Problem 59 - University Calculus: Early Transcendentals

Question

Graph the function and find its average value over the given interval. $$f(t)=(t-1)^{2} \quad \text { on } \quad[0,3]$$

Step-by-Step Solution

Answer

The average value of \( f(t) = (t-1)^2 \) over \([0,3]\) is 1.

Step 1: Graph the Function

The given function is \( f(t) = (t-1)^2 \). To graph this function, you can start by finding key points. Calculate \( f(t) \) for values like \( t = 0, 1, 2, 3 \). These results are \( f(0) = 1 \), \( f(1) = 0 \), \( f(2) = 1 \), and \( f(3) = 4 \). Plot these points on a coordinate plane and sketch a parabola opening upwards with vertex at \( (1, 0) \).

Step 2: Set Up the Integral for Average Value

The average value of a function \( f(t) \) over an interval \([a, b]\) is calculated using the formula \( \frac{1}{b-a} \int_a^b f(t) \, dt \). Here, \( a = 0 \) and \( b = 3 \), so the average value of \( f(t) = (t-1)^2 \) on the interval \([0,3]\) is \( \frac{1}{3-0} \int_0^3 (t-1)^2 \, dt \).

Step 3: Calculate the Definite Integral

The integral \( \int_0^3 (t-1)^2 \, dt \) can be solved by first expanding \((t-1)^2 = t^2 - 2t + 1 \). Now integrate each term separately: \( \int_0^3 t^2 \, dt = \left[ \frac{t^3}{3} \right]_0^3 \), \( \int_0^3 -2t \, dt = \left[ -t^2 \right]_0^3 \), and \( \int_0^3 1 \, dt = \left[ t \right]_0^3 \). Evaluate these at the limits from 0 to 3.

Step 4: Evaluate Each Term

- For \( \int_0^3 t^2 \, dt \), the result is \( \left[ \frac{t^3}{3} \right]_0^3 = \frac{27}{3} - 0 = 9 \).- For \( \int_0^3 -2t \, dt \), the result is \( \left[ -t^2 \right]_0^3 = -9 + 0 = -9 \).- For \( \int_0^3 1 \, dt \), the result is \( \left[ t \right]_0^3 = 3 - 0 = 3 \).Now add these results: \( 9 - 9 + 3 = 3 \).

Step 5: Calculate the Average Value

Substitute the result of the integral into the average value formula. The average value is \( \frac{1}{3} \times 3 = 1 \). Therefore, the average value of the function \( f(t) = (t-1)^2 \) over the interval \([0, 3]\) is 1.

Key Concepts

Definite Integral

A definite integral is a fundamental concept in calculus that serves as a tool for finding the area under a curve over a specific interval. This concept is crucial in computing the average value of a function. Consider the integral as a process of summing infinitely small pieces of area beneath a function's graph. For \[ ext{an interval } [a, b], ext{ the definite integral of } f(t) ext{, denoted by } \int_a^b f(t) \, dt, \text{ provides this total area.} \] Let’s explore the definite integral \[ \int_0^3 (t-1)^2 \, dt.\] First, we factor and expand the quadratic: \[ (t-1)^2 = t^2 - 2t + 1.\] Second, apply the rules of integration separately to each term. Remember these simple integration rules:

• The integral of \( t^n \) is \( \frac{t^{n+1}}{n+1} \).
• The integral of a constant \( c \) is \( ct \).

By evaluating the integral from 0 to 3, and summing the outcomes, you determine the aggregate area, 3, beneath the curve—a key component in calculating the average value!

Parabola Graphing

Graphing a parabola is essential to visualize the shape and direction of the quadratic function at work. A parabola represents a specific type of polynomial function, characterized by a U-shaped curve. The function\[ f(t) = (t-1)^2 \] is a typical parabola, and graphing it reveals useful insights about its behavior.To construct this graph, you identify crucial points that guide its shape:

• The vertex is at (1, 0), indicating the lowest point in the graph of \((t-1)^2\) that opens upwards.
• Choose other points such as \((0, 1), (2, 1), ext{ and } (3, 4)\), calculated by substituting their \(t\) values into \(f(t)\).

Plot these points on the coordinate plane and connect them smoothly to form an upward-opening parabola. This visualization aids in understanding where the function decreases to its minimum (at the vertex) and then increases, helping make sense of its dynamic across the interval.

Polynomial Functions

Polynomial functions encompass a broad category of algebraic functions characterized by varying degrees, like linear, quadratic, cubic and so on. A polynomial function takes the general form:\[ f(t) = a_nt^n + a_{n-1}t^{n-1} + ... + a_1t + a_0.\] Each term consists of a coefficient and a variable raised to a whole number exponent.The function\[ f(t) = (t-1)^2,\] while nested as a perfect square, is essentially a quadratic polynomial: \[ t^2 - 2t + 1.\] Key characteristics include:

• The highest degree term, \(t^2\), defines the parabola's wide U-shape graph.
• Such functions can typically have at most two roots, observed when the graph intersects the x-axis.

Understanding polynomial functions is fundamental as they appear frequently in mathematical applications and typify a vast range of curves and surfaces in different dimensions, making them an indispensable tool in modeling real-world phenomena.