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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 is -2 over the interval [0, 1].

Step by step solution

01

Understand the Function

The function given is a quadratic function: \( f(x) = 3x^2 - 3 \). We are considering its behavior on the interval \([0, 1]\). The function will be graphed, and its average value will be calculated over this interval.
02

Graphing the Function

To graph the function \( f(x) = 3x^2 - 3 \), start by identifying key points. For example:1. At \( x = 0 \): \( f(0) = 3(0)^2 - 3 = -3 \)2. At \( x = 1 \): \( f(1) = 3(1)^2 - 3 = 0 \)The function is a parabola opening upwards because the coefficient of \(x^2\) is positive. Plot these points and sketch the curve on the interval \([0, 1]\).
03

Determine the Average Value Formula

The average value of a function \( f(x) \) over an interval \([a, b]\) is given by the formula: \[ \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]For this problem, \( a = 0 \) and \( b = 1 \), hence the average value is:\[ \frac{1}{1-0} \int_{0}^{1} (3x^2 - 3) \, dx \]
04

Integrate the Function

Calculate the integral:\[ \int_{0}^{1} (3x^2 - 3) \, dx = \left[ x^3 - 3x \right]_0^1 \]Evaluate the definite integral by substituting the limits:1. Upper limit: \( (1)^3 - 3(1) = 1 - 3 = -2 \)2. Lower limit: \( (0)^3 - 3(0) = 0 \)Thus, the integral is \( -2 \).
05

Calculate the Average Value

Using the result from the integration step, plug it into the average value formula:\[ \frac{1}{1-0} (-2) = -2 \]Hence, the average value of \( f(x) = 3x^2 - 3 \) over the interval \([0, 1]\) is \(-2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Quadratic Functions
Quadratic functions play a significant role in mathematics. These functions are expressed in the form \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is not zero. In our exercise, the function is \( f(x) = 3x^2 - 3 \). This is a simple quadratic function with \( a = 3 \), \( b = 0 \), and \( c = -3 \).

Key characteristics of quadratic functions include:
  • They graph as parabolas, either opening upwards if \( a > 0 \) or downwards if \( a < 0 \).
  • The vertex of a parabola is a significant point, representing the minimum or maximum of the function.
  • The axis of symmetry can be found using the formula \( x = -\frac{b}{2a} \).
In our specific function, since \( a = 3 \) (positive), the parabola opens upwards. The vertex is the lowest point on this graph. Understanding these characteristics helps in sketching the graph accurately, particularly over specified intervals like \([0, 1]\).
The Role of Definite Integrals
The definite integral is a fundamental concept in calculus. It helps determine the net area under a curve on a specified interval. For the function \( f(x) = 3x^2 - 3 \), the definite integral from \( x = 0 \) to \( x = 1 \) will provide valuable information.

Here鈥檚 how integrals assist us:
  • They calculate the net area which may be positive or negative, depending on the position of the graph relative to the x-axis.
  • To compute this integral, we evaluate \( \int_{0}^{1} (3x^2 - 3) \, dx \), which involves finding the antiderivative and using it to determine the function's accumulated value between the limits.
  • Evaluating the upper and lower limits of this function鈥檚 integral results in a final value of \(-2\).
By integrating, we find that the average "height" of the function over the interval might actually be below the x-axis, as illustrated by our integral鈥檚 result. This result feeds into calculating the average value per the formula for average value over an interval.
Graphing Functions Efficiently
Graphing functions is a crucial skill for visualizing mathematical concepts. It involves plotting key points and understanding the shape of the function. For the function \( f(x) = 3x^2 - 3 \), graphing within the interval \([0, 1]\) is important for understanding its behavior.

Follow these steps to graph effectively:
  • Identify and plot key points: At \( x = 0 \), \( f(x) = -3 \) and at \( x = 1 \), \( f(x) = 0 \).
  • Sketch the parabola by noting its general shape - in this scenario, it's an upward-opening parabola due to the positive coefficient \( 3 \).
  • Ensure that your graph clearly represents the entire curve from \( x = 0 \) to \( x = 1 \).
Graphing gives a visual confirmation of mathematical results, such as where the function increases, decreases or stays constant. Additionally, it becomes easier to see why the average value might be negative when the function dips below the x-axis on its interval.

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Most popular questions from this chapter

Evaluate the integrals in Exercises \(13-48\) . $$ \int x^{1 / 3} \sin \left(x^{4 / 3}-8\right) d x $$

Evaluate the integrals in Exercises \(13-48\) . $$ \int \tan ^{7} \frac{x}{2} \sec ^{2} \frac{x}{2} d x $$

If \(f\) is a continuous function, find the value of the integral $$I=\int_{0}^{a} \frac{f(x) d x}{f(x)+f(a-x)}$$ by making the substitution \(u=a-x\) and adding the resulting integral to \(I .\)

Evaluate the integrals in Exercises \(13-48\) . $$ \int x^{3} \sqrt{x^{2}+1} d x $$

If you do not know what substitution to make, try reducing the integral step by step, using a trial substitution to simplify the integral a bit and then another to simplify it some more. You will see what we mean if you try the sequences of substitutions in Exercises 49 and 50 . $$ \begin{array}{l}{\int \sqrt{1+\sin ^{2}(x-1)} \sin (x-1) \cos (x-1) d x} \\\ {\text { a. } u=x-1, \text { followed by } v=\sin u, \text { then by } w=1+v^{2}} \\ {\text { b. } u=\sin (x-1), \text { followed by } v=1+u^{2}} \\\ {\text { c. } u=1+\sin ^{2}(x-1)}\end{array} $$
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