Problem 16 - Calculus

Question

If \(f(x)=3 x^{2} \sqrt{x^{3}-4},\) find the average value of \(f\) on [2,5]

Step-by-Step Solution

Answer

The average value of the function on [2, 5] is 294.

Step 1: Recall the Formula for Average Value

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

Step 2: Set Up the Definite Integral

Substitute \( f(x) = 3x^{2}\sqrt{x^{3}-4} \) into the integral:\[\int_{2}^{5} 3x^{2}\sqrt{x^{3}-4} \; dx\] This integral will be used to find the total accumulation needed for the average value formula.

Step 3: Use Substitution Method to Simplify the Integral

Let's choose \( u = x^3 - 4 \). Then, \( du = 3x^2 \, dx \), which fits well with our integral setup. Substituting these into the integral gives: \[\int \sqrt{u} \, du\]Remember to change the limits of integration. When \( x = 2 \), \( u = 2^3 - 4 = 4 \). When \( x = 5 \), \( u = 5^3 - 4 = 121 \). Thus the new integral becomes:\[\int_{4}^{121} \sqrt{u} \, du\]

Step 4: Integrate with Respect to \( u \)

The integral \( \int \sqrt{u} \, du \) is equivalent to \( \int u^{1/2} \, du \). Applying the power rule gives:\[= \frac{2}{3} u^{3/2} + C\]We will evaluate this from 4 to 121.

Step 5: Evaluate the Definite Integral

Calculate the definite integral:\[\left[ \frac{2}{3} u^{3/2} \right]_{4}^{121} = \frac{2}{3} (121)^{3/2} - \frac{2}{3} (4)^{3/2}\]Calculating each term, we find:\(121^{3/2} = (11^2)^{3/2} = 11^3 = 1331\) and \(4^{3/2} = (2^2)^{3/2} = 2^3 = 8\).Thus:\[\frac{2}{3} (1331) - \frac{2}{3} (8) = \frac{2}{3} (1331 - 8) = \frac{2}{3} \times 1323 = 882\]

Step 6: Calculate the Average Value

Now, use the result of the integral in the average value formula:\[\text{Average Value} = \frac{1}{5-2} \times 882 = \frac{882}{3} = 294\] Thus, the average value of the function \( f(x) \) on the interval \([2, 5]\) is 294.

Key Concepts

Definite Integral

A definite integral provides a way to calculate the total accumulation of a quantity represented by a function over an interval. Think of it as capturing the total "area under the curve" of a function, between two specific points.
The notation for a definite integral is \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration. These numbers indicate the start and end of the interval along the x-axis.

• The function \( f(x) \) is the integrand, and it represents the curve we want to accumulate.
• The \( dx \) symbolizes a very small change in \( x \), indicating that we're summing up an infinite number of tiny areas under the curve.

When evaluating a definite integral, you'll calculate a number that quantifies the accumulation between the two limits. Always remember that the difference between definite and indefinite integrals is the limits, which give you a concrete value instead of a general family of functions.
The definite integral played a central role in solving the exercise above, as it helped determine the total area to find the average value of the function \( f(x) \) over the interval \([2,5]\).

Integration by Substitution

Integration by substitution is a technique used to simplify complex integrals. It's akin to reversing the chain rule of differentiation. This strategy makes integration easier by converting the original variable \( x \) into a different variable, typically \( u \).
Here's how it works:

• Choose a substitution: You pick a part of the integrand to substitute with \( u \).
• Differentiate your substitution: Find \( du \) by differentiating your chosen substitution with respect to \( x \).
• Substitute and adjust limits (if necessary): Replace the chosen parts in the integral with \( u \), and adjust the bounds of integration accordingly.

In the exercise, the substitution \( u = x^3 - 4 \) greatly simplified the integral. By making this substitution, the integral became \( \int \sqrt{u} \, du \), much more straightforward than handling the original expression directly. Remember to change the limits of integration to reflect \( u \)'s new boundaries when you substitute.

Power Rule in Integration

The power rule in integration is a fundamental tool that simplifies finding the antiderivative of functions. It states that to integrate a function of the form \( x^n \), where \( n eq -1 \), you can use:
\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]

• \( n \) is any real number except \(-1\).
• \( C \) is the constant of integration, reflecting the family of all antiderivatives.

The power rule simplifies the process of integration significantly when dealing with polynomial terms. For example, in the integral \( \int u^{1/2} \, du \) from the exercise, applying the power rule yields:
\[ \frac{2}{3} u^{3/2} + C \]No longer does the function look like a complicated root; instead, it is now reduced to a simple power function.
During definite integration, we do not need the constant \( C \), thanks to the boundaries that define the specific values. This rule helped conclude the problem efficiently, allowing the straightforward calculation of definite integrals in the solution step.