Chapter 5: Problem 58
In Exercises \(55-62,\) graph the function and find its average value over the given interval. $$ f(x)=3 x^{2}-3 \quad \text { on } \quad[0,1] $$
Short Answer
Expert verified
The average value of the function on \([0, 1]\) is \(-2\).
Step by step solution
01
Identify the Function and Interval
The given function is \( f(x) = 3x^2 - 3 \), and we are asked to find the average value of this function on the interval \([0, 1]\).
02
Understand the Formula for Average Value
The formula to find the average value of a function \( f(x) \) over the interval \([a, b]\) is given by: \[ \frac{1}{b-a} \int_a^b f(x) \, dx \] In our case, \( a = 0 \) and \( b = 1 \).
03
Set Up the Integral for Average Value
We set up the integral of \( f(x) = 3x^2 - 3 \) from \( 0 \) to \( 1 \): \[ \int_0^1 (3x^2 - 3) \, dx \]
04
Compute the Integral
To compute the integral, find the antiderivative of \( 3x^2 - 3 \): - The antiderivative of \( 3x^2 \) is \( x^3 \). - The antiderivative of \(-3\) is \(-3x\).Thus, the antiderivative of \( 3x^2 - 3 \) is \( x^3 - 3x \).
05
Evaluate the Definite Integral
We now evaluate the definite integral from \( 0 \) to \( 1 \) using the antiderivative:\[ \left[ x^3 - 3x \right]_0^1 = (1^3 - 3 \times 1) - (0^3 - 3 \times 0) \]\[ = (1 - 3) - (0 - 0) \]\[ = -2 \]
06
Calculate the Average Value
Substitute the result of the integral into the average value formula:\[ \frac{1}{1 - 0} \times (-2) = -2 \]
07
Graph the Function
Graph the function \( f(x) = 3x^2 - 3 \) over the interval \([0, 1]\). The graph is a parabola opening upwards with a vertex at a point where the function achieves its minimum value and it crosses the y-axis at \( y = -3 \).
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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 used to calculate the area under a curve. It's particularly useful for finding the average value of a function over a specific interval. For instance, to determine the average value of the function \[ f(x) = 3x^2 - 3 \] over the interval \([0, 1]\), we need to find the definite integral of this function on \([0, 1]\).In general, a definite integral is represented as:\[ \int_a^b f(x) \, dx \]where \( a \) and \( b \) are the lower and upper limits of the interval, respectively. This formula computes the net area between the x-axis and the curve \( f(x) \).To find the average value of the function, follow these steps:
- Set up the integral using the limits of the interval.
- Calculate the antiderivative.
- Evaluate the integral by substituting the upper and lower limits.
- Divide by the length of the interval (\( b-a \)).
Antiderivative
The antiderivative, or indefinite integral, is a crucial part of solving problems involving definite integrals. It's essentially the reverse process of differentiation. To solve the problem of finding the average value of \( f(x) = 3x^2 - 3 \), you're required to first find its antiderivative.Here's how to find the antiderivative:
- Identify each term of the function separately. For \( 3x^2 \), the antiderivative is \( \frac{x^3}{3} \times 3 = x^3 \).
- The constant term \(-3\) becomes \(-3x\) after taking its antiderivative.
Interval Notation
Interval notation is a mathematical shorthand to describe subsets of the real number line. You'll often use it in calculus to specify ranges over which functions are integrated. In our example, \([0, 1]\) is used to indicate the interval from \(0\) to \(1\) including both endpoints.Here's a quick guide to understand interval notation:
- Square brackets \([\, ]\) indicate that an endpoint is included in the interval (closed interval).
- Parentheses \((\, )\) denote that an endpoint is not included (open interval).
Graphing Functions
Graphing a function gives a visual representation of its behavior and helps to understand the solutions to calculus problems. For the function\( f(x) = 3x^2 - 3 \),you are tasked with sketching it over the interval\([0, 1]\).Here are steps to effectively graph this quadratic function:
- Identify the key features: it's a parabola opening upwards because the coefficient of \(x^2\) is positive.
- Find the vertex, which in this case, is the minimum point when the function is expressed in its standard form.
- Determine where the function intersects with the y-axis; for this function, it crosses at \(y = -3\).