Question

If \(\mathrm{w}^{\prime}(\mathrm{t})\) is the rate of growth of a child in \(\mathrm{kg}\) per year, what does \(\int_{5}^{10} w^{\prime}(t) d t\) represent ?

Short answer

Answer: The integral \(\int_{5}^{10} w^{\prime}(t) dt\) represents the total growth in weight (in kg) of the child from age 5 to age 10.

Step-by-step solution

Note the problem as an integral

We are given that \(w'(t)\) is the rate of growth of a child in kg per year. To find the representation of \(\int_{5}^{10} w^{\prime}(t) dt\), we need to integrate the rate of growth of the child with respect to time from t = 5 to t = 10.

Apply the fundamental theorem of calculus

According to the fundamental theorem of calculus, the integral of a derivative function over an interval gives us the difference between the original function's values at the interval's endpoints. So \(\int_{5}^{10} w^{\prime}(t) dt = w(10) - w(5)\), where \(w(t)\) represents the total weight of the child at age \(t\).

Interpret the meaning of the result

Since \(w(10) - w(5)\) gives the difference in the child's weight between the ages of 10 and 5, the integral \(\int_{5}^{10} w^{\prime}(t) dt\) represents the total growth in weight (in kg) of the child from age 5 to age 10.

Key concepts

Rate of Growth

In the context of calculus, the rate of growth is a measure that tells us how quickly a quantity is changing with respect to another quantity. For instance, when studying the development of a child, we can talk about the weight of the child increasing over time. Imagine graphing the child's weight against their age; the steepness of the graph tells us how fast the child's weight is changing as they grow older.

Mathematically, this rate of growth is represented by a derivative, denoted as a function with a prime symbol, such as in the exercise with \(w'(t)\). If weight (\(w\)) is a function of time (\(t\)), then the rate of growth \(w'(t)\) is the derivative of the weight with respect to time, essentially describing how many kilograms the child is gaining each year.

Interpreting the Graph

When graphing the rate of growth, the area under the curve between two points on the graph (measured by an integral) reflects the total change in weight between those ages. If the rate of growth is constant, the graph will show a straight horizontal line, but if it's varying, the line may increase or decrease in steepness. Understanding the concept of rate of growth is crucial for interpreting the outcome of an integral that measures changes over an interval.

Fundamental Theorem of Calculus

One of the most profound principles in integral calculus is the Fundamental Theorem of Calculus. It bridges the gap between differentiation and integration, two core operations in calculus, and ties them together in a way that allows us to evaluate integrals more easily. Essentially, it states that if a function is continuous on an interval and has an antiderivative, then the definite integral of that function over that interval can be found by taking the difference between the values of the antiderivative at the endpoints of the interval.

To put it in perspective with our exercise, the function \(w'(t)\) is the derivative or rate of growth, while \(w(t)\) is the antiderivative or the actual weight function. The Fundamental Theorem tells us that integrating \(w'(t)\) from age 5 to age 10 (\( \int_{5}^{10} w'(t) dt \) ) is equivalent to finding the difference in weight (\(w(10) - w(5)\)).

In practice, this theorem simplifies the process of finding the total change over a period, avoiding the need for summing up infinite tiny changes, and provides a powerful tool for calculating areas under curves and the total accumulation of quantities.

Definite Integral Application

The concept of a definite integral has various real-world applications, particularly when it comes to quantifying total change, as shown in our textbook exercise. A definite integral is used to calculate the total accumulation of a quantity over an interval. In the scenario provided, the definite integral \( \int_{5}^{10} w'(t) dt \) represents the total increase in the child's weight from age 5 to age 10.

By completing this integral, we get a single number that sums up the cumulative effect of the continuous growth over those five years. Applications of definite integrals extend beyond just rates of growth in a child; they're used in physics for calculating work done by forces, in economics for computing total profit over time, and in engineering for assessing the total material needed for a project. The ability to convert a rate of growth into a measurable amount of change is crucial in many fields.

Importance of Units

An important consideration when dealing with definite integrals is to keep the units consistent. The rate of growth \(w'(t)\) in the exercise is measured in kilograms per year, meaning that the result of the integral will be in kilograms since we are integrating over time, measured in years. This attention to detail ensures the answers we find are not only mathematically correct but also make sense in the context of the problem.