Chapter 5: Problem 60
In Exercises \(55-62,\) graph the function and find its average value over the given interval. $$ f(t)=t^{2}-t \quad \text { on } \quad[-2,1] $$
Short Answer
Expert verified
The average value of \( f(t) = t^2 - t \) on \([-2, 1]\) is \( \frac{13}{9} \).
Step by step solution
01
Identify the Function and Interval
We have the function \( f(t) = t^2 - t \) and we need to find its average value over the interval \([-2, 1]\). To determine the average value of a continuous function \( f(x) \) over an interval \([a, b]\), we'll use the formula: \[ \text{Average value} = \frac{1}{b-a} \int_a^b f(x) \, dx. \] In our case, \( a = -2 \) and \( b = 1 \).
02
Set Up the Integral
Set up the definite integral for \( f(t) = t^2 - t \) from \( t = -2 \) to \( t = 1 \): \[ \int_{-2}^{1} (t^2 - t) \, dt. \]
03
Calculate the Integral
Calculate the integral \( \int (t^2 - t) \, dt \). The antiderivative of \( t^2 \) is \( \frac{t^3}{3} \) and for \( -t \) is \( -\frac{t^2}{2} \). Therefore, \[ \int (t^2 - t) \, dt = \frac{t^3}{3} - \frac{t^2}{2}. \]
04
Evaluate the Definite Integral
Evaluate the definite integral using the antiderivative:\[ \left[ \frac{t^3}{3} - \frac{t^2}{2} \right]_{-2}^{1} = \left( \frac{1^3}{3} - \frac{1^2}{2} \right) - \left( \frac{(-2)^3}{3} - \frac{(-2)^2}{2} \right). \] Calculate these values:\[ \left( \frac{1}{3} - \frac{1}{2} \right) = \left( \frac{2}{6} - \frac{3}{6} \right) = -\frac{1}{6}, \]\[ \left( -\frac{8}{3} - \frac{4}{2} \right) = \left( -\frac{8}{3} - 2 \right) = -\frac{8}{3} - \frac{6}{3} = -\frac{14}{3}. \]The difference is:\[ -\frac{1}{6} + \frac{14}{3} = \frac{14}{3} - \frac{1}{6} = \frac{28}{6} + \frac{1}{6} = \frac{27}{6} = \frac{13}{3}. \]
05
Compute the Average Value
To find the average value of \( f(t) \), we divide the integral result by the length of the interval:\[ \text{Average value} = \frac{1}{1 - (-2)} \times \frac{13}{3} = \frac{1}{3} \times \frac{13}{3} = \frac{13}{9}. \]
06
Graph the Function
Graph the quadratic function \( f(t) = t^2 - t \) over the interval \([-2, 1]\). The function is a parabola opening upwards. Identify key points such as the vertex and the intercepts, and sketch the graph accordingly. The function has a minimum value (vertex) somewhere in the interval, showing typical parabola characteristics.
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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 helps us calculate the area under a curve within a specific interval. Imagine you have a curve representing a function, and you want to find out the total area that exists between this curve and the horizontal axis over some interval on the x-axis.
To express this, we use notation like this: \( \int_a^b f(x) \, dx \). Here, \( f(x) \) is your function, \( a \) and \( b \) are the limits of the interval, and \( \int \) signifies that you're finding the integral. The function \( f(x) \) can be anything, like a line, curve, or even a more complex shape.
To express this, we use notation like this: \( \int_a^b f(x) \, dx \). Here, \( f(x) \) is your function, \( a \) and \( b \) are the limits of the interval, and \( \int \) signifies that you're finding the integral. The function \( f(x) \) can be anything, like a line, curve, or even a more complex shape.
- The limits \( a \) and \( b \) are where you start and stop calculating this area.
- On a graph, it looks like you are coloring the area between the x-axis and your curve from \( x = a \) to \( x = b \).
Quadratic Function
A quadratic function is a specific type of polynomial function with a degree of 2. This simply means the highest power of the variable is 2. It typically takes the form \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants.
Quadratic functions are known for their distinctive U-shaped curves called parabolas. Here’s what you should know about them:
Quadratic functions are known for their distinctive U-shaped curves called parabolas. Here’s what you should know about them:
- The graph symmetry: Every parabola is symmetric about a vertical line called the axis of symmetry. It usually passes through the vertex of the parabola.
- The vertex: This is a key point in a quadratic function. It represents the minimum or maximum value of the function, depending on whether the parabola opens upwards (\( a > 0 \)) or downwards (\( a < 0 \)).
- The roots: Also called zeroes, these are the points where the parabola intersects the horizontal axis. They can be found using the quadratic formula or factoring.
Parabola Graphing
Graphing a parabola involves plotting a quadratic function on a coordinate plane, which will result in a curve shaped like a U. To graph a parabola effectively, you need to identify certain important features of the quadratic function.
1. **Calculate the Vertex:** This is the point where the parabola is at its peak or valley. For \( f(t) = t^2 - t \), the vertex can be found by using the vertex formula \( t = -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = -1 \), giving \( t = 0.5 \).
2. **Find the Axis of Symmetry:** This is a vertical line that runs through the vertex of the parabola, effectively splitting it into two equal halves. For our function, it is \( t = 0.5 \), where \( t \) is the variable of the function.
3. **Determine Intercepts:** Calculate where the parabola intersects the y-axis (at \( t = 0 \), the y-intercept is \( f(0) = 0 \)) and, if necessary, the x-axis, using the quadratic formula or factorization. Our function hits the x-axis at \( t = 0 \) and a second point which can be calculated.
1. **Calculate the Vertex:** This is the point where the parabola is at its peak or valley. For \( f(t) = t^2 - t \), the vertex can be found by using the vertex formula \( t = -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = -1 \), giving \( t = 0.5 \).
2. **Find the Axis of Symmetry:** This is a vertical line that runs through the vertex of the parabola, effectively splitting it into two equal halves. For our function, it is \( t = 0.5 \), where \( t \) is the variable of the function.
3. **Determine Intercepts:** Calculate where the parabola intersects the y-axis (at \( t = 0 \), the y-intercept is \( f(0) = 0 \)) and, if necessary, the x-axis, using the quadratic formula or factorization. Our function hits the x-axis at \( t = 0 \) and a second point which can be calculated.
- A detailed sketch will involve marking these key points on a graph.
- Then draw a smooth curve through these points, ensuring symmetry about the axis.
- Remember, the shape of the parabola is influenced by the coefficient \( a \). If \( a > 0 \), it opens upwards, otherwise downward.