Problem 26 - Calculus for Biology and Medicine

Question

Let \(g(t)=\sin (\pi t) .\) Compute the average value of \(g(t)\) over the interval \([-1,1]\).

Step-by-Step Solution

Answer

The average value of \( g(t) = \sin(\pi t) \) over \([-1,1]\) is \( \frac{1}{\pi} \).

Step 1: Identify the Formula for Average Value

The average value of a function \( g(t) \) on an interval \([a, b]\) is given by the formula \[ \frac{1}{b-a} \int_{a}^{b} g(t) \, dt. \] Here, \(a = -1\) and \(b = 1\).

Step 2: Setup the Integral for Average Value

Using the formula from Step 1, we set up the integral as \( \frac{1}{1 - (-1)} \int_{-1}^{1} \sin(\pi t) \, dt \). This simplifies to \( \frac{1}{2} \int_{-1}^{1} \sin(\pi t) \, dt \).

Step 3: Integrate \(\sin(\pi t)\)

The integral of \( \sin(\pi t) \) is \( -\frac{1}{\pi} \cos(\pi t) \). Thus, evaluate \( \int_{-1}^{1} \sin(\pi t) \, dt = \left[-\frac{1}{\pi} \cos(\pi t)\right]_{-1}^{1} \).

Step 4: Evaluate the Definite Integral

Substitute the limits into the antiderivative: \( \left[-\frac{1}{\pi} \cos(\pi \cdot 1) \right] - \left[-\frac{1}{\pi} \cos(\pi \cdot (-1)) \right] \). This becomes \( \left[-\frac{1}{\pi} \cos(\pi) \right] + \frac{1}{\pi} \cos(-\pi) \).

Step 5: Simplify the Expression

Since \( \cos(\pi) = -1 \) and \( \cos(-\pi) = -1 \), the expression simplifies to \( \frac{1}{\pi} + \frac{1}{\pi} = \frac{2}{\pi} \).

Step 6: Find the Average Value

Multiply the result from Step 5 by \( \frac{1}{2} \) to find the average value: \( \frac{1}{2} \times \frac{2}{\pi} = \frac{1}{\pi} \).

Key Concepts

Integral Calculus Basics

Integral calculus is a branch of mathematics that focuses on finding the total size, value, or amount of a quantity. This is done using mathematical integration, which calculates the integral of functions—often yielding areas under curves or accumulated quantities.

• It involves two major concepts: the indefinite integral (or antiderivative) and the definite integral.
• An indefinite integral represents a family of functions, while a definite integral computes a numeric value over a specific interval.

Knowing how to integrate, or find the integral, is essential in various applications in physics, engineering, and beyond.
Understanding these concepts allows you to solve a wide range of real-world problems such as finding distances and areas.

Understanding Definite Integrals

A definite integral is a vital concept in calculus used to calculate the exact area under a curve within a certain interval. It brings precision to measurements that the indefinite integral can't provide.

• The definite integral of a function from one limit to another is represented as \(\int_{a}^{b} f(x) \, dx\).
• The limits \(a\) and \(b\) demarcate the starting and ending points of integration.
• The result is a numerical value that summarizes the information contained between these two points.

Definite integration is crucial for solving problems involving total quantities—like in physics for computing work done by a force over a displacement.

Exploring Trigonometric Functions

Trigonometric functions are mathematical functions that relate angles of a triangle to the lengths of the sides of the triangle. They play an essential role in various areas of mathematics and are integral in calculating wave patterns, circles, and oscillatory phenomena.

• Key trig functions include sine, cosine, and tangent, all of which are periodic and oscillate between values.
• For example, the sine function, \(\sin(x)\), oscillates between -1 and 1.
• It is often used in physics to model simple harmonic motion, such as waves and pendulums.

Understanding these functions allows you to solve complex equations involving angles and has applications ranging from engineering to computer graphics.