Problem 57 - Essential Calculus

Question

Find an equation for the function \(f\) that has the indicated derivative and whose graph passes through the given point. $$ f^{\prime}(x)=\cos \frac{x}{2}, \quad(0,3) $$

Step-by-Step Solution

Answer

The function that has the indicated derivative and passes through the given point is \(f(x) = 2\sin(\frac{x}{2}) + 3\).

Step 1: Find the Integral of the Derivative Function

To find the function \(f(x)\), we need to take the integral of the derivative function \(f'(x) = \cos \frac{x}{2}\). Using the identity that the integral of \(\cos(u)\) where \(u\) is a function of \(x\) is \(2\sin(u) + C\), we can find the integral of our function. In this case, \(u = \frac{x}{2}\), so the integral of \(f'(x)\), or \(f(x)\), is \(2\sin(\frac{x}{2}) + C\).

Step 2: Use the Given Point to Find the Constant of Integration

Now we need to find the value of \(C\). To do this, we'll use the provided point (0,3) which lies on the graph of \(f(x)\). That means when \(x = 0\), \(f(x) = 3\). By substituting these values into our function, we get: \(3 = 2\sin(0) + C\). Simplifying gives \(\sin(0) = 0\) and therefore, \(C = 3\).

Step 3: Write the Final Function

Now that we have \(C\), we can write the complete function by substituting \(C\) into \(f(x)\). This gives us: \(f(x) = 2\sin(\frac{x}{2}) + 3\).

Key Concepts

Antiderivatives

Antiderivatives, often referred to as indefinite integrals, are the reverse process of differentiation. They help us find the original function given its derivative. The goal is to determine a function whose derivative matches the given function. In our exercise, we are provided with the derivative of the function, which is \( f'(x) = \cos \frac{x}{2} \).
The task requires finding the original function, \( f(x) \).
This involves integrating the given derivative function using integration techniques and any relevant integral identities.When integrating \( \cos u \), where \( u \) is a function of \( x \), the result is \( \sin u \) plus a constant of integration. The \( dx \) in the integral ensures proper notation and indicates the variable of integration.
Hence, for \( f'(x) = \cos \frac{x}{2} \), the antiderivative is determined as \( 2\sin(\frac{x}{2}) + C \). The constant \( 2 \) arises when adjusting for the derivative \( \frac{1}{2} \) from \( \frac{x}{2} \). This step yields the general form of the original function before applying initial conditions.

Constant of Integration

Whenever you find an antiderivative, there is what we call a 'constant of integration'. This constant, denoted as \( C \), represents an infinite number of functions that the given antiderivative could be. Why the many possibilities? Because differentiation wipes out constants, going the other way can add any number.In our problem, after integrating to find \( f(x) = 2\sin(\frac{x}{2}) + C \), you will notice the presence of this constant \( C \). To identify the specific function from the family of functions represented by the antiderivative, we incorporate a given point (in this case, \( (0,3) \)).
By substituting \( x = 0 \) into our found function, we should get \( f(x) = 3 \). Thus, \( 3 = 2\sin(0) + C \) simplifies to \( C = 3 \) since \( \sin(0) = 0 \). The constant of integration is extremely vital to transition from the general solution to the specific solution.

Definite Integrals

In contrast to what we've discussed so far, definite integrals provide the accumulated value of a function between two points. While our exercise focuses on finding an antiderivative, definite integrals involve evaluating the difference of an antiderivative at two bounds.The relevance of definite integrals is deeply rooted in their ability to compute net areas under a curve over a specific interval. Mathematically, if you have a function \( g(x) \) and you wish to find the integral from \( a \) to \( b \), you would calculate \[ \int_{a}^{b} g(x) \, dx = G(b) - G(a) \] where \( G(x) \) is the antiderivative of \( g(x) \).Although definite integrals are not used in our current problem, they show how powerful calculus can be beyond finding antiderivatives.
This concept introduces an application of integration with limits, pointing to growth or change between specific instances or positions.