Question

Calculate the average value of the given functions over the specified interval: (a) \(f(x)=x^{3}\) across \([1,3]\) (b) \(f(x)=\frac{1}{x}\) across \([1,2]\) (c) \(f(t)=\sqrt{t}\) across \([0,2]\) (d) \(f(z)=z^{3}-1\) across \([-1,1]\) (e) \(f(t)=\frac{1}{t^{2}}\) across \([-3,-2]\)

Short answer

Question: Calculate the average value of the given functions over the specified intervals. (a) f(x) = x^3 on [1, 3] (b) f(x) = 1/x on [1, 2] (c) f(t) = √t on [0, 2] (d) f(z) = z^3 - 1 on [-1, 1] (e) f(t) = 1/t^2 on [-3, -2] Answer: (a) The average value of the function f(x) = x^3 on the interval [1, 3] is 10. (b) The average value of the function f(x) = 1/x on the interval [1, 2] is ln(2). (c) The average value of the function f(t) = √t on the interval [0, 2] is (2√2)/3. (d) The average value of the function f(z) = z^3 - 1 on the interval [-1, 1] is 0. (e) The average value of the function f(t) = 1/t^2 on the interval [-3, -2] is 1/6.

Step-by-step solution

(a) f(x) = x^3 on [1, 3]

For the given function, \(f(x)=x^{3}\) and the interval is [1,3]. So, a = 1, and b = 3. To find the average value, we apply the formula: 1. Find the integral of the function: \(\int x^3 dx = \frac{x^4}{4} + C\). 2. Evaluate the integral using the limits a and b: \(\frac{1}{3-1}\left[\frac{3^4}{4}-\frac{1^4}{4}\right] = \frac{1}{2}\left[\frac{81}{4}-\frac{1}{4}\right]\). 3. Simplify the expression: \(f_{avg} = \frac{1}{2}\left[\frac{80}{4}\right] = \frac{1}{2}\cdot 20 = 10\). The average value of the function \(f(x)=x^3\) over the interval [1, 3] is 10.

(b) f(x) = 1/x on [1, 2]

For the function \(f(x)=\frac{1}{x}\) and the interval [1, 2], a = 1 and b = 2. To find the average value, follow these steps: 1. Find the integral of the function: \(\int \frac{1}{x} dx = \ln|x| + C\). 2. Evaluate the integral using the limits a and b: \(\frac{1}{2-1}\left[\ln|2|-\ln|1|\right] = \ln|2|-\ln|1|\). 3. Simplify the expression: \(f_{avg} = \ln(2)\). The average value of the function \(f(x)=\frac{1}{x}\) over the interval [1, 2] is \(\ln(2)\).

(c) f(t) = √t on [0, 2]

In this case, \(f(t)=\sqrt{t}\) and the interval is [0, 2]. Let a = 0 and b = 2. Then calculate the average value using the following steps: 1. Find the integral of the function: \(\int \sqrt{t} dt = \int t^{1/2} dt = \frac{2}{3}t^{3/2} + C\). 2. Evaluate the integral using the limits a and b: \(\frac{1}{2-0}\left[\frac{2}{3}(2)^{3/2}-\frac{2}{3}(0)^{3/2}\right] = \frac{1}{2}\left[\frac{2}{3}(2\sqrt{2})\right]\). 3. Simplify the expression: \(f_{avg} = \frac{1}{2}\cdot \frac{4\sqrt{2}}{3} = \frac{2\sqrt{2}}{3}\). The average value of the function \(f(t)=\sqrt{t}\) over the interval [0, 2] is \(\frac{2\sqrt{2}}{3}\).

