Question

In Exercises \(21-26,\) construct a function of the form \(y=\int^{x} f(t) d t+C\) that satisfies the given conditions. $$\frac{d y}{d x}=\sqrt{3-\cos x},\( and \)y=4\( when \)x=-3$$

Short answer

The function that satisfies the given conditions is \(y = \int^{x} \sqrt{3 - \cos t}dt + C\), with \(C\) determined by the equation \(4 = \int^{-3} \sqrt{3 - \cos t} dt + C\).

Step-by-step solution

Identify the function

From the given, recognize that the function \(f(t) = \sqrt{3 - \cos t}\) because \(\frac{dy}{dx} = f(t)\). This function \(f(t)\) will be used in the integral.

Set up the integral

Now, it is necessary to construct \(y\) by integrating \(f(t)\). This transforms into \(y = \int^{x} \sqrt{3 - \cos t} dt + C\).

Solve for C

Since we know that \(y = 4\) when \(x = -3\), substitute these given values into our function, which gives us: \(4 = \int^{-3} \sqrt{3 - \cos t} dt + C\). This equation will allow us to solve for \(C\). Note that to actually calculate \(C\), a specific antiderivative would need to be known. Since it isn't in this case, our final answer will include the integral.

Key concepts

Antiderivative

An antiderivative, often termed as a primitive, is essentially a reverse operation of differentiation. If you're familiar with the process of finding the derivative of a function, understanding antiderivatives is about thinking of that process in reverse. An antiderivative of a function 'f' is a different function 'F' such that the derivative of 'F' is the original function 'f'. In other words, if you were to take the derivative of 'F', you would end up back at 'f'.

In the case of the exercise provided, we consider the function given by its derivative, which is \(f(t) = \sqrt{3 - \cos t}\). To construct the original function 'y' from its derivative \(\frac{dy}{dx}\), we seek the antiderivative of \(f(t)\). This involves integrating the function \(f(t)\) with respect to 't'.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of integration and differentiation, serving as one of the most important principles in calculus. It actually consists of two parts: the first part provides an antiderivative for continuous functions, while the second part allows the computation of definite integrals.

The first part of the theorem states that if a function is continuous over an interval and there is a function 'F' that is an antiderivative of 'f' on that interval, then \(F'(x) = f(x)\). The second part states that if 'F' is an antiderivative of a continuous function 'f' on the interval \[a, b\], then the definite integral of 'f' from 'a' to 'b' is given by \(F(b) - F(a)\).

For the exercise, this theorem is used to connect the given rate of change \(f(t) = \sqrt{3 - \cos t}\) with the accumulation function 'y', through integration, to construct the function based on the condition that \(y = 4\) when \(x = -3\).

Indefinite Integral

The indefinite integral, represented by the integral sign with no specified limits, is another way of referring to the antiderivative. It is used to find a general form of the original function without any boundaries. The outcome of an indefinite integral includes a constant of integration, 'C', because the antiderivation process adds an arbitrary constant - any number could be there and still satisfy the condition that differentiation of the result yields the original function.

Following the indefinite integral of \(f(t) = \sqrt{3 - \cos t}\), we write the general solution to our exercise as \(y = \int \sqrt{3 - \cos t} dt + C\). In this case, 'C' can be determined by using the additional condition given. When we are told that \(y = 4\) for \(x = -3\), we use this information to solve for 'C', adding concrete value to our otherwise general solution.