Question

In Exercises \(21-24\) , solve the initial value problem using the Fundamental Theorem. (Your answer will contain a definite integral.) \(F^{\prime}(x)=e^{\cos x}\) and \(F(2)=9\)

Short answer

The function \(F(x)\) that solves the initial value problem is \(F(x) = \int e^{\cos x} dx + 9\).

Step-by-step solution

Integrate \(F'(x)\)

The formula \(F'(x) = e^{\cos x}\) is provided. To get \(F(x)\), we need to integrate \(F'(x)\). That gives us \(F(x) = \int e^{\cos x} dx + C\), where \(C\) is the constant of integration. However, we can't simplify the integral \(\int e^{\cos x} dx\) any further, so we'll leave it as such.

Use the initial condition to find \(C\)

We're told that \(F(2) = 9\). This means that when we substitute \(x = 2\) into our formula for \(F(x)\), the result should be \(9\). Plugging in these values gives \(9 = \int_2^2 e^{\cos x} dx + C\). Notice that the definite integral from any number to itself is always zero, so we get that \(9=C\).

Write the function \(F(x)\)

Now that we have \(C = 9\), we can substitute this back into our formula for \(F(x)\), obtaining that \(F(x) = \int e^{\cos x} dx + 9\). This is the equation of the function \(F\) that satisfies the initial value problem.

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation with that of integration. It's essentially the bridge that connects these two primary operations in calculus. It has two parts: the first part states that if we have a continuous function, the integral of its derivative over an interval is equal to the difference between the values of the original function at the endpoints of the interval. This means, symbolically, if we have a function f(x) that is continuous and its derivative f'(x), and F(x) is the antiderivative of f(x), then:

\[\begin{equation}F(b) - F(a) = \right.\int_a^b f'(x) \,dx\right.\end{equation}\]where a and b define the interval. The second part, which is particularly useful in solving the initial value problems, tells us that the integral of a function from a to x can be used to define another function F(x), which is the antiderivative of the original function, with F'(x) = f(x). In the context of an initial value problem like the exercise provided, this theorem allows us to derive the function F(x) by integrating the derivative F'(x), and then apply the initial condition to solve for any constants.

Definite Integral

A definite integral is a concept that represents the area under the curve of a function, bounded between two specific values, often referred to as the limits of integration. In mathematics, when we write

\[\begin{equation}\int_a^b f(x) \,dx\right.\end{equation}\]we are referring to the definite integral of the function f(x) from point a to b. The value of a definite integral is a number that represents the total accumulation of the quantity defined by f(x) over the interval [a, b].

In the context of the initial value problem given in the exercise, understanding definite integrals helps us to find the specific value of the antiderivative function at a point given the initial conditions. It is worth noting that the definite integral from any number to the same number is always zero, as there is no 'width' of an interval to measure the area under the curve. That's why in the solution, the definite integral from 2 to 2 is zero, which simplifies the process of finding the constant of integration.

Integration of Exponential Functions

The integration of exponential functions involves finding the antiderivative of functions that have the exponential function e raised to a power. The exponential function e^x has a unique property in calculus: it is its own derivative. This makes finding the antiderivative quite straightforward:

\[\begin{equation}\int e^x \,dx = e^x + C\right.\end{equation}\]where C is the constant of integration. However, integrating exponential functions becomes more challenging when the exponent is a function of x, such as e^{g(x)}. In general, we would need to use the method of substitution if g(x) has a simple derivative we can work with. In our given exercise, though, the integral

\[\begin{equation}\int e^{\cos x} \,dx\right.\end{equation}\]does not have an elementary antiderivative. Therefore, we keep it in its integral form in the final solution. In situations where the exponential function is more complex, like in our exercise, recognizing that not all integrals have closed-form solutions is vital. Sometimes, expressing the function as an indefinite integral is the most precise way to represent the antiderivative.