Problem 8 - Calculus Single Variable

Question

In each of Exercises \(1-12,\) calculate the average value of the given function on the given interval. $$ f(x)=x^{1 / 2}-x^{1 / 3} \quad I=[0,64] $$

Step-by-Step Solution

Answer

Integral calculation error, redo the integration steps.

Step 1: Understanding the Average Value Formula

The average value of a function \( f(x) \) on the interval \([a, b]\) is given by the formula:\[\text{Average value} = \frac{1}{b-a} \int_a^b f(x) \, dx\]where \( a = 0 \) and \( b = 64 \) for this problem.

Step 2: Set Up the Integral

We need to set up the integral of \( f(x) = x^{1/2} - x^{1/3} \) over the interval \([0, 64]\):\[\int_0^{64} \left( x^{1/2} - x^{1/3} \right) \, dx\]

Key Concepts

Definite Integral

A definite integral is a fundamental concept in calculus that calculates the accumulation of quantities, such as area under a curve. It's often symbolized as \( \int_a^b f(x) \, dx \), where \( f(x) \) is your function, and \( [a, b] \) is the interval over which integration occurs. In this context, the definite integral provides a method to find the overall accumulated value of a function between two points. To solve a definite integral, you'll convert the mathematical expression into its anti-derivative form (a related concept known as indefinite integration), evaluate it at the upper and lower bounds, and then subtract the results. This process effectively captures the total "area" or value described by the function within the specified domain of \( [a, b] \). When computing the average value of a function on an interval, the role of the definite integral is crucial because it measures the entire "height" or accumulation of the function across the given range.

Interval Notation

Interval notation is a way to represent a set of numbers along a number line. It uses brackets and parentheses to show which numbers are included in or excluded from a set. - If a number is included in the interval, a square bracket \([ \text{or} ]\) is used.- If a number is not included in the set, a parenthesis \(( \text{or} )\) is used. Understanding this helps in calculus when setting limits on integrals. For example, in our exercise, the interval \([0, 64]\) uses square brackets. This means both 0 and 64 are part of the interval where we're interested in evaluating the function. Such an interval indicates that all numbers from 0 to 64 including both endpoints are considered in our calculations, so they form the boundaries for our definite integral.

Function Evaluation

Function evaluation is a basic yet crucial skill in mathematics. It involves substituting a number into a function to get a result. If you have a function \( f(x) \), and you want to know its value at \( x = c \), you substitute \( c \) into the function for \( x \). For instance, if we want to evaluate \( f(x) = x^{1/2} - x^{1/3} \), at any point \( x = a \), you simply compute \( a^{1/2} - a^{1/3} \). When dealing with definite integrals, the fundamental theorem of calculus allows us to evaluate the function's accumulated effect between two points. Here, "function evaluation" during integration turns into "upper and lower bound substitution" after integration, translating anti-derivative results back into numerical values by plugging and substituting the bounds of the interval.

Integration Techniques

To solve integrals, especially when calculating the average value of functions, you may use various integration techniques. Each technique provides a unique approach suited to particular forms of functions or conditions. - **Basic Antiderivatives**: This involves reversing differentiation, finding a function whose derivative is the given function. For instance, \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) is a fundamental rule you'd use to integrate basic power functions.- **Substitution**: Useful when you can simplify a function by substituting part of \( f(x) \) with a new variable, making the integral easier to solve.- **Integration by Parts**: Relevant for integrals where direct antiderivatives are challenging, applying the formula \( \int u \, dv = uv - \int v \, du \).In the problem \( \int_0^{64} (x^{1/2} - x^{1/3}) \, dx \), basic antiderivatives are directly applied to integrate the terms \( x^{1/2} \) and \( x^{1/3} \) separately. Such techniques are essential in finding out general solutions that are evaluated over desired intervals to get definite values.