Question

Limits of sums Use the definition of the definite integral to evaluate the following definite integrals. Use right Riemann sums and Theorem 5.1 $$\int_{1}^{5}(1-x) d x$$

Short answer

Question: Evaluate the definite integral $$\int_{1}^{5}(1-x) d x$$ using right Riemann sums and Theorem 5.1. Answer: -8

Step-by-step solution

step 1: Partition the interval

Divide the interval [1, 5] into n equal subintervals, where the width of each subinterval is given by: $$\Delta x = \frac{b-a}{n} = \frac{5-1}{n} = \frac{4}{n}$$

step 2: Find the function value at the right endpoint

For each right endpoint (denoted by x_i), evaluate the function (1-x): $$f_i = 1 - x_i$$

step 3: Calculate the Riemann sum

Using right Riemann sums, the sum will be: $$R_n = \sum_{i=1}^{n} f_i * \Delta x = \sum_{i=1}^{n} (1-x_i)\left(\frac{4}{n}\right) $$ We know that $$x_i = a +i*\Delta x = 1 + i\left(\frac{4}{n}\right) $$ Therefore, $$R_n = \frac{4}{n}\sum_{i=1}^{n}(1-(1+i\left(\frac{4}{n}\right)))$$

step 4: Find the limit of the sum

Theorem 5.1 states that the definite integral is equal to the limit of the Riemann sums as the number of partitions approaches infinity: $$\int_{1}^{5}(1-x) d x = \lim_{n \to \infty}R_n = \lim_{n \to \infty}\frac{4}{n}\sum_{i=1}^{n}(1-(1+i\left(\frac{4}{n}\right)))$$ By simplifying and calculating the limit using standard rules for algebraic manipulation and calculus, we arrive at the definite integral value: $$\int_{1}^{5}(1-x) d x = -8$$

Key concepts

Riemann Sums

Understanding Riemann sums is essential to grasp the concept of definite integrals. Riemann sums are a technique used to approximate the area under a curve—the integral—by dividing the area into small rectangles or trapezoids. To form a Riemann sum, first, we divide the interval of interest into smaller subintervals. Then, we calculate the area of each subinterval by multiplying the width of the subinterval by the value of the function at a chosen point within the subinterval, which can be the left endpoint, right endpoint, or the midpoint.

The Riemann sum is then the sum of the areas of all these rectangles, which can be expressed as:
\[ R_n = \text{sum of all rectangles} = \sum_{i=1}^{n} f(x_i) \cdot \Delta x \] where \( f(x_i) \) represents the function's value at a specific point on the subinterval, and \( \Delta x \) is the width of each subinterval. If we increase the number of subintervals \(n\), the rectangles become thinner, and the sum becomes a better approximation of the area under the curve.

Limits of Sums

The limit of sums plays a pivotal role in defining the definite integral. It involves taking the Riemann sum and increasing the number of subintervals to infinity while simultaneously decreasing their width to zero. Mathematically, this means finding the limit of the Riemann sum as \( n \) approaches infinity:
\[ \lim_{n \to \infty} R_n \] This process ensures that we accurately measure the entire area under the curve without overestimating or underestimating by too much. The more subintervals we use, the closer our sum gets to the actual value of the definite integral. It is this limit of Riemann sums that equates to the exact value of the definite integral, providing a bridge between summation and integration.

Definite Integral Evaluation

Evaluating definite integrals is a central operation in calculus. A definite integral gives the net area under a curve for a given interval on the x-axis. It is represented as \[ \int_{a}^{b} f(x) dx \] where \(a\) and \(b\) are the lower and upper bounds of the integral, respectively. To evaluate this integral, one can use analytic techniques involving antiderivatives, or by approximating the value using Riemann sums and then finding the limit as described earlier. In the context of our exercise, the function \( f(x) = 1 - x \) is integrated over the interval from 1 to 5. By applying the techniques of Riemann sums and limits, one arrives at a specific numerical value, representing the exact net area under the curve between these two bounds.

Theorem 5.1 in Calculus

Theorem 5.1, sometimes referred to as the Fundamental Theorem of Calculus, links the process of differentiation with integration. It essentially states that if \( f \) is a continuous function on the interval \([a, b]\), and \( F \) is an antiderivative of \( f \), then the definite integral of \( f \) from \( a \) to \( b \) is given by evaluating \( F \) at the bounds and subtracting:
\[ \int_{a}^{b} f(x) dx = F(b) - F(a) \] In the case of evaluating limits of sums, it allows us to conclude that the definite integral of a function across an interval can be found by taking the limit of the Riemann sum as \( n \) approaches infinity. This theorem provides the theoretical foundation for evaluating definite integrals using the methods of limits and Riemann sums, as demonstrated in the workout of our textbook problem.