Problem 48 - 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 $$\int_{1}^{5}(1-x) d x$$

Step-by-Step Solution

Answer

In order to evaluate the definite integral $\int_{1}^{5}(1-x)dx$ using the definition of the definite integral and right Riemann sums, we followed these steps: 1. Stated the definition of the definite integral. 2. Divided the interval [1, 5] into equally spaced subintervals. 3. Calculated the right Riemann sums for the function. 4. Evaluated the limit as the number of subintervals approached infinity using Theorem 5.1. 5. Derived the value of the integral, which was determined to be 12.

Step 1: Definition of the definite integral

According to the definition of the definite integral, the definite integral of a function f(x) on the interval [a, b] is given by: $$\int_{a}^{b} f(x) d x = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_{i})(x_{i} - x_{i-1})$$ where \(x_i - x_{i-1}\) is the width of each subinterval and \(x_i\) is the right endpoint of each subinterval we use to evaluate the function in computing the Riemann sums.

Step 2: Divide the interval [1, 5] evenly

Start by dividing the interval [1, 5] into n equally spaced subintervals. The width of each subinterval is given by: $$\Delta x = \frac{(b-a)}{n} = \frac{(5-1)}{n} = \frac{4}{n}$$ So each subinterval has width \(x_{i} - x_{i-1} = \Delta x = \frac{4}{n}\).

Step 3: Right Riemann sums for the function

For each subinterval, the right endpoint is \(x_i\). The height of the i-th rectangle is \(f(x_i) = 1 - x_i\). Thus, the area of the i-th rectangle is given by: $$f(x_i) \Delta x = (1 - x_i) \frac{4}{n}$$ The right Riemann sums for the entire interval [1, 5] is the sum of the areas of all rectangles: $$\sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} (1 - x_i) \frac{4}{n}$$

Step 4: Evaluate the limit as n approaches infinity

According to Theorem 5.1, the definite integral is the limit of the right Riemann sums as the number of subintervals (n) approaches infinity: $$\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x = \lim_{n \to \infty} \sum_{i=1}^{n} (1 - x_i) \frac{4}{n}$$ To compute this limit, rewrite \(x_i\) as \(x_1 + (i-1) \Delta x\): $$\lim_{n \to \infty} \sum_{i=1}^{n} \left(1 - (1 + (i-1) \frac{4}{n})\right) \frac{4}{n} = \lim_{n \to \infty} \sum_{i=1}^{n} (5 - \frac{4i}{n})\frac{4}{n}$$

Step 5: Determine the limit expression and the value of the integral

Now, we can compute the limit expression and the value of the integral: $$\int_{1}^{5}(1-x) d x = \lim_{n \to \infty} \sum_{i=1}^{n} (5 - \frac{4i}{n})\frac{4}{n} = \lim_{n \to \infty}( 20 - 4 \sum_{i=1}^n \frac{i}{n} \frac{4}{n})$$ To determine the limit, apply the properties of summation and the limit of sums rule: $$\int_{1}^{5}(1-x) d x = \lim_{n \to \infty}( 20 - 16 \frac{ \sum_{i=1}^n i}{n^2})$$ Now, we evaluate this limit by substituting the formula for the sum of the first n integers \(\sum_{i=1}^n i = \frac{n(n+1)}{2}\): $$\int_{1}^{5}(1-x) d x = \lim_{n \to \infty}( 20 - 16 \frac{n(n+1)/(2)}{n^2}) = 20 - 16 \lim_{n \to \infty} \frac{n^2 + n}{2n^2}$$ $$\int_{1}^{5}(1-x) d x = 20 - 16 \cdot \frac{1}{2} = 20 - 8 = 12$$ The value of the definite integral \(\int_{1}^{5}(1-x) d x\) is 12.

Key Concepts

Riemann Sums

Riemann sums are a powerful tool for approximating the area under a curve, which is essentially the basis of integration. When working with Riemann sums, you divide the area under a curve into small, manageable rectangles whose total area estimates the integral. In our problem, we used right Riemann sums with a function on the interval

• The interval is divided into \( n \) subintervals, and \( x_i \) is the right endpoint of each subinterval.
• The function's value at these right endpoints, \( f(x_i) \), forms the height of each rectangle.
• The width of each rectangle is \( \Delta x = \frac{b-a}{n} \).

As \( n \) becomes very large, the sum of the areas of these rectangles converges to the exact value of the definite integral. This process is fundamentally what we're doing with Riemann sums.

Theorem 5.1

Theorem 5.1 is a foundational concept in calculus that connects Riemann sums with definite integrals. It effectively states that as the number of intervals, \( n \), approaches infinity, the Riemann sum becomes equivalent to the definite integral. This theorem allows us to take

• The step-by-step process of summing rectangles from Riemann sums, applying limits, and determining an exact area representation under a curve.
• Providing a rigorous definition of the definite integral using limits.

By using this theorem, you can evaluate integrals by calculating the limit of the appropriate Riemann sum as \( n \to \infty \). It emphasizes the precision and applicability of calculus for solving area-related problems.

Limits

Limits are a fundamental concept in calculus, crucial for understanding both Riemann sums and definite integrals. The limit tells us how a function behaves as it approaches a particular value. In the context of definite integrals:

• The limit of the Riemann sum is taken as \( n \to \infty \).
• This process ensures that the approximation of the area under a curve becomes exact.
• It allows us to bypass direct calculations of infinitesimally small rectangles.

When dealing with our definite integral, transitioning our sum to the limit is what allows us to transform an approximation into precise mathematical truth. Mastering limits is key to mastering calculus.

Integration Techniques

Integration techniques are varied and help simplify or directly solve definite integrals. Using the definition of the integral, like in our exercise, is one direct approach. Here are some techniques:

• Substitution: Used when a change of variable can simplify the integral.
• Integration by Parts: Useful for products of functions.
• Partial Fractions: Decomposes fractions into simpler parts to integrate each part more easily.
• Defining with Riemann Sums and Theorem 5.1: Applicable for precise calculations from first principles, as executed here.

The choice of technique depends on the form of the function and the context of the problem, demonstrating the flexibility and depth of integration in calculus.