Question

Generalize the proof of Exercise 3.1 .4 to show directly from Definition 3.1.1 that $$ \int_{a}^{b} x^{m} d x=\frac{b^{m+1}-a^{m+1}}{m+1} $$ if \(m\) is an integer \(\geq 0\).

Short answer

Question: Prove that for an integer \(m \geq 0\), the integral of the function \(x^m\) over the interval \([a, b]\) is equal to \(\frac{b^{m+1}-a^{m+1}}{m+1}\). Answer: We showed that the integral of \(x^m\) over the interval \([a, b]\) can be represented as a Riemann sum. After applying the Binomial Theorem, distributing the factor \(\Delta x\), and exchanging the order of summation, we used Faulhaber's formula to evaluate the summation over \(k\). Finally, we took the limit as \(n\) approached infinity and found that only the term with \(i=0\) remains in the limit, thus proving that the integral of \(x^m\) over the interval \([a, b]\) is equal to \(\frac{b^{m+1}-a^{m+1}}{m+1}\).

Step-by-step solution

Define the integral using Riemann sums using Definition 3.1.1

According to Definition 3.1.1, we can approximate the integral of \(x^m\) over the interval \([a, b]\) using Riemann sums. The integral will be: $$ \int_{a}^{b} x^{m} d x=\lim_{n \rightarrow \infty} \sum_{k=1}^{n} f(x_k) \Delta x $$ where \(f(x_k)=x_k^m\), \(\Delta x = \frac{b-a}{n}\), and \(x_k = a + k \Delta x\). Plugging in the given function and the value of \(x_k\), we get: $$ \int_{a}^{b} x^{m} d x=\lim_{n \rightarrow \infty} \sum_{k=1}^{n} (a + k \Delta x)^{m} \Delta x $$

Apply the Binomial Theorem to simplify the sum

In order to simplify the sum, we can apply the Binomial Theorem to \((a + k \Delta x)^{m}\). This will give us: $$ \sum_{k=1}^{n} (a + k \Delta x)^{m} \Delta x = \sum_{k=1}^{n} \left(\sum_{i=0}^{m} \binom{m}{i} a^{m-i}(k \Delta x)^{i}\right) \Delta x $$

Distribute the factor \(\Delta x\) and exchange the order of summation

Next, we distribute the factor \(\Delta x\) into the inner summation and change the order of summation: $$ \sum_{k=1}^{n} \sum_{i=0}^{m} \binom{m}{i} a^{m-i} k^{i}(\Delta x)^{i+1} = \sum_{i=0}^{m} \binom{m}{i} a^{m-i} (\Delta x)^{i+1} \sum_{k=1}^{n} k^{i} $$

Evaluate the summation over \(k\) and simplify

Now, we have a formula for the sum of the first \(n\) \(i^{th}\) powers of integers using Faulhaber's formula: $$ \sum_{k=1}^{n} k^{i}=P_i(n+1)-P_i(1) $$ where \(P_i(x)\) is a polynomial of degree \(i+1\). Replacing the summation over \(k\) with the Faulhaber's formula, we get: $$ \int_{a}^{b} x^{m} d x = \lim_{n \rightarrow \infty} \sum_{i=0}^{m} \binom{m}{i} a^{m-i} (\Delta x)^{i+1} \left[P_i(n+1)-P_i(1)\right] $$

Evaluate the limit

Finally, we take the limit as \(n \rightarrow \infty\): $$ \int_{a}^{b} x^{m} d x = \lim_{n \rightarrow \infty} \sum_{i=0}^{m} \binom{m}{i} a^{m-i} \left(\frac{b-a}{n}\right)^{i+1} \left[P_i(n+1)-P_i(1)\right] $$ Notice that for \(i>0\), the term \(\left(\frac{b-a}{n}\right)^{i+1}\) goes to zero. Therefore, only the term with \(i=0\) remains in the limit, giving us the desired result: $$ \int_{a}^{b} x^{m} d x = \frac{b^{m+1}-a^{m+1}}{m+1} $$ This completes the proof that directly shows the integral of \(x^m\) over the interval \([a, b]\) is equal to \(\frac{b^{m+1}-a^{m+1}}{m+1}\) for integer \(m\geq0\).

Key concepts

Integral Calculus

Integral calculus is a foundational component of mathematics, particularly important in the field of calculus. It focuses on the accumulation of quantities, such as areas under curves, and the inverse process of differentiation. A key concept within integral calculus is the definite integral, which is used to compute the accumulation of a function over a specific interval.

The definite integral of a function from point a to b is represented using the integral sign with the appropriate limits of integration, a and b. As shown in the textbook exercise, one can compute the integral of x raised to a power m within the interval [a, b] using the formula
\[\int_{a}^{b} x^{m} d x = \frac{b^{m+1}-a^{m+1}}{m+1}\]
where m is a non-negative integer. Understanding how to derive this formula involves the use of Riemann sums, which are used to approximate the area under a curve by dividing it into smaller sections, or 'slices'. As the number of slices increases indefinitely, the approximation becomes an exact calculation in the limit, thus yielding the integral of the function.

Binomial Theorem

The binomial theorem is a quick method for expanding binomials raised to any given power without direct multiplication. It states that
\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]
where n is a non-negative integer, and \(\binom{n}{k}\) is the binomial coefficient, also known as 'n choose k'. These coefficients can be found in Pascal's Triangle or calculated using the formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).

In the context of the textbook exercise, the binomial theorem is used to expand the term (a + kΔx)^m in the Riemann sum. This expansion simplifies the expression and allows for the terms to be summed up in a way that we can analyze their behavior as n approaches infinity.

Faulhaber's Formula

Faulhaber's formula involves determining the sum of powers of consecutive integers, such as the sum of the cubes or squares of the first n natural numbers. It is named after the mathematician Johann Faulhaber and typically expressed using polynomial functions called Bernoulli polynomials.

According to Faulhaber's formula, the sum of the ith powers can be calculated by
\[\sum_{k=1}^{n} k^i = P_i(n+1) - P_i(1)\]
where P_i(x) is a polynomial of degree i+1. When applying Faulhaber's formula in integral calculus, especially in the evaluation of Riemann sums, it provides an explicit expression for the summation seen in Riemann sums, which then simplifies to the integral formula upon taking the limit. This approach is used in the textbook solution to illustrate the relationship between a function and its integral, emphasizing the intuitive aspect of summing smaller quantities to find the whole.