Question

In Exercises \(21-24\) , solve the initial value problem using the Fundamental Theorem. (Your answer will contain a definite integral.) \(G^{\prime}(s)=\sqrt[3]{\tan s}\) and \(G(0)=4\)

Short answer

\(G(s) = \int \sqrt[3]{\tan{s}} \,ds + 4\)

Step-by-step solution

Integrate using the Fundamental theorem of Calculus

The Fundamental Theorem of Calculus states that if a function has a derivative, its antiderivative can be found by integrating the derivative. Therefore, we integrate \(G'(s)\) to get the original function. This is done as follows: \[G(s) = \int G'(s) \,ds = \int \sqrt[3]{\tan{s}} \,ds\] Here, we introduce a constant of integration C such that: \[G(s) = \int \sqrt[3]{\tan{s}} \,ds + C\] . The farther calculation of the integral \(\int \sqrt[3]{\tan{s}} \,ds\) is not general possible by elementary functions.

Determine the constant of integration

We are given the initial condition \(G(0) = 4\). Using this, we can find the value of the constant of integration C. Substituting \(s=0\) and \(G(0)=4\) into \(G(s) = \int \sqrt[3]{\tan{s}} \,ds + C\), we obtain: \[4 = \int \sqrt[3]{\tan{0}} \,ds + C \rightarrow 4 = \int 0 \,ds + C \rightarrow 4 = C\]. Thus the constant of integration is \(C=4\).

State the final solution

Given that \(C=4\), we can now write the final solution of the initial value problem as: \[G(s) = \int \sqrt[3]{\tan{s}} \,ds + 4\] Thus, this is the function \(G(s)\) satisfying the given conditions.

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone concept in calculus that connects differentiation with integration, two of the main operations in the subject. It ensures that if a function is continuous on an interval, then it has an antiderivative in this interval, and the definite integral of a function over that interval can be calculated using its antiderivatives. More elegantly put, it allows us to evaluate the definite integral of a function from point a to point b by knowing its antiderivative, and applying the evaluation from b to a.

Let's consider the function represented by the derivative denoted as \(G'(s)\) in our initial value problem. According to the Fundamental Theorem of Calculus, to acquire the original function \(G(s)\), we can integrate this derivative. So, we set up the integral, \[G(s) = \int G'(s) \,ds\], and proceed to calculate this integral. What this theorem guarantees is that the integration of the derivative retrieves the original function up to a constant—the constant of integration, designated as \(C\).

Antiderivative

An antiderivative, also known as an indefinite integral, is essentially the reverse of finding a derivative. If you have a function \(f(x)\), an antiderivative is another function \(F(x)\) such that \(F'(x) = f(x)\). It means that differentiating \(F(x)\) gives us back the function \(f(x)\). In context to our initial value problem, where we are given the derivative of a function \(G'(s)\), the antiderivative would be the original function \(G(s)\) that we seek. The process of finding \(G(s)\) involves integrating \(G'(s)\), which in the problem is \(\sqrt[3]{\tan s}\).

When we set out to integrate the derivative, we are essentially searching for all functions \(G(s)\) whose derivative is \(\sqrt[3]{\tan s}\). Now, this family of functions includes the original function plus a constant of integration, because differentiating a constant always yields zero, which is why the constant is not 'seen' in the derivative. Therefore, the antiderivative we're looking for takes the form \(G(s) = \int \sqrt[3]{\tan{s}} \,ds + C\).

Constant of Integration

When we integrate a function, we need to add a constant of integration because the process of differentiation discards any constant term the original function may have had. This constant is crucial, especially when solving an initial value problem, which provides an additional piece of information to find the specific value of the constant.

In an initial value problem like the one we are examining, the constant of integration is determined by substituting a known point of the function into the antiderivative we have found. For our exercise, we use the initial condition \(G(0) = 4\) to find the constant of integration \(C\). By plugging in \(s = 0\) into the antiderivative and setting it equal to 4, we're essentially finding the particular value of \(C\) that makes the antiderivative comply with the given condition. In this case, since \(\sqrt[3]{\tan{(0)}} \) is 0, we find that \(C = 4\), resolving the initial value problem for the specific constant that meets the condition \(G(0) = 4\).

Hence, the constant is not just any arbitrary 'add-on'; it is a definitive term that provides completeness to the solution. Every unique initial condition will produce a different constant of integration, differentiating one potential solution from another, yet all are part of the same family of antiderivatives.