Problem 73 - Calculus of a Single Variable

Question

Limits and Integrals In Exercises 73 and 74 , evaluate the limit and sketch the graph of the region whose area is represented by the limit. $$ \lim _{ \| \rightarrow 0} \sum_{i=1}^{n}\left(x_{i}-x_{i}^{2}\right) \Delta x, \text { where } x_{i}=\frac{i}{n} \text { and } \Delta x=\frac{1}{n} $$

Step-by-Step Solution

Answer

The limit, which represents the area under the curve of the function \(x - x^{2}\) from 0 to 1, evaluates to \(1/6\).

Step 1: Recognize the Limit as Integral

The limit of a Riemann sum can be recognized as a definite integral. This limit is given by: \[ \lim _{ \|\Delta x\| \rightarrow 0} \sum_{i=1}^{n} f(x_i) \Delta x \] which is equivalent to \[ \int_{a}^{b} f(x) dx \] where \(a\) and \(b\) are the lower and upper limits of the function respectively.

Step 2: Identify the Function

The function under this integral is identified as \(f(x) = x - x^2 \). This is derived from \(x_{i} - x_{i}^{2}\) where \(x_i = i/n\).

Step 3: Identify Limits of Integration

The limits of integration,\( a \) and \( b \), correspond to the smallest and largest values of \( x_i \). Given \(x_i = i/n\) where \( i = 1, 2,..., n \), we can see that the smallest value of \(x_i\) is when \( i = 1 \), giving \(x_i = 1/n\), and goes up to when \( i = n \), giving \(x_i =1\). So \( a = 0 \) and \( b = 1 \).

Step 4: Compute the Integral

Now we evaluate the integral: \[ \int_{0}^{1} (x - x^2) dx \] It is firstly necessary to find the antiderivative of \(x - x^2\), which is \(\frac{x^2}{2} - \frac{x^3}{3}\). Using the Fundamental Theorem of Calculus, subtract the antiderivative at the lower limit from the antiderivative at the upper limit. So, it computes to \([(\frac{1^2}{2} - \frac{1^3}{3})] - [(\frac{0^2}{2} - \frac{0^3}{3})] = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}\)

Step 5: Sketch the Graph

We're asked to sketch the graph of the region whose area is represented by the limit (i.e., the integral). We graph \(y = x - x^2\) on the domain \([0, 1]\). It is a downward opening parabola with roots at \(x = 0\) and \(x = 1\). The area beneath the curve and above the x-axis between these two points is represented by the definite integral.

Key Concepts

Definite Integral

In calculus, a definite integral is a mathematical tool used to calculate the area under a curve on a graph, usually between two specified points known as the limits of integration. This involves evaluating the integral of a function over an interval, which leads to determining the net area. For the function in our problem, the integral takes the form of \[ \int_{0}^{1} (x - x^2) \, dx \]This results in finding the area beneath the curve of the function \(x - x^2\) as it stretches from \(x = 0\) to \(x = 1\). By understanding how this integral operates, we can solve complex real-world problems such as calculating distances or the quantity of material needed for a shape.

Limits of Integration

The concept of limits of integration is integral to solving definite integrals as it determines the range over which we evaluate the integral. In the exercise problem, we recognize the limits of integration as \(a = 0\) and \(b = 1\). These values define the interval

• Where the calculation starts at \(x = 0\)
• Ends at \(x = 1\), encompassing the entire relevant domain of our function \(f(x) = x - x^2\).

Given the Riemann sum framework, these limits correspond to the endpoints of our subdivisions along the x-axis. The accuracy of our integral depends heavily on the precision of these limits – starting and stopping at the right points ensures accurate area calculation.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus serves as a bridge between the processes of differentiation and integration. It's a fundamental insight that connects the

• concept of finding an antiderivative of a function
• actual process of evaluating a definite integral to find the area under the curve.

This theorem states that if \(F\) is an antiderivative of \(f\) over the interval \([a, b]\), then:\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a)\]For our exercise, the theorem allows us to efficiently evaluate:\[ \int_{0}^{1} (x - x^2) \, dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1\]Computation using this theorem provides the difference between the antiderivative values at the upper and lower limits, yielding \(\frac{1}{6}\). It simplifies finding the exact area under the curve.

Antiderivative

To solve a definite integral, an antiderivative needs to be determined from the given function. An antiderivative is essentially a reverse operation of differentiation. When differentiating a function, we find its rate of change. However, an antiderivative works backwards to find the original function that was differentiated.

• For example, the antiderivative of \(x\) is \(\frac{x^2}{2}\)
• For \(x^2\), it becomes \(\frac{x^3}{3}\).

So for the function \(f(x) = x - x^2\), its antiderivative is computed as:\[ \frac{x^2}{2} - \frac{x^3}{3}\]Finding this expression is crucial as it allows us to apply the Fundamental Theorem of Calculus to acquire the final result. The ease of this calculation portrays the importance of mastering antiderivatives in calculus.