Chapter 4: Problem 41
In each of Exercises \(41-48\), use the given information to find \(F(c)\). $$ F^{\prime}(x)=3 x^{2}, \quad F(2)=-4, \quad c=3 $$
Short Answer
Expert verified
\( F(c) = 15 \)
Step by step solution
01
Understand the Problem
We need to find the value of the function \( F(c) \), given the derivative \( F'(x) = 3x^2 \), an initial condition \( F(2) = -4 \), and \( c = 3 \). This requires us to find \( F(x) \) by integrating \( F'(x) \).
02
Integrate the Derivative
The given derivative is \( F'(x) = 3x^2 \). To find \( F(x) \), we need to integrate \( 3x^2 \): \[ F(x) = \int 3x^2 \, dx \]The integral of \( 3x^2 \) is \( x^3 \) plus a constant \( C \), so:\[ F(x) = x^3 + C \]
03
Apply the Initial Condition
We use the initial condition \( F(2) = -4 \) to find the constant \( C \). Substitute \( x = 2 \) into the equation:\[ F(2) = 2^3 + C = -4 \]\[ 8 + C = -4 \]Solving for \( C \), we get \( C = -12 \).
04
Write the Function
Substitute the value of \( C \) back into the equation for \( F(x) \):\[ F(x) = x^3 - 12 \]
05
Calculate \( F(c) \)
Now that we have \( F(x) = x^3 - 12 \), we can find \( F(c) \) by substituting \( c = 3 \):\[ F(3) = 3^3 - 12 \]\[ F(3) = 27 - 12 \]\[ F(3) = 15 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integral Calculus
Integral calculus is a branch of mathematics that helps us find functions when their derivatives are given. The main idea is to reverse the process of differentiation. When you're given a derivative, like in our exercise with \( F'(x) = 3x^2 \), we need to integrate it to find the original function \( F(x) \). The integration process involves determining a function whose derivative matches the one we have. Integrating \( 3x^2 \) means calculating \( \int 3x^2 \, dx \). This leads to the integral \( x^3 + C \), where \( C \) is a constant. This process is like undoing the derivative, helping us "go backward" to find the original equation.
- Integrals offer ways to calculate areas under curves.
- They are essential in solving real-world problems involving accumulations, like total distances or quantities.
Initial Conditions
Initial conditions are key to solving integrals when you find an unknown constant. In calculus, when we integrate a function like \( 3x^2 \), we get \( F(x) = x^3 + C \). But without an initial condition, \( C \) could be virtually any number. In our exercise, we know \( F(2) = -4 \). This tells us exactly where the curve of such a function touches a specific point on the graph when \( x = 2 \). Applying the initial condition, we substitute \( x = 2 \) into our equation, giving \( 8 + C = -4 \). Solving for \( C \) results in \( C = -12 \).
- Initial conditions help specify our function in context, removing ambiguity.
- They allow us to solve precisely for constants, refining solutions.
Constant of Integration
The constant of integration, denoted usually as \( C \), emerges from the indefinite integral of a function. Every time you integrate, you must include this constant because derivatives "forget" constant terms. When you differentiate something with respect to \( x \), constants vanish. Thus, any function derived initially could differ by some constant yet have the same derivative. In our example, though the integration of \( 3x^2 \) gives us \( x^3 \), we write it as \( x^3 + C \) to acknowledge all potential original functions. To pinpoint the actual constant, we rely on specific initial conditions. The constant of integration accounts for all those possible translations of a curve that differentiate identically.
- Whenever integrating, remember to add \( C \) to reflect full generality.
- Solving for \( C \) personalizes the solution to specific conditions.