Problem 51 - Calculus Early Transcendentals

Question

Use the definition of the definite integral to evaluate the following definite integrals. Use right Riemann sums and Theorem 5.1 $$\int_{1}^{4}\left(x^{2}-1\right) d x$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral $$\int_{1}^{4}\left(x^{2}-1\right) d x$$ using right Riemann sums and Theorem 5.1. Answer: The definite integral evaluates to 9.

Step 1: Identify the function and the interval

The required function to integrate is \(f(x) = x^{2} -1\) and the interval [a,b] is [1,4].

Step 2: Calculate Delta x

To calculate \(\Delta x\), use the formula \(\Delta x = \frac{b-a}{n}\), where a=1 and b=4. Therefore, \(\Delta x = \frac{4-1}{n} = \frac{3}{n}\).

Step 3: Use Theorem 5.1 to set up the integral

Theorem 5.1 states: $$\int_{a}^{b}f(x) dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(a+i\Delta x)\Delta x$$ Apply it to our function \(f(x) = x^2 -1\) and the interval [1,4]: $$\int_{1}^{4}\left(x^{2}-1\right) d x = \lim_{n \to \infty} \sum_{i=1}^{n} f(1+i\Delta x)\Delta x$$

Step 4: Replace f(x) and Delta x

Substitute \(f(x) = x^2 -1\) and \(\Delta x = \frac{3}{n}\) in the equation: $$\int_{1}^{4}\left(x^{2}-1\right) d x = \lim_{n \to \infty} \sum_{i=1}^{n} [(1 + i \frac{3}{n})^2 - 1](\frac{3}{n})$$

Step 5: Simplify and evaluate the limit

Simplify and then evaluate the limit as n approaches infinity: $$\int_{1}^{4}\left(x^{2}-1\right) d x = \lim_{n \to \infty} \sum_{i=1}^{n} \left[\left(\frac{3i}{n}\right)^2 + 2\left(\frac{3i}{n}\right) \right] \left(\frac{3}{n}\right) $$ This sum leads to: $$\int_{1}^{4}\left(x^{2}-1\right) d x = \lim_{n \to \infty} \left[\frac{9}{n^3}\sum_{i=1}^{n} i^2 + \frac{6}{n^2}\sum_{i=1}^{n} i\right]$$ As n goes to infinity, we know that \(\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\) and \(\sum_{i=1}^{n} i = \frac{n(n+1)}{2}\), so $$\int_{1}^{4}\left(x^{2}-1\right) d x = \lim_{n \to \infty} \left[\frac{9}{n^3}\cdot\frac{n(n+1)(2n+1)}{6} + \frac{6}{n^2}\cdot\frac{n(n+1)}{2}\right]$$ Now, take the limit as n approaches infinity. The definite integral evaluates to: $$\int_{1}^{4}\left(x^{2}-1\right) d x = 9$$

Key Concepts

Riemann Sum

A Riemann Sum is a method for approximating the total area under a curve on a given interval by dividing it into small rectangles. We calculate these areas' sum to approximate the integral. Each rectangle's height is determined by the function's value at a certain point within the subinterval.

• A Riemann Sum is expressed as \(\sum_{i=1}^{n} f(x_i) \Delta x\), where \(\Delta x\) is the width of each subinterval.
• The values \(x_i\) can represent different sampling points, such as left, right, or midpoints of subintervals.
• In our example, we use right Riemann sums which means the height of each rectangle is determined by the function value at the right end of each subinterval.

By using more and more partitions, making \(\Delta x\) smaller, the Riemann Sum becomes a better estimate of the definite integral.

Limit of a Sum

The Limit of a Sum is a crucial concept in defining the Definite Integral in calculus. As we increase the number of subintervals \(n\) to infinity, the width \(\Delta x\) of each subinterval approaches zero. This process turns the Riemann Sum into an exact value.

• The limit is symbolically represented as \(\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x\).
• This expression indicates that as the number of rectangles increases, the approximation of the area under the curve becomes more accurate.
• In our example, calculating the limit allows the Riemann sum to transform into the precise value of the definite integral.

By using the limit, we ensure the approximation becomes the exact integral \(\int_{a}^{b} f(x) \; dx\). This approach is the cornerstone for transitioning from discrete sums to continuous integrals.

Theorem 5.1

Theorem 5.1 is a fundamental theorem that helps us move from the practical calculation of sums to the analytical evaluation of integrals. It provides a rigorous foundation by showing how definite integrals are connected to limits of Riemann Sums.

• The theorem states \(\int_{a}^{b}f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(a+i\Delta x)\Delta x\).
• This formula combines the process of taking the limit of a Sum with the definition of definite integrals.
• The use of this theorem helps convert multiplication of sums into integral forms, aligning with the concept of infinity and continuity in calculus.

Understanding Theorem 5.1 allows mathematicians and students to move seamlessly from theoretical approximations to exact integral evaluations.

Calculus

Calculus is a branch of mathematics that focuses on the concepts of change and motion. It provides tools for analyzing quantities that vary continuously as opposed to algebra, which deals with fixed quantities.

• Calculus is divided into Differential Calculus and Integral Calculus.
• The definite integral, as explored in this exercise, is a significant concept in Integral Calculus, involving the calculation of areas, volumes, and other quantities.
• Fundamental Theorems of Calculus, like Theorem 5.1, link the concept of differentiation and integration.

With calculus, we analyze and understand dynamic and changeable aspects of the world, forming the basis for advanced topics and problems in engineering, physics, economics, and many other fields.