Question

In Exercises \(29-34,\) find all possible functions \(f\) with the given derivative. $$f^{\prime}(x)=x$$

Short answer

The all possible functions \(f(x)\) with the given derivative are \(f(x) = \frac{x^2}{2} + C\), where \(C\) is any real number.

Step-by-step solution

Apply the Integral

To find the original function from the derivative, apply integration to the given derivative \(f^{\prime}(x)=x\). Hence, the integral of \(f^{\prime}(x)\) with respect to \(x\) is written as \(\int f^{\prime}(x) \, dx = \int x \, dx\).

Evaluate the Integral

The integral of \(x\) with respect to \(x\) is \(\frac{x^2}{2}\). This follows the general power rule for integration, \(\int x^n \, dx = \frac{x^{n+1}}{n+1}\) where \(n\) is any real number. So, \(\int x \, dx = \frac{x^2}{2}\).

Include the Integration Constant

The process of integration introduces an arbitrary constant, denoted by \(C\), because any constant value's derivative is zero. Therefore, the most general form of the function \(f(x)\) that could have led to \(f^{\prime}(x)=x\) is \(f(x) = \frac{x^2}{2} + C\), where \(C\) is any real number.

Key concepts

Indefinite Integral

In calculus, when we're trying to find a function whose derivative is given, we're dealing with an indefinite integral. An indefinite integral, represented by the symbol \( \int \), is essentially the reverse operation of differentiation. It's like trying to trace back to the origin when you only have the path. More technically, it's the set of all antiderivatives of a function. For instance, in our exercise, we're asked to find the function \( f \) whose derivative \( f’(x)=x \). By integrating \( f’(x) \), we're looking for all possible functions that, when differentiated, would yield \( x \). This is particularly handy when analyzing motion where we know the velocity and need to find the position, or when dealing with areas under curves.

Power Rule for Integration

An integral rule that frequently comes to the rescue is the power rule for integration. If you recall algebra, you might remember that exponents and powers tend to play a major role, and the same is true when we reverse the operation in calculus. The power rule for integration tells us that to integrate a power of \( x \), \( x^n \), we increase the exponent by one and then divide by the new exponent, as long as \( n eq -1 \). That brings us to the formula \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( C \) represents the integration constant. In our exercise, we apply this rule with \( n=1 \), so the integral of \( x \) with respect to \( x \) is \( \frac{x^2}{2} \). This simple rule makes a wide array of integrals manageable, lifting a heavy burden off the shoulders of anyone learning calculus.

Integration Constant

You may wonder why we always tack on a '+' and a 'C' at the end of every indefinite integral. This little \( C \) that appears in our solution stands for the integration constant. Here's the thing: if you add a constant to a function and then differentiate it, the constant disappears. In other words, a constant doesn't have a visible effect on the rate of change represented by the derivative. This means that when we're going in reverse—from the derivative back to the original function—there might have been a constant hanging around that we didn't know about. Because of this, we include \( C \) in our answer to cover all our bases and acknowledge that we're dealing with a family of functions, any of which could be the original function with different constants added. In our exercise, this renders the most general function as \( f(x) = \frac{x^2}{2} + C \).

Fundamental Theorem of Calculus

At the heart of calculus lies the fundamental theorem of calculus, and it's quite a big deal. This theorem beautifully bridges the gap between differentiation and integration, two principal concepts in calculus. It tells us that if a function \( f \) is continuous over an interval, then the function that charts the area under the curve of \( f \) (the integral) from a fixed point to variable point \( x \) is actually the antiderivative of \( f \). Essentially, it's saying that integration and differentiation are inverse processes. This theorem is the basis for calculating exact areas, solving differential equations, and much more. It's profound because it gives our exercise a solid ground to stand on—after finding \( f(x) \) by integrating \( f’(x) \), we can say that differentiating \( f(x) \) will give us \( f’(x) \) back, closing the loop of this mathematical conversation.