Chapter 4: Problem 2
Find the average value of the function on the given interval. $$ f(x)=5 x^{2} ; \quad[1,4] $$
Short Answer
Expert verified
The average value is 35.
Step by step solution
01
Understanding the Problem
To find the average value of a function on a given interval, we'll use the formula for the average value of a function, which involves integrating the function over the given interval and dividing by the interval's length. For this problem, the function is \( f(x) = 5x^2 \) and the interval is \([1, 4]\).
02
Formula for Average Value
The formula for the average value of a function \( f \) on the interval \([a, b]\) is given by \[ \text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \] where \( a = 1 \) and \( b = 4 \) for this problem.
03
Setting Up the Integral
Substitute the given function \( f(x) = 5x^2 \) into the integral. Setup the integral as \[ \int_{1}^{4} 5x^2 \, dx \]
04
Calculating the Integral
Evaluate the integral \( \int 5x^2 \, dx \). The antiderivative of \( 5x^2 \) is \( \frac{5}{3}x^3 \). Thus, the integral from 1 to 4 is \[ \left[ \frac{5}{3}x^3 \right]_1^4 = \frac{5}{3}(4^3) - \frac{5}{3}(1^3) \].
05
Computing the Definite Integral
Calculate \( \frac{5}{3}(4^3) - \frac{5}{3}(1^3)\). First compute \( 4^3 = 64 \) and \( 1^3 = 1 \). Thus, \[ \frac{5}{3}(64) - \frac{5}{3}(1) = \frac{320}{3} - \frac{5}{3} = \frac{315}{3} = 105 \].
06
Using the Average Value Formula
Using the formula \[ \text{Average Value} = \frac{1}{4-1} \cdot 105 \], simplify to \[ \frac{105}{3} = 35 \].
07
Conclusion
Thus, the average value of the function \( f(x) = 5x^2 \) on the interval \([1, 4]\) is 35.
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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 represents the accumulation of a quantity over an interval on the real line. This area under the curve is crucial for finding the average value of a continuous function over a certain interval. For example, when evaluating the definite integral of a function like \( f(x) = 5x^2 \) over an interval \([1, 4]\), the integral gives us the total area accumulated between the curve \( 5x^2 \) and the x-axis, from \( x = 1 \) to \( x = 4 \).
- To calculate the definite integral, we identify the antiderivative, which is a function whose derivative is the integrand.
- We then apply the limits of integration to this antiderivative to find the net signed area or total accumulation.
- In mathematical terms, for a continuous function \( f(x) \), this can be expressed as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.
Antiderivative
An antiderivative of a function is essentially the reverse of taking a derivative. While a derivative tells us the rate of change of a function, the antiderivative helps us rebuild the original function. In the context of calculating the average value of a function, finding the antiderivative is an essential step in evaluating a definite integral.For \( f(x) = 5x^2 \), its antiderivative is \( F(x) = \frac{5}{3}x^3 \). This means that if you were to differentiate \( \frac{5}{3}x^3 \), you'd regain the original function \( 5x^2 \).
- To find an antiderivative, you often increase the power of \( x \) by one and divide by the new power.
- Don't forget the constant of integration when computing general antiderivatives, although it is usually unnecessary for definite integrals.
- Use the antiderivative to compute the definite integral by evaluating it at the bounds of the given interval.
Interval Notation
Interval notation is a simple way of representing a segment of the number line and is particularly useful in calculus for specifying the boundaries over which a function is integrated or evaluated. For example, the interval notation \([1, 4]\) designates all the numbers from 1 to 4, including the endpoints. This is crucial in our problem because it dictates the start and end points for the definite integral.
- Square brackets, as in \([a, b]\), indicate that the endpoints \(a\) and \(b\) are included in the interval (closed interval).
- If a parenthesis is used, such as \((a, b)\), this indicates that the endpoints are not included (open interval).
- Interval notation is not only employed to specify limits of integration but is essential in domains and ranges of functions.