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Chapter 5: Problem 107

The following problems extend and augment the material presented in the text. 107\. dollars \((A

Short Answer

Expert verified
\( N = \frac{a}{-b+1} (B^{-b+1} - A^{-b+1}) \)

Step by step solution

01

Identify the Integral

The given integral is \( N = \int_{A}^{B} a x^{-b} \, dx \). Here, \( a \) and \( b \) are constants, and \( b eq 1 \).
02

Apply the Power Rule of Integration

Using the power rule for integration \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) (for \( n eq -1 \)), apply it to \( x^{-b} \). This gives: \[ \int x^{-b} \, dx = \frac{x^{-b+1}}{-b+1} + C \]
03

Incorporate Constant 'a' into Integration

Since the integral has a constant multiplier \( a \), factor it out of the integral: \[ N = a \int_{A}^{B} x^{-b} \, dx = a \left[ \frac{x^{-b+1}}{-b+1} \right]_{A}^{B} \]
04

Evaluate the Definite Integral

Calculate the definite integral from \( A \) to \( B \) using the antiderivative found in Step 3: \[ = a \left( \frac{B^{-b+1}}{-b+1} - \frac{A^{-b+1}}{-b+1} \right) \]
05

Simplify the Expression

Simplify the expression for \( N \): \[ N = \frac{a}{-b+1} \left( B^{-b+1} - A^{-b+1} \right) \] This is the required evaluation of the definite integral.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule of Integration
The power rule of integration is a fundamental concept in calculus that helps find antiderivatives of functions.
When you integrate a power of a variable, such as \( x^n \), the power rule is your tool. A simple version of the rule is:
  • If \( n eq -1 \), then \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration.
In the example from our exercise, \( x^{-b} \) is integrated using this rule:
  • For \( x^{-b} \), \( n = -b \), so \( \int x^{-b} \, dx = \frac{x^{-b+1}}{-b+1} + C \).
It is crucial that \( n eq -1 \) because otherwise, we'd use the natural logarithm. Hence, \( b eq 1 \) for \( x^{-b} \).The power rule makes integration straightforward as it essentially reverses the process of differentiation.
Constant Multiplier in Integrals
Handling constants in an integral is easily done by the rule of constant multipliers.
Suppose you have an integral of the form \( c \times f(x) \), where \( c \) is a constant. You can factor the constant out of the integral:
  • \( \int c f(x) \, dx = c \int f(x) \, dx \)
This rule simplifies calculations because you can focus on integrating \( f(x) \) first. In our exercise, the given function is \( a x^{-b} \). Here, \( a \) is a constant, allowing you to rewrite the integral as:
  • \( N = a \int x^{-b} \, dx \)
This separation makes it easier to apply the power rule solely on \( x^{-b} \). Once the antiderivative is found, multiply it by the constant \( a \) to obtain the result.
Evaluating Definite Integrals
Understanding how to evaluate definite integrals is essential for finding specific numerical results from functions
over an interval \( [A, B] \).The evaluation of definite integrals involves several steps:
  • First, find the antiderivative using the methods discussed, like the power rule.
  • Next, use the evaluation notation \( \left. F(x) \right|_A^B = F(B) - F(A) \), where \( F(x) \) is the antiderivative of the function.
  • Subtract the value of the antiderivative at \( A \) from its value at \( B \) to get the result.
For the integral in our exercise of form \( N = \int_{A}^{B} a x^{-b} \, dx \):
  • Calculate \( \left[ \frac{x^{-b+1}}{-b+1} \right]_A^B = \frac{B^{-b+1}}{-b+1} - \frac{A^{-b+1}}{-b+1} \).
  • Multiply by the constant \( a \) to complete the computation.
This simplifies into the formula: \( N = \frac{a}{-b+1} \left( B^{-b+1} - A^{-b+1} \right) \), providing a precise value over \([A, B]\).

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