Problem 64 - Calculus

Question

From Special Sum Formulas \(1-4\) you might guess that $$ 1^{m}+2^{m}+3^{m}+\cdots+n^{m}=\frac{n^{m+1}}{m+1}+C_{n} $$ where \(C_{n}\) is a polynomial in \(n\) of degree \(m .\) Assume that this is true (which it is) and, for \(a \geq 0\), let \(A_{a}^{b}\left(x^{m}\right)\) be the area under the curve \(y=x^{m}\) over the interval \([a, b]\). (a) Prove that \(A_{0}^{b}\left(x^{m}\right)=\frac{b^{m+1}}{(m+1)}\). (b) Show that \(A_{a}^{b}\left(x^{m}\right)=\frac{b^{m+1}}{m+1}-\frac{a^{m+1}}{m+1}\).

Step-by-Step Solution

Answer

(a) \( A_{0}^{b}(x^m) = \frac{b^{m+1}}{m+1} \). (b) \( A_{a}^{b}(x^m) = \frac{b^{m+1}}{m+1} - \frac{a^{m+1}}{m+1} \).

Step 1: Understand the Problem

The problem is asking us to prove two formulas for calculating the area under a curve for a polynomial function. We are given that the sum of powers formula is related to the area, and we need to use calculus to derive the required expressions.

Step 2: Recall the Definition of Area under a Curve

The area under the curve of a function \( y = x^m \) from \( a \) to \( b \) is given by the definite integral, \( \int_{a}^{b} x^m \, dx \). We will use this definition to find the area in both parts (a) and (b).

Step 3: Prove Part (a) Formula

To prove the formula \( A_{0}^{b}(x^m) = \frac{b^{m+1}}{m+1} \), we calculate the integral from \( 0 \) to \( b \): \[ \int_{0}^{b} x^m \, dx = \left[ \frac{x^{m+1}}{m+1} \right]_{0}^{b} = \frac{b^{m+1}}{m+1} - \frac{0^{m+1}}{m+1} = \frac{b^{m+1}}{m+1}. \] Thus, we have successfully derived the formula for the area from 0 to b.

Step 4: Prove Part (b) Formula

For part (b), we need to find \( A_{a}^{b}(x^m) \). This is done by calculating \( \int_{a}^{b} x^m \, dx \): \[ \int_{a}^{b} x^m \, dx = \left[ \frac{x^{m+1}}{m+1} \right]_{a}^{b} = \frac{b^{m+1}}{m+1} - \frac{a^{m+1}}{m+1}. \] This shows that the area under the curve from \( a \) to \( b \) is the difference between the area from 0 to \( b \) and the area from 0 to \( a \).

Key Concepts

Definite Integral

The concept of a definite integral is a cornerstone in calculus, primarily used to find the area under a curve. To understand this better, imagine you're trying to find the space beneath a curve of a function within a specific range on the x-axis.
Definite integrals help in calculating this area precisely. They are represented as:

• \( \int_{a}^{b} f(x) \, dx \)

where:

• \(a\) and \(b\) are the lower and upper limits, defining the interval.
• \(f(x)\) is the function whose area under the curve you're interested in.

The fundamental theorem of calculus links the concept of a definite integral to antiderivatives. It provides a way to evaluate the integral using the formula:

• \( \int_{a}^{b} f(x) \ dx = F(b) - F(a) \)

where \(F(x)\) is the antiderivative of \(f(x)\), which is a function such that \(F'(x) = f(x)\). This methodology makes it simple to calculate areas without manually summing up small segment areas.

Polynomial Function

A polynomial function is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. It looks like a smooth curve when plotted on a graph, and it can be represented as:

• \( f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \)

Here, \(a_n, a_{n-1}, ..., a_1, a_0\) are constant coefficients, and \(n\) is the highest power, called the degree of the polynomial.

Polynomial functions are notably straightforward to integrate or differentiate, which makes them pivotal in calculus. When working with definite integrals of polynomial functions, you use the same rules to find the antiderivative as with any other function type. For example, the integral of a polynomial function \(x^m\) is:

• \( \int x^m \ dx = \frac{x^{m+1}}{m+1} + C \)

where \(C\) is the constant of integration. This property is vital when calculating areas under polynomial curves using definite integrals.

Area Under a Curve

Finding the area under a curve, especially for polynomial functions, is a common application of definite integrals in calculus. This area represents the integral of the function along the interval between two points on the x-axis.
When given a function like \(y = x^m\), finding the area from \(a\) to \(b\) involves calculating:

• \( A_{a}^{b}(x^m) = \int_{a}^{b} x^m \, dx \)

This calculates the "net area," which considers spaces above the x-axis as positive and below as negative.

For example, if you need to find the area under the curve from \(0\) to \(b\), it simplifies to:

• \( \frac{b^{m+1}}{m+1} \)

If you consider a different starting point \(a\) instead, the area adjusts to remove the section from \(0\) to \(a\), calculated as:

• \( \frac{b^{m+1}}{m+1} - \frac{a^{m+1}}{m+1} \)

These expressions help in understanding the distribution and measure of areas bounded by polynomial curves on a given interval.