Chapter 7: Problem 4
In each of Exercises \(1-12,\) calculate the average value of the given function on the given interval. $$ f(x)=3 x^{2}-6 x+1 \quad I=[3,7] $$
Short Answer
Expert verified
The average value of the function on the interval is 50.
Step by step solution
01
Write the Average Value Formula
The average value of a function \( f(x) \) over the interval \( [a, b] \) is given by the formula: \[ \text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]In this exercise, \( a = 3 \) and \( b = 7 \).
02
Set Up the Integral
Replace \( a \) and \( b \) with 3 and 7 and set up the integral for the function \( f(x) = 3x^2 - 6x + 1 \): \[ \text{Average Value} = \frac{1}{7-3} \int_{3}^{7} (3x^2 - 6x + 1) \, dx \]
03
Simplify the Expression
Simplify the expression \( \frac{1}{7-3} = \frac{1}{4} \), so the integral becomes:\[ \frac{1}{4} \int_{3}^{7} (3x^2 - 6x + 1) \, dx \]
04
Integrate the Function
Integrate \( 3x^2 - 6x + 1 \):\[ \int (3x^2 - 6x + 1) \, dx = x^3 - 3x^2 + x + C \]where \( C \) is the constant of integration. We don’t need to consider \( C \) since we're evaluating a definite integral.
05
Evaluate the Definite Integral
Evaluate the definite integral from 3 to 7:\[ \left[ x^3 - 3x^2 + x \right]_{3}^{7} = (7^3 - 3 \times 7^2 + 7) - (3^3 - 3 \times 3^2 + 3) \]Calculate each part separately:- At \( x = 7 \): \( 7^3 - 3 \times 7^2 + 7 = 343 - 147 + 7 = 203 \)- At \( x = 3 \): \( 3^3 - 3 \times 3^2 + 3 = 27 - 27 + 3 = 3 \)Thus, the integral evaluates to: \( 203 - 3 = 200 \).
06
Finish Calculating the Average Value
Take the result from the integral and multiply by \( \frac{1}{4} \):\[ \text{Average Value} = \frac{1}{4} \times 200 = 50 \]
07
Conclusion
Therefore, the average value of the function \( f(x) = 3x^2 - 6x + 1 \) on the interval \( [3,7] \) is 50.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Definite Integrals
Definite integrals are a fundamental concept in calculus. They are used to calculate the net area under a curve between two points on a graph. This calculation provides significant insights into many real-world problems, such as finding the total accumulated value of something over an interval.
When you see the notation \( \int_{a}^{b} f(x) \, dx \), it means we are calculating the integral of function \( f(x) \) from \( x = a \) to \( x = b \). This is known as a "definite integral." Here’s what each part signifies:
When you see the notation \( \int_{a}^{b} f(x) \, dx \), it means we are calculating the integral of function \( f(x) \) from \( x = a \) to \( x = b \). This is known as a "definite integral." Here’s what each part signifies:
- \( a \) and \( b \) are the lower and upper bounds of the interval respectively.
- \( f(x) \) is the function being integrated.
- The result gives the net area under the \( f(x) \) curve from \( a \) to \( b \).
Exploring Integration Techniques
Integration techniques are methods used to solve integrals, which are not always straightforward. In our example, we integrated a polynomial function, which is relatively simple using basic integration rules.
For beginners, it's essential to understand:
For beginners, it's essential to understand:
- Power Rule for Integration: To integrate \( x^n \), you use \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) where \( n eq -1 \).
- Linearity of Integration: This means you can integrate a sum or difference of functions by integrating each term separately, i.e., \( \int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx \).
Understanding Polynomial Functions
Polynomial functions are algebraic expressions involving terms with non-negative integer exponents. They are expressed in the form \( a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 \), with coefficients \( a_n, a_{n-1}, a_1, \) and \( a_0 \) as constants.
In our exercise, the function \( f(x) = 3x^2 - 6x + 1 \) is a quadratic polynomial. These features make them a significant area of study due to their simplicity and wide application in modeling real-world situations. Key properties include:
In our exercise, the function \( f(x) = 3x^2 - 6x + 1 \) is a quadratic polynomial. These features make them a significant area of study due to their simplicity and wide application in modeling real-world situations. Key properties include:
- Degree of the Polynomial: The highest exponent of \( x \) in the polynomial. Here, it's 2, making \( f(x) \) a quadratic polynomial.
- Coefficients: Numbers multiplying the variable terms, such as 3, -6, and 1 in our polynomial.
- Roots: Values of \( x \) where the polynomial equals zero. Determining these can be critical in solving equations and graphing.