Problem 69 - Essential Calculus

Question

In Exercises 69 and 70 , evaluate the limit and sketch the graph of the region whose area is represented by the limit. \(\lim _{\|\Delta\| 0} \sum_{i=1}^{n}\left(x_{i}-x_{i}^{2}\right) \Delta x,\) where \(x_{i}=i / n\) and \(\Delta x=1 / n\)

Step-by-Step Solution

Answer

The limit value is \(\frac{1}{6}\). The sketch of the graph is a downward-opening parabola with roots at 0 and 1. The area under the graph between these points corresponds to the limit value steps.

Step 1: Interpreting the limit

The expression \(\lim _{\|\Delta\| 0} \sum_{i=1}^{n}\left(x_{i}-x_{i}^{2}\right) \Delta x,\) represents the limit of Riemann sums that approach the definite integral of the function f(x) = x - x^2 as the partition of the interval becomes finer and finer. In this case, the interval is [0,1] because the values of \(x_i = i/n\) range between 0 and 1 for \(i = 1, ..., n\). Likewise, \(\Delta x = 1/n\) defines the width of each partition in the limit.

Step 2: Evaluating the limit

The limit of the Riemann sums is equal to the definite integral of the function x-x^2 over the interval [0,1]. For the limit, we can write \(\int _0 ^1 (x-x^2) dx\). Calculating this integral requires us to find the antiderivative of the function x - x^2, apply the Fundamental theorem of Calculus, and subtract the values at the endpoints of the interval. We get \[\frac{1}{2}x^2 - \frac{1}{3}x^3 \Biggr\rvert_0 ^1 = \frac{1}{2}*1^2 - \frac{1}{3}*1^3 - (\frac{1}{2}*0^2 - \frac{1}{3}*0^3) = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}\]

Step 3: Sketching the graph

Now, we sketch the graph of f(x) = x - x^2. This is a quadratic function with roots at x = 0 and x = 1. The vertex is at x = 0.5, which is the midpoint of the roots. The graph opens downwards because the coefficient of \(x^2\) is negative. The area under the graph between x = 0 and x = 1 represents the value of the integral we have computed.

Key Concepts

Definite Integral

The concept of a definite integral is essential in calculus, as it quantifies the accumulation of quantities, such as areas under curves. When we see a limit of Riemann sums like \[\lim _{\|\Delta\| 0} \sum_{i=1}^{n}\left(x_{i}-x_{i}^{2}\right) \Delta x,\]it represents the area beneath a curve from one point to another on the x-axis. In the original exercise, we're integrating the function \(f(x) = x - x^2\) over the interval \([0, 1]\).

The definite integral of a function \( f(x) \) across an interval \([a, b]\) is written as \( \int_a^b f(x) \, dx \). This is the main tool we use to calculate the exact accumulation over a region. By using all points within the interval to form small rectangles under the curve, the Riemann sum estimation becomes more precise as the number of partitions increases. Eventually, this process yields the exact value of the integral, which is what we achieved in the example with the limit process.

Quadratic Function

Quadratic functions are polynomials of degree 2, with a general form \( ax^2 + bx + c \). In our exercise, the function \( f(x) = x - x^2 \) is a specific quadratic function where \( a = -1 \), \( b = 1 \), and \( c = 0 \).

Key characteristics of quadratic functions include:

• Roots: The points where the function intersects the x-axis. Here, the roots are \(x = 0\) and \(x = 1\).
• Vertex: The highest or lowest point on the graph. In \( f(x) = x - x^2 \), the vertex is at \( x = 0.5 \), which is the midpoint of these roots.
• Opening Direction: Since the coefficient \( a \) is negative, the parabola opens downwards.

Quadratic functions are integral to many calculus problems because their parabola shape can model real-world phenomena like projectile motion and optimization issues.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation with integration, providing a method to evaluate definite integrals. It states that if \( F \) is an antiderivative of \( f \) on an interval \([a, b]\), then:

\[\int_a^b f(x) \, dx = F(b) - F(a)\]

In the original solution, we used this theorem to evaluate the integral of \( x - x^2 \) over \([0, 1]\). First, we found the antiderivative of \( x - x^2 \) to be \( F(x) = \frac{1}{2}x^2 - \frac{1}{3}x^3 \).
Then, using the Fundamental Theorem, we calculated:

• At \( x = 1 \): \( \frac{1}{2}(1)^2 - \frac{1}{3}(1)^3 = \frac{1}{2} - \frac{1}{3} \)
• At \( x = 0 \): \( \frac{1}{2}(0)^2 - \frac{1}{3}(0)^3 = 0 \)
• Result: Thus, the definite integral is \( \frac{1}{6} \).

This theorem not only demonstrates the inverse relationship between differentiation and integration but also simplifies the process of evaluating definite integrals by reducing it to function evaluation at boundary points.