Chapter 5: Problem 59
Graph the function and find its average value over the given interval. $$f(t)=(t-1)^{2} \quad \text { on } \quad[0,3]$$
Short Answer
Expert verified
The average value of \( f(t) = (t-1)^2 \) over \([0,3]\) is 1.
Step by step solution
01
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 \( (1, 0) \).
02
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 \( \frac{1}{3-0} \int_0^3 (t-1)^2 \, dt \).
03
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.
04
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: \( 9 - 9 + 3 = 3 \).
05
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( ct \).
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 \(f(t)\).
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.