Problem 6 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

The amount of a certain chemical in a mixture varies with time. If \(g(t)=5 e^{-t}\) is the number of grams of the chemical at time \(t\), what is the average number of grams of the chemical in the mixture on the time interval \([0,1] ?\)

Step-by-Step Solution

Answer

The average amount of the chemical in the mixture in the interval \([0,1]\) is approximately 1.82 grams.

Step 1: Notate the Given Variables

From the problem, the given function is \(g(t)=5e^{-t}\), and the time range is given as [0,1]. So a=0 and b=1.

Step 2: Implement the Average Value Formula

The formula for the average value of a function \(f(t)\) over the interval [a,b] is \(\frac {1}{b-a} ∫_{a}^{b} f(t) dt\). Substitute the given function, limits, and calculate the integral. The average value \(A = \frac {1}{1-0}∫_{0}^{1} 5e^{-t} dt = ∫_{0}^{1} 5e^{-t} dt\).

Step 3: Solve the Integral

To solve the integral, use the rule ∫e^{kt} dt = \(\frac{1}{k}e^{kt}\) for k ≠ 0. Let \(I = 5 ∫_{0}^{1} e^{-t} dt\). Substituting the values, \(I = [ -5e^{-t}]_{0}^{1} = -5(e^{-1} - e^{0})\).

Step 4: Simplify and Compute the Expression

Substituting \(e^{0}\) by 1, \(I = -5(\frac{1}{e} -1)\). Simplify it to obtain the average amount of the substance during the interval.

Key Concepts

Definite Integral

In calculus, the concept of a definite integral is foundational for understanding how to compute the accumulation of quantities, such as areas under a curve or the total value of a function across an interval. The definite integral of a function represents the net area bounded by the function's graph, the x-axis, and the vertical lines at the endpoints of the interval. It's calculated using the limits of integration, which in this case are the boundaries of the time interval under consideration, [0,1].

For the given function, representing the concentration of a chemical over time, we seek to find the total amount of the substance within this time frame. Calculating the definite integral of the function from 0 to 1 allows us to do precisely this. We would express the integral as \[ \int_{0}^{1} g(t) \, dt \], where \( g(t) = 5e^{-t} \). By solving this integral, one determines the precise amount of the chemical that is present over the entire interval.

Exponential Functions

Exponential functions are a significant category of mathematical functions that play a key role in modeling growth and decay processes. In the given problem, the function \( g(t) = 5e^{-t} \) is an exponential function where the base is the mathematical constant \( e \), which is approximately equal to 2.718. This particular function models the decay of the chemical concentration over time.

In exponential decay, the rate of change decreases over time, which is depicted by the negative exponent in the function. As \( t \) increases, \( e^{-t} \) becomes smaller, indicating that there is less of the chemical in the mixture as time goes on. Understanding the behavior of exponential functions is essential for correctly interpreting the changes in concentration and for computing integrals that involve such functions.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration, two cornerstone ideas in calculus. It states that if a function is continuous over an interval and has an antiderivative on that interval, then the definite integral of the function can be evaluated by using the values of its antiderivative at the endpoints.

In practical terms, this theorem allows us to evaluate the integral involved in our problem in Step 3. To solve for the integral \( I = 5 \int_{0}^{1} e^{-t} dt \), we find the antiderivative of \( 5e^{-t} \), which yields \( -5e^{-t} \). Applying the Fundamental Theorem of Calculus, we evaluate this antiderivative at the endpoints of our interval, 0 and 1. Substraction of these values gives us the net area under the curve, which, when multiplied by \( \frac{1}{b-a} \) as per the average value formula, provides the average concentration of the chemical in the time range given.