Question

In Exercises \(11 - 14 ,\) use NINT to find the average value of the function on the interval. At what point (s) in the interval does the function assume its average value? $$y = ( x - 1 ) ^ { 2 } , \quad [ 0,3 ]$$

Short answer

The average value of the function \( y = (x - 1)^2 \) on the interval \( [0, 3] \) is \( \frac {2}{3} \). The function takes on this average value at points \( x = \frac {1}{2} \) and \( x = \frac {3}{2} \).

Step-by-step solution

Calculate the average value of the function

First, find the average value of the function \( y = (x - 1)^2 \) on the interval \( [0, 3] \) using the formula of average value of a function. This gives:\( \frac {1}{3 - 0} \int_{0}^{3} (x - 1)^2 dx = \frac {1}{3} [ \frac {x^3}{3} - 2 * \frac {x^2}{2} + x]_{0}^{3} = \frac {1}{3} * (\frac {27}{3} - 6*3 + 3) = \frac {2}{3} \)

Find the point \( c \) on the interval \( [0, 3] \) where the function takes on the average value

Now, set \( y = (x - 1)^2 = \frac {2}{3} \) and solve for \( x \) to find \( c \).Solving the equation we get \( (x - 1)^2 = \frac {2}{3} \). When solving we get two possible values for \( x \), namely \( x = \frac {1}{2} \) and \( x = \frac {3}{2} \).After checking, both these points \( x = \frac {1}{2} \) and \( x = \frac {3}{2} \) lie on the interval \( [0, 3] \). Therefore, the function takes on its average value at these points.

Key concepts

Definite Integral

The definite integral is a foundational concept in calculus, representing the accumulation of quantities as they continuously change. It can be visualized as the area under a curve on a graph of a function, bounded by two points on the x-axis.

When we calculate the definite integral of a function over an interval, such as \( [0,3] \) for \( y = (x - 1)^2 \), we are essentially adding up tiny increments of the function's value over this range. Mathematically, we write this as \( \int_{0}^{3} (x - 1)^2 dx \), and the process involves finding the antiderivative of the function, then subtracting its value at the lower limit from its value at the upper limit.

In the context of our exercise, the definite integral is used to compute the total sum of the function's values over the interval before we can determine its average value. To perform this integration, we apply the power rule and obtain the antiderivative \( \frac{x^3}{3} - x^2 + x \), after simplifying the integral's evaluation from \(x=0\) to \(x=3\).

Calculus

Calculus is a branch of mathematics that studies how things change and includes two major areas: differential calculus and integral calculus. Differential calculus concerns the rate of change of quantities, while integral calculus focuses on the accumulation of quantities, as we've seen with the definite integral.

The underlying principle behind calculus is the concept of limits, which allows us to handle quantities that are infinitesimally small. By using calculus, we can solve a variety of problems that involve optimization, motion, and area under curves, to name just a few applications.

In exercises like ours, calculus is crucial because it gives us the tools needed to find the average value of a function. This process hinges on the ability to integrate a function and understand the behavior of the function over given intervals. It's a powerful mathematical tool that demonstrates the interplay of geometric representation and algebraic manipulation to yield meaningful numerical results.

Mean Value Theorem

The Mean Value Theorem for integrals states that for a continuous function \( f \) over a closed interval \( [a,b] \) there exists at least one point \( c \) in \( [a,b] \) such that \( f(c) \) is equal to the average value of \( f \) over that interval.

This theorem has a deep connection with our current exercise, where we are tasked to find the average value of the function \( y = (x - 1)^2 \) and identify the point(s) where the function assumes this average value on the interval \( [0,3] \). After calculating the average value, we used the result \( \frac{2}{3} \) to find the corresponding point(s) \( c \), which implies that there exists at least one \( c \) such that the function's value at \( c \) is the same as the average value of the function over the interval. As we found, the function takes on its average value at \( x = \frac{1}{2} \) and \( x = \frac{3}{2} \), both lying within the given interval, hence satisfying the Mean Value Theorem.