Problem 43 - University Calculus: Early Transcendentals

Question

Find a formula for the Riemann sum obtained by dividing the interval \([a, b]\) into \(n\) equal subintervals and using the right-hand endpoint for each \(c_{k} .\) Then take a limit of these sums as \(n \rightarrow \infty\) to calculate the area under the curve over \([a, b]\). \(f(x)=x+x^{2}\) over the interval [0,1]

Step-by-Step Solution

Answer

The area under the curve is \(\frac{5}{6}\).

Step 1: Divide the Interval into n Equal Subintervals

The interval \([a, b]\) is \([0, 1]\). We divide this interval into \(n\) equal subintervals. Each subinterval will have a width of \(\Delta x = \frac{b-a}{n} = \frac{1-0}{n} = \frac{1}{n}\).

Step 2: Define the Right-Hand Endpoint

For each subinterval \([x_{k-1}, x_k]\), \(x_k = a + k\Delta x = k \cdot \frac{1}{n}\) where \(k = 1, 2, ..., n\).

Step 3: Express the Riemann Sum

The Riemann sum using the right-hand endpoints is given by \(S_n = \sum_{k=1}^{n} f(x_k) \Delta x\). Here, \(x_k = \frac{k}{n}\) and \(f(x) = x + x^2\), hence \(f(x_k) = \frac{k}{n} + \left( \frac{k}{n} \right)^2\). Thus, the Riemann sum becomes \(S_n = \sum_{k=1}^{n} \left( \frac{k}{n} + \left( \frac{k}{n} \right)^2 \right) \cdot \frac{1}{n}\).

Step 4: Simplify the Riemann Sum

Simplify \(S_n = \frac{1}{n^2} \sum_{k=1}^{n} k + \frac{1}{n^3} \sum_{k=1}^{n} k^2\). Use the formulas for the sums: \(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\) and \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\).

Step 5: Calculate the Sums

Substitute these formulas into the Riemann sum: \[ S_n = \frac{1}{n^2} \cdot \frac{n(n+1)}{2} + \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} \]. This simplifies to \[ S_n = \frac{n+1}{2n} + \frac{(n+1)(2n+1)}{6n^2} \].

Step 6: Take the Limit as n Approaches Infinity

Take the limit \( \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( \frac{n+1}{2n} + \frac{(n+1)(2n+1)}{6n^2} \right) \). Simplifying each term, it becomes \( \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \).

Key Concepts

Definite Integral

The concept of a definite integral is essentially the idea of calculating the total area under a curve over a specific interval. For a function \( f(x) \) over an interval \([a, b]\), the definite integral represents the accumulation of quantities, such as area, along the curve from point \(a\) to point \(b\).
In mathematical terms, the definite integral of \( f(x) \) from \( a \) to \( b \) is denoted as \( \int_{a}^{b} f(x) \, dx \). This notation represents the limit of the sum of areas of rectangles under the curve as the width of these rectangles becomes infinitely small. Calculating definite integrals is a fundamental concept in calculus, leading to an understanding of various natural and mathematical phenomena.

Limit Process

The limit process is crucial when working with integrals, particularly with Riemann sums. To find the exact area under a curve, we must consider the limit of a Riemann sum as the number of subintervals \( n \) approaches infinity.
The more subintervals we use, the more accurate our approximation until it converges to the exact value of the integral. This concept can be beautifully visualized by imagining the gradual sharpening of a blurry image; with each step, the image becomes clearer, resolving into its exact form.

• Starting with dividing the interval into \( n \) parts.
• Taking the sum of areas of rectangles under the curve.
• Letting \( n \) approach infinity to make the rectangles infinitely thin.

This infinite increase in the number of rectangles helps in getting the precise value of the integral.

Polynomial Functions

Polynomial functions are expressions consisting of variables and coefficients, combined using operations of addition, subtraction, multiplication, and non-negative integer exponents. An example of a polynomial function is \( f(x) = x + x^2 \), which represents a simple quadratic polynomial.
Working with polynomials in calculus often involves implementing Riemann sums and integrals to find areas under their curves. These functions are fundamental as they frequently appear in mathematical modeling.

• Easy to differentiate and integrate.
• Widely applicable in various calculus problems.

The exercise demonstrates this by calculating the area under the polynomial \( x + x^2 \) over the interval \([0, 1]\).

Subintervals

To approximate the area under a curve using Riemann sums, the given interval \([a, b]\) is divided into smaller parts called subintervals. The more subintervals used, the more accurate the Riemann sum will be in approximating the true area.
For example, dividing the interval \([0, 1]\) into \( n \) subintervals results in each having width \( \Delta x = \frac{1}{n}\). The endpoint of each subinterval is used to compute the height of the rectangle representing the area under the curve. Typically, the right-hand endpoint is preferred for simplicity in calculations. This step-by-step method builds up mathematical intuition about how limits and functions behave.