Question

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

Short answer

The average value of the function is 35.

Step-by-step solution

Understand the concept of average value of a function

The average value of a continuous 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 problem, \( f(x) = 5x^2 \), \( a = 1 \), and \( b = 4 \).

Set up the integral for the given function

Substitute the function \( f(x) = 5x^2 \) and the interval \([1, 4]\) into the formula for the average value:\[\text{Average value} = \frac{1}{4-1} \int_{1}^{4} 5x^2 \, dx\] This simplifies to:\[\text{Average value} = \frac{1}{3} \int_{1}^{4} 5x^2 \, dx\]

Calculate the indefinite integral \( \int 5x^2 \, dx \)

Find the antiderivative of the function \( 5x^2 \). The antiderivative is:\[\int 5x^2 \, dx = \frac{5}{3}x^3 + C\] Here, \( C \) is the constant of integration, which we will evaluate when applying the definite limits.

Evaluate the definite integral \( \int_{1}^{4} 5x^2 \, dx \)

Use the antiderivative to evaluate the integral from 1 to 4:\[\left[ \frac{5}{3}x^3 \right]_{1}^{4} = \frac{5}{3}(4^3) - \frac{5}{3}(1^3)\] Calculate each part:\[\frac{5}{3}(4^3) = \frac{5}{3}(64) = \frac{320}{3}\]\[\frac{5}{3}(1^3) = \frac{5}{3}(1) = \frac{5}{3}\]Thus, the definite integral is:\[\frac{320}{3} - \frac{5}{3} = \frac{315}{3}\simeq 105\]

Compute the average value

Using the result from the definite integral, substitute back into the average value formula:\[\text{Average value} = \frac{1}{3} \times 105 = 35\] So the average value of the function \( f(x) = 5x^2 \) over the interval [1, 4] is 35.

Key concepts

Definite integral

The definite integral is a fundamental concept in calculus that allows us to find the area under the curve of a function over a specific interval. This provides a way to reconcile the seemingly infinite nuances of a function with concrete measurements. In mathematical terms, a definite integral of a function \( f(x) \) from \( a \) to \( b \) is denoted as \( \int_{a}^{b} f(x) \, dx \). This symbol represents the sum of all the infinite tiny areas beneath the curve of \( f(x) \) between the points \( x = a \) and \( x = b \).
To compute this, we typically use the antiderivative of the function, often called the integral without limits. This involves calculating the difference between the values of the antiderivative evaluated at \( b \) and \( a \). When you see the integration limits attached to the integrand \( \int_{a}^{b} \), it turns it into a definite value representing an area.

Antiderivative

An antiderivative, also known as an indefinite integral, is essentially the "reverse" of a derivative. If you think of differentiation as a process of finding the slope of a function, using a derivative, finding an antiderivative is like retracing your steps to get back the original function.
In the context of our exercise, when we find the antiderivative of \( 5x^2 \), we determine it to be \( \frac{5}{3}x^3 + C \). Here \( C \) represents the constant of integration. This means that any function of the form \( \frac{5}{3}x^3 + C \) could have \( 5x^2 \) as its derivative.
The goal when solving definite integrals is to use this antiderivative to evaluate the specific area under the function's curve between two points. Consequently, the concept of the antiderivative is critical when finding average values of functions over intervals.

Continuous function

A continuous function is a type of function where small changes in the input lead to small changes in the output. Mathematically, a function \( f(x) \) is said to be continuous at a point \( x_0 \) if the limit of \( f(x) \) as \( x \) approaches \( x_0 \) is equal to \( f(x_0) \). This means there are no sudden jumps or breaks in the graph of the function.
In order to properly calculate an average value with integrals, it's typically required that the function be continuous over the interval you're interested in. This ensures that there are no gaps or irregularities that could skew the results. For our function \( f(x) = 5x^2 \), it is continuous because it's a polynomial, and all polynomial functions are continuous everywhere. This continuity allows us to calculate the definite integral and the average value without any issues.

Average value formula

The concept of an average value of a function over an interval is similar to finding an average in statistics. Instead of adding up numbers and dividing by the number of entries, you integrate the function over the interval and divide by the length of the interval.
The formula for the average value of a function \( f(x) \) on an interval \([a, b]\) is:

• \( \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \)

This formula essentially tells us to sum up all the values of \( f(x) \) over the interval and then "spread" these values out evenly across the interval, much like how a mean works.
Applying this to the function \( f(x) = 5x^2 \) on the interval \( [1, 4] \), we used the formula to find that the average value is 35. This means that, on average, the function has a value of 35 when evaluated from 1 to 4.