Problem 32 - Applied Calculus

Question

Find an antiderivative \(F(x)\) with \(F^{\prime}(x)=\) \(f(x)\) and \(F(0)=0\). Is there only one possible solution? $$ f(x)=e^{x} $$

Step-by-Step Solution

Answer

The antiderivative is \( F(x) = e^x - 1 \), and it is unique for \( F(0)=0 \).

Step 1: Find the General Antiderivative

The antiderivative of a function involves finding a function whose derivative is the given function. Here, we are given the derivative \( f(x) = e^x \). The antiderivative of \( e^x \) is the function itself because the derivative of \( e^x \) is still \( e^x \). Therefore, a general antiderivative of \( f(x) = e^x \) is \( F(x) = e^x + C \), where \( C \) is a constant of integration.

Step 2: Use the Initial Condition

We are given the condition \( F(0) = 0 \). Substitute \( x = 0 \) into the general antiderivative \( F(x) = e^x + C \). This gives us \( F(0) = e^0 + C = 1 + C = 0 \). Solving for \( C \), we find \( C = -1 \).

Step 3: Determine the Specific Antiderivative

Substituting back \( C = -1 \) into the general solution, we find the specific antiderivative \( F(x) = e^x - 1 \). This function satisfies both \( F'(x) = e^x \) and \( F(0) = 0 \).

Key Concepts

Calculus

Calculus is a branch of mathematics that studies how things change. It has two main parts: differentiation and integration. Differentiation is about finding rates of change, while integration is about finding quantities when rates of change are known.
In the given exercise, we are asked to find an antiderivative, which is part of integral calculus. The antiderivative, also called an indefinite integral, is essentially the opposite of taking a derivative. If you know the rate at which something changes, calculus allows you to find the original function that describes it.
For the function we have, which is \(f(x) = e^x\), we need to find a function \(F(x)\) such that its derivative \(F'(x)\) equals \(f(x)\). In this case, because the derivative \(\frac{d}{dx}(e^x) = e^x\), the function \(F(x) = e^x + C\) is an antiderivative, where \(C\) is some constant.

Initial Condition

An initial condition in math provides specific information about a function at a particular point. This is crucial when finding a specific solution from a family of solutions, like when determining a particular antiderivative from its general form.
In the exercise, the initial condition given is \(F(0) = 0\). This means that at \(x = 0\), the function \(F(x)\) must equal 0. We use this to find the specific value of the constant \(C\) in the antiderivative.
By plugging in \(x = 0\) into the general antiderivative \(F(x) = e^x + C\), we set \(F(0) = 1 + C\). Since \(F(0)\) needs to be 0, solving \(1 + C = 0\) gives \(C = -1\). This specific value of \(C\) helps us find the specific solution rather than just any solution from an infinite set.

Constant of Integration

The constant of integration \(C\) is a fundamental concept when dealing with antiderivatives. It appears because of the indefinite nature of integration in calculus. When you take a derivative of a constant value, it becomes 0. Thus, when finding an antiderivative, any constant added to the function could also be valid.
Without more information, the constant of integration allows the antiderivative to represent a family of possible functions. Each different value of \(C\) represents a different member of this family.
In the exercise provided, starting with the general antiderivative \(F(x) = e^x + C\), the initial condition \(F(0) = 0\) guides us to find \(C = -1\). This makes the specific function \(F(x) = e^x - 1\), which is the particular solution to the given problem. The constant of integration is essential since it accounts for the vertical shift possible in a function's graph.