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. $$ f^{\prime \prime}(x)=\sqrt{x} $$

Short answer

Antidifferentiate twice to get \( f(x) = \frac{4}{15}x^{5/2} + C_1 x + C_2 \).

Step-by-step solution

Antidifferentiate the Second Derivative to Find the First Derivative

To find the first derivative, we need to integrate the second derivative. Given \( f''(x) = \sqrt{x} \), the first derivative \( f'(x) \) is obtained by integrating \( \sqrt{x} \) with respect to \( x \). Recall that \( \sqrt{x} = x^{1/2} \), so the integral is \( \int x^{1/2} \, dx \). Using the power rule for integration, add 1 to the exponent and divide by the new exponent: \( \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} + C_1 = \frac{2}{3}x^{3/2} + C_1 \).

Antidifferentiate the First Derivative to Find the Original Function

Now, integrate the first derivative to find the original function \( f(x) \). We have \( f'(x) = \frac{2}{3}x^{3/2} + C_1 \). Integrating this with respect to \( x \), consider each term separately. The first term, \( \frac{2}{3}x^{3/2} \), integrates to \( \frac{2}{3} \times \frac{x^{5/2}}{5/2} = \frac{2}{3} \times \frac{2}{5}x^{5/2} = \frac{4}{15}x^{5/2} \). For the second term, \( C_1 \), integrate as \( C_1x \). The integral of the first derivative is thus \( f(x) = \frac{4}{15}x^{5/2} + C_1 x + C_2 \), where \( C_2 \) is another constant of integration.

Key concepts

Integration

Integration is a fundamental concept in calculus, closely related to differentiation. It is essentially the process of finding the antiderivative or the area under a curve. This technique helps us reverse the process of differentiation.

When we integrate a function, we are working to find a new function whose derivative gives us the original function. In terms of the exercise, integrating the second derivative, \( f''(x) \), allows us to find the first derivative, \( f'(x) \). Continuing further with another integration gives us the original function, \( f(x) \).

Integration involves using rules and techniques such as the power rule, substitution, and integration by parts. Each method helps deal with different forms or complexities of functions. It is a crucial tool for solving real-world problems such as calculating areas, volumes, and other related rates.

Arbitrary Constants

Arbitrary constants appear in integration because antidifferentiation, or integration, is not a precise reversal of differentiation. Whenever we differentiate a constant, it becomes zero. Hence, when we antidifferentiate, there is no way to determine what constant was initially present. This is why each time you integrate a function, you must add a constant, typically denoted as \( C \).

For the given exercise, integrating the second derivative \( f''(x) \) results in the first derivative \( f'(x) \) with an arbitrary constant \( C_1 \). When we integrate \( f'(x) \) to find \( f(x) \), another arbitrary constant \( C_2 \) appears. These constants cannot be combined or simplified since they represent different integrations with potentially different original values.

In applications, these constants can often be determined if additional information is known, such as initial conditions or specific values of the function. This allows the solution to be tailored to a specific problem or scenario.

Power Rule

The power rule is a straightforward technique used in both differentiation and integration. It allows us to easily find antiderivatives of specific functions, particularly polynomials. For a function \( x^n \), the power rule for integration is key.

The rule states that when you integrate \( x^n \), you add 1 to the exponent and then divide by the new exponent. Mathematically, \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration. It's important to note that this rule does not apply when \( n = -1 \) because it would cause division by zero; instead, the integral becomes \( \ln|x| \).

In the exercise provided, the power rule was used to integrate \( f''(x) = \sqrt{x} \), which is \( x^{1/2} \). Applying the power rule results in \( \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} + C_1 \). This simplification helps make integration a routine process for polynomial terms.

Definite Integral

The concept of a definite integral extends the idea of integration to finding the exact numerical area under a curve between two specified points. Unlike indefinite integrals, which include arbitrary constants, definite integrals evaluate to a specific number.

A definite integral is usually represented as \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) are the lower and upper limits of the integral. The main advantage of definite integrals is that they provide a precise value which can be crucial in practical applications, such as determining displacement or accumulation in physics and engineering.

While the exercise at hand involves indefinite integration due to the presence of arbitrary constants, it is essential to understand that definite integration follows a slightly different approach and does not produce arbitrary constants. Instead, it uses the Fundamental Theorem of Calculus to compute the accumulated area directly between two points. This connection underscores the versatile power of integration in calculus.