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Magazine Chapter 5: Problem 37
Find the average value of the function on the given interval. $$ y=x \text { on }[0,4] $$
Short Answer
Expert verified
The average value of the function on the interval is 2.
Step by step solution
01
Understand the Formula for Average Value
The average value of a continuous function \( f(x) \) on a closed interval \([a, b]\) is given by the formula: \[\text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx\] Here, \( f(x) = x \), \( a = 0 \), and \( b = 4 \).
02
Set Up the Integral
Substitute \( f(x) = x \), \( a = 0 \), and \( b = 4 \) into the formula for average value. The integral becomes: \[ \text{Average value} = \frac{1}{4-0} \int_{0}^{4} x \, dx \] This simplifies to: \[ \text{Average value} = \frac{1}{4} \int_{0}^{4} x \, dx \]
03
Calculate the Integral
Find the integral \( \int_{0}^{4} x \, dx \). The antiderivative of \( x \) is \( \frac{x^2}{2} \). Evaluate from 0 to 4: \[ \int_{0}^{4} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{4} = \frac{4^2}{2} - \frac{0^2}{2} = \frac{16}{2} = 8 \]
04
Calculate the Average Value
Use the result from the integral to find the average value. Substitute \( \int_{0}^{4} x \, dx = 8 \) into the formula: \[ \text{Average value} = \frac{1}{4} \times 8 = 2 \]
05
Conclusion: Interpret the Result
The average value of the function \( y = x \) on the interval \([0, 4]\) is \( 2 \), meaning that over the interval, the function has an average y-value of 2.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Continuous Function
A continuous function is a type of function that has no interruptions or breaks in its graph. Imagine you draw the function without lifting your pencil. That's continuance! In mathematical terms, a function is continuous on an interval if it is continuous at every point in that interval.
Why is continuity important for finding the average value of a function? Simply put, if a function is continuous, it's smoothly defined across the interval, ensuring that the integral (area under the curve) can be calculated precisely. In our exercise:
Why is continuity important for finding the average value of a function? Simply put, if a function is continuous, it's smoothly defined across the interval, ensuring that the integral (area under the curve) can be calculated precisely. In our exercise:
- The function \( f(x) = x \) is continuous over the interval \([0, 4]\).
- This continuity allows the use of definite integrals to find averages accurately.
Definite Integral
The definite integral is a fundamental concept in calculus. It finds the total accumulation of a quantity, such as area, volume, or mass, from one point to another.
Think of the definite integral as the sum of infinitely small quantities, creating a total over an interval. For our function \( y=x \), we calculated the integral over \([0, 4]\) to find the area under the curve:
Think of the definite integral as the sum of infinitely small quantities, creating a total over an interval. For our function \( y=x \), we calculated the integral over \([0, 4]\) to find the area under the curve:
- The integral \( \int_{0}^{4} x \, dx \) provides the total area from \( x = 0 \) to \( x = 4 \).
- This integral is solved by finding the antiderivative of the function.
- Our result was 8, which helped us find the average value.
Antiderivative
The antiderivative, or the indefinite integral, is a function whose derivative gives the original function. Essentially, it's the reverse process of differentiation.
In our problem, to integrate \( f(x) = x \), we needed its antiderivative. The antiderivative of \( x \) is \( \frac{x^2}{2} \). This function, when differentiated, would bring back \( x \).
In our problem, to integrate \( f(x) = x \), we needed its antiderivative. The antiderivative of \( x \) is \( \frac{x^2}{2} \). This function, when differentiated, would bring back \( x \).
- To solve the integral, we evaluate the antiderivative at the bounds \([0, 4]\).
- This means calculating \( \left[ \frac{x^2}{2} \right]_{0}^{4} \), resulting in the value 8.
Interval
The concept of an interval in mathematics describes a continuous range of values. These can be open or closed.
A closed interval, like in our example \([0, 4]\), includes both endpoints, 0 and 4. This ensures the function is evaluated precisely at these values. Intervals guide us in both calculating integrals and understanding the range over which a function's behavior is analyzed.
A closed interval, like in our example \([0, 4]\), includes both endpoints, 0 and 4. This ensures the function is evaluated precisely at these values. Intervals guide us in both calculating integrals and understanding the range over which a function's behavior is analyzed.
- For \( f(x) = x \), the interval tells us where to start and stop the integral.
- It also helps in finding the average value, as the interval length divides the total integral to get the average.
