Question

\(f^{\prime \prime}(x)\) is given. Find \(f(x)\) by antidifferentiating twice. Note that in this case your answer should involve two arbitrary constants, one from each antidifferentiation. For example, if \(f^{\prime \prime}(x)=x\), then \(f^{\prime}(x)=x^{2} / 2+C_{1}\) and \(f(x)=\) \(x^{3} / 6+C_{1} x+C_{2} .\) The constants \(C_{1}\) and \(C_{2}\) cannot be combined because \(C_{1} x\) is not a constant. $$ f^{\prime \prime}(x)=-2 x+3 $$

Short answer

\(f(x) = -\frac{x^3}{3} + \frac{3x^2}{2} + C_1x + C_2\)

Step-by-step solution

Identify the second derivative

The given second derivative of the function is \( f^{\prime \prime}(x) = -2x + 3 \). Our task is to find the function \( f(x) \) by integrating this second derivative twice.

First Integration to Find f'

Integrate \( f^{\prime \prime}(x) = -2x + 3 \) with respect to \( x \) to find \( f^{\prime}(x) \).\[f^{\prime}(x) = \int (-2x + 3) \, dx = -x^2 + 3x + C_1\]where \( C_1 \) is an arbitrary constant of integration.

Second Integration to Find f

Integrate \( f^{\prime}(x) = -x^2 + 3x + C_1 \) with respect to \( x \) to find \( f(x) \).\[f(x) = \int (-x^2 + 3x + C_1) \, dx = -\frac{x^3}{3} + \frac{3x^2}{2} + C_1x + C_2\]where \( C_2 \) is another arbitrary constant of integration.

Combine and Simplify the Function

Combine all parts to write the final form of the function:\[f(x) = -\frac{x^3}{3} + \frac{3x^2}{2} + C_1x + C_2\]This represents the original function \( f(x) \), which includes both constant terms \( C_1 \) and \( C_2 \) from the integrations.

Key concepts

calculus

Calculus is a branch of mathematics that explores continuous change. It is used to understand how things evolve over time, such as the changing rate of a car's speed or the growth of a population. Calculus is divided into two main parts: differential calculus and integral calculus. Differential calculus deals with the concept of the derivative, which represents how a function changes as its input changes. On the other hand, integral calculus focuses on the concept of integration, which is used to find quantities such as area under a curve.

In this exercise, we are tasked with recovering the original function by antidifferentiation, which means applying integral calculus. Antidifferentiation involves finding a function whose derivative gives us the original function we started with. Therefore, when given a second derivative, as in this problem, we aim to work backward by integrating twice.

Overall, calculus offers powerful tools for modeling and solving problems involving change and accumulation, making it essential in fields ranging from physics to economics.

arbitrary constants

When we perform antidifferentiation, arbitrary constants are added to the resulting expressions. These constants arise because integrations provide a family of functions rather than a unique solution.

In the given problem, we encounter two arbitrary constants, denoted as \( C_1 \) and \( C_2 \), resulting from the two rounds of integration. The inclusion of these constants is crucial because:

• Each antidifferentiation step adds an arbitrary constant. After the first integration, we introduce \( C_1 \), and after the second, \( C_2 \).
• They account for the general nature of antiderivatives—since derivatives remove constants, the original constant terms in our function are not determinable without additional conditions.

In this particular example, \( C_1 \) appears alongside a term involving \( x \), while \( C_2 \) stands alone as an additive constant. Because \( C_1x \) is a variable-dependent term, it cannot be combined with \( C_2 \) when forming the final expression for \( f(x) \).

integrals

Integrals are fundamental to calculus and serve as the tool for antidifferentiation. They are used to determine the accumulation of quantities, such as area under a curve, total distance traveled, or any other quantity that accumulates over time. In essence, integrals help reconstruct the whole from its infinitesimally small parts.

When finding integrals, the basic operation is to identify the antiderivative. This process was applied twice in the problem at hand. The first integration of \( f''(x) = -2x + 3 \) resulted in the antiderivative \( f'(x) = -x^2 + 3x + C_1 \). Subsequent integration of this expression yielded the function \( f(x) = -\frac{x^3}{3} + \frac{3x^2}{2} + C_1x + C_2 \).

Integrals can be classified as either definite or indefinite:

• Indefinite integrals, as used here, include an arbitrary constant of integration and represent a family of functions.
• Definite integrals supply a specific numerical value, defined by limits of integration, and precisely compute the accumulated quantity over that interval.

The application of integrals, particularly indefinite ones, is crucial for solving problems that involve recovering original functions from their derivatives.