Problem 2 - Calculus

Question

Find the average value of the function on the given interval. $$ f(x)=5 x^{2} ; \quad[1,4] $$

Step-by-Step Solution

Answer

The average value is 35.

Step 1: Understanding the Problem

To find the average value of a function on a given interval, we'll use the formula for the average value of a function, which involves integrating the function over the given interval and dividing by the interval's length. For this problem, the function is \( f(x) = 5x^2 \) and the interval is \([1, 4]\).

Step 2: Formula for Average Value

The formula for the average value of a function \( f \) on the interval \([a, b]\) is given by \[ \text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \] where \( a = 1 \) and \( b = 4 \) for this problem.

Step 3: Setting Up the Integral

Substitute the given function \( f(x) = 5x^2 \) into the integral. Setup the integral as \[ \int_{1}^{4} 5x^2 \, dx \]

Step 4: Calculating the Integral

Evaluate the integral \( \int 5x^2 \, dx \). The antiderivative of \( 5x^2 \) is \( \frac{5}{3}x^3 \). Thus, the integral from 1 to 4 is \[ \left[ \frac{5}{3}x^3 \right]_1^4 = \frac{5}{3}(4^3) - \frac{5}{3}(1^3) \].

Step 5: Computing the Definite Integral

Calculate \( \frac{5}{3}(4^3) - \frac{5}{3}(1^3)\). First compute \( 4^3 = 64 \) and \( 1^3 = 1 \). Thus, \[ \frac{5}{3}(64) - \frac{5}{3}(1) = \frac{320}{3} - \frac{5}{3} = \frac{315}{3} = 105 \].

Step 6: Using the Average Value Formula

Using the formula \[ \text{Average Value} = \frac{1}{4-1} \cdot 105 \], simplify to \[ \frac{105}{3} = 35 \].

Step 7: Conclusion

Thus, the average value of the function \( f(x) = 5x^2 \) on the interval \([1, 4]\) is 35.

Key Concepts

Definite Integral

A definite integral is a fundamental concept in calculus that represents the accumulation of a quantity over an interval on the real line. This area under the curve is crucial for finding the average value of a continuous function over a certain interval. For example, when evaluating the definite integral of a function like \( f(x) = 5x^2 \) over an interval \([1, 4]\), the integral gives us the total area accumulated between the curve \( 5x^2 \) and the x-axis, from \( x = 1 \) to \( x = 4 \).

• To calculate the definite integral, we identify the antiderivative, which is a function whose derivative is the integrand.
• We then apply the limits of integration to this antiderivative to find the net signed area or total accumulation.
• In mathematical terms, for a continuous function \( f(x) \), this can be expressed as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.

This definite integral results in a number that has various interpretations such as area, accumulated change, or total amount depending on the context of the problem.

Antiderivative

An antiderivative of a function is essentially the reverse of taking a derivative. While a derivative tells us the rate of change of a function, the antiderivative helps us rebuild the original function. In the context of calculating the average value of a function, finding the antiderivative is an essential step in evaluating a definite integral.For \( f(x) = 5x^2 \), its antiderivative is \( F(x) = \frac{5}{3}x^3 \). This means that if you were to differentiate \( \frac{5}{3}x^3 \), you'd regain the original function \( 5x^2 \).

• To find an antiderivative, you often increase the power of \( x \) by one and divide by the new power.
• Don't forget the constant of integration when computing general antiderivatives, although it is usually unnecessary for definite integrals.
• Use the antiderivative to compute the definite integral by evaluating it at the bounds of the given interval.

This process requires careful attention to detail and is a skill practiced frequently in calculus to solve real-world problems involving rates and quantities.

Interval Notation

Interval notation is a simple way of representing a segment of the number line and is particularly useful in calculus for specifying the boundaries over which a function is integrated or evaluated. For example, the interval notation \([1, 4]\) designates all the numbers from 1 to 4, including the endpoints. This is crucial in our problem because it dictates the start and end points for the definite integral.

• Square brackets, as in \([a, b]\), indicate that the endpoints \(a\) and \(b\) are included in the interval (closed interval).
• If a parenthesis is used, such as \((a, b)\), this indicates that the endpoints are not included (open interval).
• Interval notation is not only employed to specify limits of integration but is essential in domains and ranges of functions.

Understanding and correctly applying interval notation is key to accurately describe function behaviors and set up integrals.