Question

Find the mean value of each of the following functions across the interval stated: (a) \(f(t)=\sin t,[0, \pi]\) (b) \(f(t)=t^{3},[1,5]\) (c) \(f(t)=\mathrm{e}^{3 t},[0,2]\) (d) \(f(t)=20+2 \sin \pi t,[0,0.5]\)

Short answer

Question: Find the mean value of each function across the given interval. (a) \(f(t)=\sin t, [0, \pi]\) Answer: \(M \approx 0.6366\) (b) \(f(t)=t^{3},[1,5]\) Answer: \(M = 39\) (c) \(f(t)=\mathrm{e}^{3 t},[0,2]\) Answer: \(M \approx 122.92\) (d) \(f(t)=20+2 \sin \pi t,[0,0.5]\) Answer: \(M = 20\)

Step-by-step solution

(a) \(f(t)=\sin t, [0, \pi]\)#

Apply the mean value theorem formula: \(M = \frac{1}{\pi - 0}\int_{0}^{\pi}\sin(t) dt\) Now, integrate \(\sin(t)\) over the interval \([0, \pi]\): \(\int_{0}^{\pi}\sin(t) dt = \Big[-\cos(t)\Big]_{0}^{\pi} = (-\cos(\pi) - (-\cos(0))) = 2\) Finally, calculate the mean value: \(M = \frac{2}{\pi} \approx 0.6366\)

(b) \(f(t)=t^{3},[1,5]\)#

Apply the mean value theorem formula: \(M = \frac{1}{5 - 1}\int_{1}^{5} t^3 dt\) Now, integrate \(t^3\) over the interval \([1, 5]\): \(\int_{1}^{5}t^3 dt = \Big[\frac{1}{4}t^4\Big]_{1}^{5} = \frac{1}{4}(5^4 - 1^4) = \frac{624}{4}\) Finally, calculate the mean value: \(M = \frac{624}{4(5 - 1)} = \frac{624}{16} = 39\)

(c) \(f(t)=\mathrm{e}^{3 t},[0,2]\)#

Apply the mean value theorem formula: \(M = \frac{1}{2 - 0}\int_{0}^{2} \mathrm{e}^{3t} dt\) Now, integrate \(\mathrm{e}^{3t}\) over the interval \([0, 2]\): \(\int_{0}^{2}\mathrm{e}^{3t} dt = \Big[\frac{1}{3}\mathrm{e}^{3t}\Big]_{0}^{2} = \frac{1}{3}(\mathrm{e}^{6} - \mathrm{e}^{0})\) Finally, calculate the mean value: \(M = \frac{\frac{1}{3}(\mathrm{e}^{6} - 1)}{2} = \frac{\mathrm{e}^{6} - 1}{6} \approx 122.92\)

(d) \(f(t)=20+2 \sin \pi t,[0,0.5]\)#

Apply the mean value theorem formula: \(M = \frac{1}{0.5 - 0}\int_{0}^{0.5} (20+2\sin(\pi t)) dt\) Now, integrate \(20 + 2\sin(\pi t)\) over the interval \([0, 0.5]\): \(\int_{0}^{0.5} (20 + 2\sin(\pi t)) dt = 20\int_{0}^{0.5} dt + 2\int_{0}^{0.5} \sin(\pi t) dt\) \(= 20\Big[t\Big]_{0}^{0.5} + 2\Big[-\frac{1}{\pi}\cos(\pi t)\Big]_{0}^{0.5} = 20(0.5) - 2(\frac{1}{\pi} - \frac{1}{\pi}) = 10\) Finally, calculate the mean value: \(M = \frac{10}{0.5} = 20\) Thus, the mean values are: (a) \(M \approx 0.6366\) (b) \(M = 39\) (c) \(M \approx 122.92\) (d) \(M = 20\)

Key concepts

Integral Calculus

Integral calculus is central to understanding how to calculate the mean value of a function over an interval. This mathematical field focuses on integration—the process of finding the integral of a function. An integral can be thought of as the "total amount" of some quantity, accumulated over a specific range. This plays a crucial role when determining the average value of a function with the Mean Value Theorem for Integrals.

To find the mean value of a function, you first need to find its definite integral over the given interval. A definite integral is typically represented as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) denote the interval's endpoints. The operation of integration sums up infinitely small quantities of the function along the interval, giving us an accumulated measure of the function's behavior. Subsequently, this accumulated measure is divided by the length of the interval to find the average value.

• Integration helps in measuring areas under curves, representing functionalities like total distance, total population, or total growth.
• The concept of the definite integral is utilized to calculate exact values, such as areas and average values.
• Through integral calculus, we apply the systematic rules of integration to solve meaningful real-world problems.

Function Analysis

Understanding function analysis is vital when applying the Mean Value Theorem for Integrals. This concept involves examining the characteristics and behaviors of mathematical functions. Each function can express a specific relationship between input and output values. Analyzing this relationship helps in predicting the behavior of the function across different intervals, which is a key step in applying the Mean Value Theorem.

With function analysis, you evaluate properties such as continuity, derivative existence, and integrability. The Mean Value Theorem for Integrals requires a function to be continuous over the closed interval \([a, b]\) to ensure an average value exists. Here are fundamental aspects of analyzing functions in the context of integral calculus:

• Continuity Check: Verifying that the function is continuous means there are no breaks, jumps, or holes in the interval.
• Differentiability: A critical step while performing integration if further analysis on derivative existence is needed.
• Behavioral Insight: Understanding a function's growth, decay, or oscillation offers insights into how the average is derived over the interval.

Definite Integrals

Definite integrals are a core component of calculating the mean value of functions over specified intervals. They provide a precise measure of the accumulated quantity of a function. In the context of the Mean Value Theorem for Integrals, the definite integral \( \int_{a}^{b} f(x) \, dx \) is divided by the length of the interval \((b - a)\) to find the mean value.

A mean value essentially tells how a function behaves "on average" across an interval. This succinct summary relies on converting the integral's accumulated total into an average value. Here's how definite integrals and mean values are mutually interconnected:

• Integral Calculation: Determines the total sum that the mean value will depend on.
• Interval Division: By dividing the integral by the interval's length, the overall average can be computed.
• Practical Applicability: Great for understanding averages in contexts ranging from physics to economics.