Problem 64 - Calculus Early Transcendentals: Pearson New International Edition

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 goal for Part (a)

For part (a), we have to prove that the area under the curve of \( y = x^m \) from \( x = 0 \) to \( x = b \) is given by the formula \( A_0^b(x^m) = \frac{b^{m+1}}{m+1} \). This is essentially finding the definite integral of \( x^m \) from \( 0 \) to \( b \).

Step 2: Set up the integral for Part (a)

The area under the curve \( y = x^m \) from \( x = 0 \) to \( x = b \) is calculated by the integral:\[ A_0^b(x^m) = \int_0^b x^m \, dx. \]

Step 3: Integrate for Part (a)

To perform the integration, apply the power rule for integrals: \[ \int x^m \, dx = \frac{x^{m+1}}{m+1} + C, \] where \( C \) is the constant of integration, which will be cancelled when taking the definite integral.

Step 4: Calculate the definite integral for Part (a)

Evaluate the definite integral: \[ \left[ \frac{x^{m+1}}{m+1} \right]_0^b = \frac{b^{m+1}}{m+1} - \frac{0^{m+1}}{m+1} = \frac{b^{m+1}}{m+1}. \]Thus, we have proved that \( A_0^b(x^m) = \frac{b^{m+1}}{m+1} \).

Step 5: Understand the goal for Part (b)

For part (b), show that the area under the curve from \( x = a \) to \( x = b \) is \( A_a^b(x^m) = \frac{b^{m+1}}{m+1} - \frac{a^{m+1}}{m+1} \). This also involves definite integrals.

Step 6: Set up the integral for Part (b)

The area under the curve \( y = x^m \) from \( x = a \) to \( x = b \) is calculated by the integral:\[ A_a^b(x^m) = \int_a^b x^m \, dx. \]

Step 7: Integrate for Part (b)

Just like in Step 3, apply the power rule: \[ \int x^m \, dx = \frac{x^{m+1}}{m+1} + C. \]

Step 8: Calculate the definite integral for Part (b)

Evaluate the definite integral:\[ \left[ \frac{x^{m+1}}{m+1} \right]_a^b = \frac{b^{m+1}}{m+1} - \frac{a^{m+1}}{m+1}. \]This shows that \( A_a^b(x^m) = \frac{b^{m+1}}{m+1} - \frac{a^{m+1}}{m+1} \).

Key Concepts

Definite Integral

A definite integral represents the area under the curve of a given function over a specific interval. In calculus, we use the definite integral to calculate this area precisely. Here, a definite integral transforms the summation of infinite small areas under the curve into a tangible value. For instance, to find the area under the curve of the function \( y = x^m \) from \( x = 0 \) to \( x = b \), we set up the integral:

• \( \int_0^b x^m \, dx \)

This integral shows us how the area is accumulated from \( x = 0 \) to \( x = b \). The outcome of this calculation is a numeric value representing the total area under the curve in this interval. This form of integration is practical in various fields, such as physics and engineering, where determining areas or quantities over an interval is essential.
In our context, it's important to find the definite integral to see exactly how much space is encapsulated beneath a curve, or what these partitions sum up to as a whole.

Power Rule

The power rule is a fundamental concept in calculus used for integrating and deriving power functions efficiently. It's a nifty shortcut that simplifies the process of integration by providing a formula:

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

Here, \( C \) is the constant of integration, representing any constant term you can add. When handling definite integrals, \( C \) does not affect the result since it gets eliminated when computing the bounds. The power rule significantly reduces the complexity involved in finding the integral of x raised to any power. This is particularly useful when calculating areas under polynomial graphs, such as the exercise presented, where solving involves raising \( x \) to the power \( m \), and then adding one to it.
Applying the power rule simplifies the integration process, so whether you're working on indefinite integrals or determining precise areas using definite integrals, it remains an invaluable tool.

Polynomial

A polynomial is a mathematical expression involving variables raised to whole number exponents and multiplied by coefficients. Polynomials can take various forms, depending on the number of terms and the degree of the expression. Typically, they look something like this:

• \( a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \)

The degree of a polynomial is dictated by the highest power of the variable present in the expression. In many calculus problems, especially involving integration like the given exercise, we often deal with polynomial functions because they are straightforward to integrate and model numerous real-world scenarios.
Because polynomials are smooth and continuous, integrating them allows us to calculate areas and solve differential equations efficiently. Furthermore, understanding the behavior of polynomials is essential because it is the cornerstone for approximating more complex functions using series and understanding fundamental calculus concepts. Thus, mastering polynomials and their integration is crucial for tackling a broad range of calculus problems.