) is an antiderivative of \(f\), we mean the derivative of \(F\) is \(f\). In the exercise, knowing that \(F(x)\) is an antiderivative allows us to express differences in evaluations of \(F\) at different points in terms of the original function \(f\). This setup helps connect the initial problem to integral calculus, highlighting how \(F(b) - F(a)\) can represent a sum that translates to an integral.

Riemann Sum

A Riemann sum is a method of approximating the total area under a curve, or the integral of a function. It works by dividing the area into rectangles, whose heights are defined by the value of the function at specific points, often \(c_i\), in each subinterval. In our example, the sum \(\sum_{i=1}^{n} f(c_i) (x_i - x_{i-1})\) is constructed as a Riemann sum. Each term represents the area of a rectangle, and as the partition gets finer, the sum approaches the exact area under \(f\) from \(a\) to \(b\). Thus, the Riemann sum provides a practical step toward calculating the definite integral, as capturing this concept is vital for understanding how sums and integrals are related.

Definite Integral

The definite integral of a function represents the total accumulation of quantities, such as area, over an interval \[a, b\]. It is written as \(\int_{a}^{b} f(x) \, dx\). This integral is closely linked with antiderivatives through the Fundamental Theorem of Calculus, which states that if \(F\) is an antiderivative of \(f\), then \(F(b) - F(a) = \int_{a}^{b} f(x) \, dx\). In this problem, showing that \(F(b) - F(a)\) equals a Riemann sum paves the way to the final conclusion using this theorem. The process involves substituting finer partitions into the sum, which refined results into an expression equivalent to the integral, thus solving the Evaluation Theorem.