Question

In Exercises \(29-32,\) express the desired quantity as a definite integral and evaluate the integral using Theorem \(2 .\) Find the calories burned by a walker burning 300 calories per hour between \(6 : 00\) P.M. and \(7 : 30\) P.M.

Short answer

The walker burned a total of 450 calories.

Step-by-step solution

Convert Given Time To Hours

1 hour 30 minutes is equivalent to 1.5 hours.

Express Quantity As A Definite Integral

We can express the total number of calories burned as a definite integral from 0 to 1.5 (the time duration in hours) of 300 (the rate at which calories are burned) dt.

Evaluate The Integral

Using Theorem 2, we can calculate the definite integral as follows: \[\int_0^{1.5} 300 dt = 300 \int_0^{1.5} dt = 300 [t]_0^{1.5} = 300 * (1.5 - 0)\]

Key concepts

Calories Burned Calculation

The calories burned by someone during a physical activity can be calculated by understanding the rate at which calories are burned and the duration of the activity. In our exercise, the walker burns calories at a constant rate of 300 calories per hour. To find the total calories burned over a specific time period—here, between 6:00 P.M. and 7:30 P.M.—we need to perform a simple calculation. We consider the rate of calorie expenditure per unit of time, and multiply it by the duration of the activity in the same time units.

For a clearer understanding, imagine the calorie burning as a function of time, where the rate doesn't change over the considered interval. If we represent this scenario graphically, it results in a rectangle where one side is the duration and the other is the constant rate of calorie burn. The area of this rectangle would give us the total calories burned, which is analogous to finding the definite integral of a constant function over an interval.

Theoretical Application of Integrals

Integrals have extensive applications in theoretical and applied mathematics, particularly when it comes to calculating quantities that accumulate over a period. In our exercise context, integrals are used to calculate the total effect of a rate of change over time—specifically, the total calories burned. The function representing the calorie burn rate can be integrated over the time interval of the activity to find the cumulative calories burned.

Consider the act of walking as causing the burner of calories to flow at a constant rate, analogous to the flow of water from a faucet into a bathtub. The integral, in this context, represents the total volume of water (or calories burned) collected in the bathtub (or body) over the flowing period. This illustrates the theoretical application of integrals as a tool for aggregating quantities over a given interval.

Integral Evaluation

Evaluating a definite integral is the process of calculating the total accumulation of a function over a certain interval. In our walker's scenario, we are given a constant function representing the calorie burning rate, making the evaluation straightforward. The integral of a constant function is simply the product of the constant value and the width of the interval over which we integrate.

The steps to evaluate the definite integral involve setting up the integral with the appropriate limits, which are the start and end points of the time interval, and then calculating the result. This is a vital concept in calculus that is utilized frequently in real-world scenarios, as it helps to quantify accumulated change in various contexts.

Theorem Application in Calculus

Theorems in calculus, such as the Fundamental Theorem of Calculus, play a crucial role in linking the concept of differentiation with integration. When we refer to 'Theorem 2', we invoke the principles laid out by such theorems to simplify and calculate definite integrals. In our example, the application of the theorem streamlines the process of finding the total calories burned by providing a formulaic approach to evaluate the integral of a function over a specified interval.

By applying the theorem to evaluate the definite integral, we bypass the need for more complex integration techniques, especially when dealing with simple functions, such as a constant function in the exercise. Having a firm grasp of these theorems reduces the amount of computation and understanding required to solve calculus problems, making them more accessible and manageable for students.