Problem 30 - Essential Calculus Early Transcendentals

Question

Find \(f\) $$f^{\prime \prime}(x)=8 x^{3}+5, \quad f(1)=0, \quad f^{\prime}(1)=8$$

Step-by-Step Solution

Answer

\( f(x) = \frac{2}{5}x^5 + \frac{5}{2}x^2 + x - \frac{39}{10} \).

Step 1: Integrate the Second Derivative

To find the function \( f(x) \), we start by integrating the second derivative \( f''(x) = 8x^3 + 5 \) with respect to \( x \) to obtain the first derivative \( f'(x) \). \[ \int (8x^3 + 5) \, dx = 2x^4 + 5x + C_1\]Here, \( C_1 \) is the constant of integration.

Step 2: Use Initial Condition for the First Derivative

We utilize the initial condition \( f'(1) = 8 \). Substitute \( x = 1 \) and \( f'(1) = 8 \) into the equation from Step 1:\[2(1)^4 + 5(1) + C_1 = 8\]Simplifying this gives:\[2 + 5 + C_1 = 8 \]\[C_1 = 1\]Thus, \( f'(x) = 2x^4 + 5x + 1 \).

Step 3: Integrate the First Derivative

Next, integrate \( f'(x) = 2x^4 + 5x + 1 \) to find \( f(x) \):\[ \int (2x^4 + 5x + 1) \, dx = \frac{2}{5}x^5 + \frac{5}{2}x^2 + x + C_2\]Where \( C_2 \) is another constant of integration.

Step 4: Use Initial Condition for the Function

Employ the initial condition \( f(1) = 0 \). Substitute \( x = 1 \) and \( f(1) = 0 \) into the equation from Step 3:\[\frac{2}{5}(1)^5 + \frac{5}{2}(1)^2 + 1 + C_2 = 0\]Simplifying this gives:\[\frac{2}{5} + \frac{5}{2} + 1 + C_2 = 0\]\[\frac{4}{10} + \frac{25}{10} + \frac{10}{10} + C_2 = 0\]\[\frac{39}{10} + C_2 = 0\]\[C_2 = -\frac{39}{10}\]Hence, \( f(x) = \frac{2}{5}x^5 + \frac{5}{2}x^2 + x - \frac{39}{10} \).

Key Concepts

Integration

In calculus, integration is a fundamental concept. It is the reverse process of differentiation. When you are given a function, such as the second derivative, integrating it helps find the original function or the antiderivative.

• For example, given the second derivative \( f''(x) = 8x^3 + 5 \), to find \( f(x) \), integrate this expression with respect to \( x \).
• Integration involves finding a function whose derivative would give you the original function, in this case, \( f'(x) \) from \( f''(x) \).

Therefore, when you integrate \( 8x^3 + 5 \), the result is \( 2x^4 + 5x + C_1 \), where \( C_1 \) is an integration constant. This constant is crucial because it reflects that different functions can have the same derivative, differing only by a constant.

Initial Conditions

Initial conditions are specific values given for a function and its derivatives at certain points. They are essential to find the particular solutions of differential equations or integrals.

• In this problem, you have the initial conditions \( f(1) = 0 \) and \(f'(1) = 8 \).
• These conditions allow us to determine the values of integration constants such as \( C_1 \) and \( C_2 \).

Without these initial conditions, only a general solution can be found. By plugging these initial conditions into the integrated forms, you can solve for the unknown constants, ensuring that the solution meets the given conditions.

Differential Equations

Differential equations involve derivatives and describe how a function changes. These occur extensively in fields like physics and engineering to model various phenomena.

• The original exercise involves a second-order differential equation: \( f''(x) = 8x^3 + 5 \).
• The goal is to work backward from this derivative to find the original function \( f(x) \).

To solve a differential equation like this, you must first integrate the expression given. Initial conditions help tailor the generic integrated form of the function to satisfy these specific circumstances. Thus, you can translate a general solution into one that fits the problem precisely.

Indefinite Integrals

Indefinite integrals are an essential concept in calculus. They represent the antiderivative of a function and include an arbitrary constant, which differentiates them from definite integrals.

• An indefinite integral looks like \( \int f(x) \, dx = F(x) + C \).
• Here, \( F(x) \) is the antiderivative, and \( C \) is the constant of integration.

In the exercise, when you integrate \( f''(x) = 8x^3 + 5 \), the resulting indefinite integral is \( \int (8x^3 + 5) \, dx = 2x^4 + 5x + C_1 \). This methodology allows us to extract the antiderivative, leading towards the function itself. Therefore, indefinite integrals are crucial for transitioning from derivatives back to their parent functions, with constants being key to solving initial condition problems.