(d) f(z) = z^3 - 1 on [-1, 1]

The given function is \(f(z) = z^3 - 1\) and the interval is [-1, 1]. Define a = -1 and b = 1. Calculate the average value using the following steps: 1. Find the integral of the function: \(\int (z^3 - 1) dz = \frac{z^4}{4} - z + C\). 2. Evaluate the integral using the limits a and b: \(\frac{1}{1-(-1)}\left[\left(\frac{1^4}{4}-1\right)-\left(\frac{(-1)^4}{4}-(-1)\right)\right] = \frac{1}{2}\left[\frac{1}{4}-1-\frac{1}{4}+1\right]\). 3. Simplify the expression: \(f_{avg} = \frac{1}{2}\left[0\right] = 0\). The average value of the function \(f(z)=z^3-1\) over the interval [-1, 1] is 0.

(e) f(t) = 1/t^2 on [-3, -2]

The given function for this part is \(f(t)= \frac{1}{t^2}\) and the interval is [-3, -2]. Assign a = -3 and b = -2, and then calculate the average value as follows: 1. Find the integral of the function: \(\int \frac{1}{t^2} dt = \int t^{-2} dt = -t^{-1} + C=-\frac{1}{t}+C\). 2. Evaluate the integral using the limits a and b: \(\frac{1}{-2-(-3)}\left[-\frac{1}{-2}-\left(-\frac{1}{-3}\right)\right] = -\left[-\frac{1}{2}+\frac{1}{3}\right]\). 3. Simplify the expression: \(f_{avg} = -\left[-\frac{1}{6}\right] = \frac{1}{6}\). The average value of the function \(f(t)=\frac{1}{t^2}\) over the interval [-3, -2] is \(\frac{1}{6}\).

Key concepts

Integral Calculus

Integral calculus is a fundamental part of mathematics that deals with the concept of integration, which is essentially about finding the total accumulation of quantities. Imagine you're accumulating water in a tank; integration helps measure the total volume. It is the mathematical tool used to calculate areas, volumes, and other quantities when a function describes how these quantities change.
For example, when tasked with finding the average value of a function over a certain interval, it's the principle of integration that allows us to sum up all the small pieces: we integrate the function over the interval and then divide by the interval's width.
Using integration, we handle problems related to calculating areas under curves, determining distances by accumulating small intervals, and finding the volume of shapes bounded by curves. Understanding these basic aspects of integral calculus is crucial in tackling more advanced calculus problems.

Definite Integrals

Definite integrals serve a special purpose in integral calculus because they allow us to evaluate the accumulation of a function between specific limits.
When you see an expression like \( \int_a^b f(x) \, dx \), it is instructing us to find the area under the curve of \( f(x) \) from \( x = a \) to \( x = b \).
In practical terms, a definite integral computes the accumulated change of a function between two points. It's like calculating how much distance you've traveled between two landmarks along a road.
Each of the exercises in the original problem utilized definite integrals. By leveraging the limits and evaluating these integrals, we can then calculate the average values of the provided functions over their respective intervals, showcasing the power of definite integrals in solving real-world problems.

Calculus Problems

Solving calculus problems is all about understanding and applying the fundamental principles of calculus. These problems often require breaking complex functions into simpler parts, finding derivatives or integrals, and then assembling the results to derive a solution.
Take, for instance, the average value problem given in the exercise. To solve this, we start by identifying the function and the interval. Then, using integration, we determine the total area or accumulation over that interval. Lastly, dividing this sum by the total length of the interval gives us the average value.
Tackling such problems requires practice in recognizing which calculus principles apply to which scenarios and developing a workflow to solve them efficiently. This approach not only helps in solving textbook problems but also in applying these concepts to real-world situations, like physics or engineering applications.

Mathematical Analysis

Mathematical analysis is essentially the underpinning of calculus. It provides the rigorous framework and methodologies for developing and proving the theorems that calculus relies on.
In the realm of calculating average function values, mathematical analysis helps validate the steps by ensuring that we adhere to the right limits and correctly evaluate functions. For instance, using limits, analysis proves the methodology behind calculating integrals and average values.
Understanding mathematical analysis is key to appreciating why calculus works. It helps us delve deeper into the nature of functions and their behavior as they approach certain points or infinity. Theoretical insights from analysis provide the tools needed for crafting elegant solutions to seemingly complex calculus problems, offering a deeper comprehension of not just how but why these methods work.