Problem 7 - Calculus Single Variable

Question

In each of Exercises \(1-12,\) calculate the average value of the given function on the given interval. $$ f(x)=(x-1)^{1 / 2} \quad I=[2,5] $$

Step-by-Step Solution

Answer

The average value of the function on the interval is \( \frac{14}{9} \).

Step 1: Understand the Formula for Average Value

The average value of a continuous function \( f(x) \) on the interval \( [a, b] \) is given by the formula: \[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx. \] For this exercise, the function \( f(x) = (x-1)^{1/2} \) is defined on the interval \([2, 5]\).

Step 2: Set Up the Integral

Substitute the function \( f(x) \) and the interval into the formula. This gives us: \[ \text{Average value} = \frac{1}{5-2} \int_{2}^{5} (x-1)^{1/2} \, dx. \] This simplifies to: \[ \frac{1}{3} \int_{2}^{5} (x-1)^{1/2} \, dx. \]

Step 3: Integrate the Function

To integrate \( (x-1)^{1/2} \), use the substitution \( u = x-1 \), then \( du = dx \). Change the limits of integration: when \( x=2, u=1 \) and when \( x=5, u=4 \). The integral becomes: \[ \int_{1}^{4} u^{1/2} \, du. \] This integral can be solved using the power rule: \[ \int u^{n} \, du = \frac{u^{n+1}}{n+1} + C. \] Applying this rule, we get: \[ \frac{u^{3/2}}{3/2} \bigg|_{1}^{4} = \frac{2}{3} (u^{3/2}) \bigg|_{1}^{4}. \]

Step 4: Evaluate the Integral

Evaluate the definite integral from Step 3: \[ \frac{2}{3} [(4)^{3/2} - (1)^{3/2}] = \frac{2}{3} [8 - 1] = \frac{2}{3} \times 7 = \frac{14}{3}. \]

Step 5: Compute the Average Value

Now substitute the result of the integral back into the formula for average value: \[ \text{Average value} = \frac{1}{3} \times \frac{14}{3} = \frac{14}{9}. \]

Key Concepts

Definite Integral

A definite integral is a fundamental concept in calculus that represents the accumulation of quantities, such as area under a curve, over a specific interval. When you see an integral with limits, those limits are part of a definite integral.
In our example, calculating the average value of the function involves evaluating a definite integral. The integral \[ \int_{2}^{5} (x-1)^{1/2} \, dx \]
represents the area under the curve \( (x-1)^{1/2} \) from \( x = 2 \) to \( x = 5 \).
You're essentially finding a weighted average of the function's values over that interval.

• Definite integrals can be thought of as providing the "sum" of infinitely many small values of the function, with each small value multiplied by a tiny width.
• Limits of the integral \([a, b]\) help to specify where the "sum" should occur.

Power Rule Integration

The power rule for integration is a straightforward technique to find antiderivatives of power functions. If you know how to differentiate, this is quite similar but reversed.
The power rule states: If \( f(x) = x^n \) (where \( n \) is not -1), then its integral is \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C. \]
This rule allows you to find the antiderivative (or integral) of functions like \( u^{1/2} \), by increasing the power by 1 and dividing by the new power.
In our exercise, we apply the power rule to integrate the expression \( \int u^{1/2} \, du \), resulting in

• \( \frac{u^{3/2}}{3/2} + C \).

This forms a crucial part of completing the definite integral once the power rule's result is evaluated within the bounds.

Substitution Method

In calculus, the substitution method is like changing variables to make an integral easier to solve. Think of it as making a hard problem into an easier one with a clever trick.
The substitution method, also known as "u-substitution," involves setting \( u = g(x) \) to simplify the integrand.
For the given exercise, we substituted \( u = x - 1 \). By doing this:

• Transform the integral \( \int (x-1)^{1/2} \, dx \) into \( \int u^{1/2} \, du \).
• Don't forget to change the limits of integration accordingly: when \( x = 2, u = 1 \) and when \( x = 5, u = 4 \).

The goal of substitution is to rewrite the integrand in terms of \( u \) and \( du \), simplifying the process significantly. It is often combined perfectly with the power rule to evaluate integrals efficiently.