Problem 4 - Calculus: Early Transcendentals

Question

Find the average value of the function on the given interval. \( g(t) = \dfrac{t}{\sqrt{3 + t^2}} \) , \( [1, 3] \)

Step-by-Step Solution

Answer

The average value is \( 2\sqrt{3} - 2 \).

Step 1: Understanding the Average Value Formula

The average value of a function \( g(t) \) over a closed interval \([a, b]\) is given by the formula: \[ \text{Average} = \frac{1}{b-a} \int_a^b g(t) \, dt \].Here, \( a = 1 \) and \( b = 3 \). So, our task is to compute:\[ \frac{1}{3-1} \int_1^3 \frac{t}{\sqrt{3 + t^2}} \, dt \].

Step 2: Setting up the Integral

Substitute the function \( g(t) \) into the integral from the formula:\[ \int_1^3 \frac{t}{\sqrt{3 + t^2}} \, dt \]. This integral now represents the area under the curve \( \frac{t}{\sqrt{3 + t^2}} \) from \( t = 1 \) to \( t = 3 \).

Step 3: Choosing a Substitution

To evaluate the integral, use the substitution method. Let \( u = 3 + t^2 \). Then, differentiate to find \( du = 2t \, dt \) or equivalently, \( \frac{1}{2}du = t \, dt \).

Step 4: Changing Limits of Integration

We need to change the limits of integration to correspond with \( u = 3 + t^2 \). Initially:- When \( t = 1 \), \( u = 3 + 1^2 = 4 \).- When \( t = 3 \), \( u = 3 + 3^2 = 12 \).So, the new limits for \( u \) are from 4 to 12.

Step 5: Computing the Integral

Rewrite the integral in terms of \( u \):\[ \int_4^{12} \frac{1}{\sqrt{u}} \cdot \frac{1}{2} du \].This simplifies to \( \frac{1}{2} \int_4^{12} u^{-1/2} \, du \).The antiderivative of \( u^{-1/2} \) is \( 2u^{1/2} \), so the integral becomes:\[ \frac{1}{2} \times 2 \left[ u^{1/2} \right]_4^{12} = \left[ u^{1/2} \right]_4^{12} \].Evaluate this to obtain:\[ \sqrt{12} - \sqrt{4} = 2\sqrt{3} - 2 \].

Step 6: Finding the Average Value

Plug the result of the integral back into the average value formula:\[ \text{Average} = \frac{1}{3-1} (2\sqrt{3} - 2) = 2\sqrt{3} - 2 \].This represents the average value of the function \( g(t) \) over the interval \([1, 3]\).

Key Concepts

Integral Calculus

Integral calculus is like finding the area under a curve. When you see an integral symbol \( \int \), it signifies that we are summing up an infinite number of tiny areas to find the total area between the curve and the x-axis. This process is crucial in many fields, like physics and engineering, where you want to know accumulations, such as distances, areas, and more.
The integral is defined by two components:

• The integrand \( g(t) \), which is the function you want to integrate
• The limits of integration \([a, b]\), depicting the range over which the integration occurs

In our example, we are finding the average value of a function using this concept, which involves calculating the integral of \( g(t) = \frac{t}{\sqrt{3 + t^2}} \) from \( t = 1 \) to \( t = 3 \). This allows us to find how much space the function takes up over this interval.

Substitution Method

The substitution method is a handy technique for making integrals simpler. Think of it as changing the coordinates of a map for easier navigation. By substituting \( u = 3 + t^2 \), we essentially transform a complicated integral into a more manageable one.
This method involves:

• Choosing a substitution that simplifies the integrand
• Finding the derivative of the substitution \( du \/ dt \)
• Adjusting the limits of integration based on the new variable \( u \)

In our example, after substituting and rearranging, we achieve a simpler form \( \int \frac{1}{\sqrt{u}} \cdot \frac{1}{2} \, du \). The tricky original function \( \frac{t}{\sqrt{3 + t^2}} \) becomes easier to integrate, showcasing the power of substitution.

Antiderivative

The antiderivative is essentially the reverse process of differentiation. If differentiation is about finding the rate of change of a function, antiderivatives tell us which function has this rate of change.
When integrating, you are often required to find the antiderivative to solve an integral. In this scenario:

• We found the antiderivative of \( u^{-1/2} \), which is \( 2u^{1/2} \).

This discovery allowed us to evaluate the definite integral from 4 to 12, simplifying our integral to \( \sqrt{12} - \sqrt{4} \). Without finding the antiderivative, solving integrals would be a much more complex task.

Closed Interval

A closed interval is simply a range of numbers that includes its endpoints. In calculus, working within a closed interval \([a, b]\) means we're looking at all the values from \( a \) to \( b \), including \( a \) and \( b \) themselves.
For the average value of a function, the concept of a closed interval is crucial because:

• It defines the specific range over which you compute the integral
• It influences the calculation, as the endpoints contribute to the final result

For our function \( g(t) \), the closed interval \([1, 3]\) indicates we consider the function values and area from \( t = 1 \) to \( t = 3 \), a key step in applying the average value formula.