Question

It can be shown by mathematical induction ( see Appendix 2 ) that $$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$$ Use this fact to give a formal proof that $$\int_{0}^{1} x^{2} d x=\frac{1}{3}$$ by following the steps given in the next column. (a) Partition \([0,1]\) into \(n\) subintervals of length 1\(/ n .\) Show that the RRAM Riemann sum for the integral is \(\sum_{k=1}^{n}\left(\left(\frac{k}{n}\right)^{2} \cdot \frac{1}{n}\right)\) (b) Show that this sum can be written as $$\frac{1}{n^{3}} \cdot \sum_{k=1}^{n} k^{2}$$ (c) Show that the sum can therefore be written as $$\frac{(n+1)(2 n+1)}{6 n^{2}}$$ (d) Show that \(\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\left(\frac{k}{n}\right)^{2} \cdot \frac{1}{n}\right)=\frac{1}{3}\) (e) Explain why the equation in part (d) proves that \(\int_{0}^{1} x^{2} d x=\frac{1}{3}\)

Short answer

The integral \( ∫_{0}^{1} x^{2} dx \) equals \(\frac{1}{3}\) as can be proven using the definition of Riemann sums and the property of \(\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}\), when \(n\) approaches infinity.

Step-by-step solution

Part a: Partition and Riemann sum

Firstly, divide the interval [0,1] into \(n\) equal parts. This will yield \(n\) subintervals of length \(1/n\). By the definition of the right Riemann sum, each term in the sum can be represented by \(f(x_i) * Δx_i\), where \(f(x_i)\) is the function at the right end-point of each subinterval and \(Δx_i\) is the width of the subinterval. Here, \(f(x_i) = x^2\), which would be \((\frac{k}{n})^2\), and \(Δx_i\) = \(\frac{1}{n}\), yielding \(\sum_{k=1}^{n} ((\frac{k}{n})^2 * \frac{1}{n})\).

Part b: Simplify the sum

Notice in part (a), all terms in the sum have \(\frac{1}{n^3}\) as the common factor. This means the sum can be written as \(\frac{1}{n^3} * \sum_{k=1}^{n} k^2\).

Part c: Substitute series sum

Then substitute the sum \(\sum_{k=1}^{n} k^2\) with \(\frac{n(n+1)(2n+1)}{6}\), according to the fact given in the question. This leads to \(\frac{1}{n^3} * \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6n^2}\).

Part d: Take the limit as n approaches infinity

To find the integral from 0 to 1 for \(x^2\), evaluate the limit as \(n\) tends to infinity for the sum \(\sum_{k=1}^{n} ((\frac{k}{n})^2 * \frac{1}{n}) = \frac{(n+1)(2n+1)}{6n^2}\). Simplifying, the limit is \(\frac{1}{3}\).

Part e: Explain the proof

As \(n\) approaches infinity, the right Riemann sum \(\sum_{k=1}^{n} ((\frac{k}{n})^2 * \frac{1}{n}) \) approximates the exact area under the curve \(x^2\) between 0 and 1. Therefore, since the limit is \(\frac{1}{3}\), the integral \( ∫_{0}^{1} x^{2} dx \) is equal to \(\frac{1}{3}\).

Key concepts

Mathematical Induction

Mathematical induction is a powerful proof technique used in mathematics to show that a statement is true for all natural numbers. It is structured in two main parts: the base case and the inductive step.

In the base case, you prove that the statement is true for the first natural number, often 1. In the inductive step, you assume the statement is true for an arbitrary natural number, say 'n', and then prove it must also be true for 'n+1'. If both these steps are satisfied, mathematical induction allows us to conclude that the statement holds for all natural numbers.

For example, in proving that the sum of the squares of the first 'n' natural numbers is given by the formula \[\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}\], we would first verify it for 'n=1' and then proceed to assume it holds for some 'n', and show that it must then hold for 'n+1'. This creates a sort of 'domino effect' ensuring the formula is valid for each natural number proceeding from our base case.

Definite Integral of x Squared

The definite integral of a function over a specified interval can be interpreted as the net area under the curve of that function between two points. For the function \(x^2\), we are particularly interested in finding the area under its curve from \(x=0\) to \(x=1\).

The analytical way to find this area is by calculating the definite integral, which is written as \[\int_{0}^{1} x^{2} dx\]. This integral represents the limit of the sums of areas of rectangles under the curve as the number of rectangles approaches infinity and their width approaches zero.

To obtain this integral, we can express the area as a sum of the areas of 'n' rectangles (a Riemann sum). As the number of rectangles increases, the approximation improves, which mathematically means taking the limit as 'n' goes to infinity, resulting in the integral of \(x^2\) over this interval being exactly \(\frac{1}{3}\).

Limits of Sequence

The concept of a limit of a sequence is fundamental in calculus, especially when dealing with infinite processes. Essentially, a limit asks what value a sequence approaches as its index grows without bound.

In the context of Riemann sums and integrals, the limit as the number of subintervals, 'n', approaches infinity, is essential. It represents the transition from a discrete sum of individual areas to the continuous total area under a curve. Mathematically, we write this idea as \[\lim_{n \rightarrow \infty} \sum_{k=1}^{n} a_k = L\] where \(a_k\) is the 'k-th' term in the sequence and 'L' is the value the sequence approaches.

This concept allows us to convert a finite sum of squared terms into an accurate representation of the integral of \(x^2\), by showing that as 'n' becomes very large, the sum approaches \(\frac{1}{3}\), which is precisely the area under the curve of \(x^2\) from 0 to 1.

Area Under a Curve

The area under a curve is a key concept in calculus and refers to the measurement of the region bounded by the graph of a function, the x-axis, and the vertical lines corresponding to the boundary of the interval.

Riemann sums, as used in the exercise, are a method to approximate this area by dividing the region into smaller rectangles and summing their areas. When the function is positive over the interval, the definite integral is exactly equal to the area under the curve.

For \(x^2\), the area under the curve from 0 to 1 is the same as the definite integral \[\int_{0}^{1} x^{2} dx\]. The exercise showed this using Riemann sums and taking a limit. The more rectangles we use (as 'n' approaches infinity), the closer the sum of the areas of these rectangles gets to the true area under the curve, concluding that \(\int_{0}^{1} x^{2} dx = \frac{1}{3}\) is indeed the exact area.