Understanding the Specific Antiderivative of f
When you integrate a function f(x), you obtain a family of functions known as antiderivatives. Each member of this family differs by a constant, often written as C. While the general antiderivative tells you how to reverse differentiation, a specific antiderivative (or particular antiderivative) pinpoints the exact function you need by applying an additional condition—usually a value at a particular point. This process is essential in solving initial value problems that arise in physics, engineering, and many other fields where a unique solution is required But it adds up..
The formula for the specific antiderivative of f can be expressed as
[ F_{\text{specific}}(x)=\int f(x),dx ;-; \frac{\displaystyle\int f(x),dx\Big|_{x=x_0}}{1};+;f(x_0) ]
or more simply
[ F_{\text{specific}}(x)=\int f(x),dx ;+;C, ]
where the constant C is determined from the given condition, such as F(a)=k or f(b)=m. Below, we break down the reasoning, step‑by‑step, and illustrate the method with concrete examples.
Steps to Find the Specific Antiderivative
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Integrate the given function
Compute the indefinite integral (\int f(x),dx). This yields a general antiderivative (F(x)+C). Use standard integration rules—power rule, exponential rule, trigonometric integrals, substitution, integration by parts, etc.—as needed. -
Identify the condition that defines “specific”
Look for an initial value such as (F(x_0)=k) or a point ((x_0, y_0)) that the antiderivative must pass through. This condition will allow you to solve for the constant C. -
Solve for the constant
Substitute the known point into the general antiderivative expression:[ k = F(x_0) + C \quad\Longrightarrow\quad C = k - F(x_0). ]
This step often involves algebraic manipulation and careful evaluation of any transcendental functions (e.g., (\sin), (\ln)) The details matter here..
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Write the specific antiderivative
Replace C in the general solution with the numeric value obtained. The resulting function is the specific antiderivative that satisfies the given condition Less friction, more output.. -
Verify the solution (optional but recommended)
Plug the specific antiderivative back into the original condition to confirm it holds true. Additionally, differentiate the result to ensure it reproduces the original f(x) Small thing, real impact..
Scientific Explanation: Why the Constant Matters
The presence of the integration constant C stems from the fact that differentiation eliminates constant terms. If (F(x)) is an antiderivative of f, then any function (F(x)+C) also differentiates back to f(x), because (\frac{d}{dx}C = 0). As a result, the indefinite integral represents an entire family of curves that are vertical translations of each other Small thing, real impact..
In practical applications, only one of these curves is relevant. Take this case: when modeling the position of a moving object given its velocity v(t), the constant C corresponds to the initial position. Specifying that position at time (t_0) yields a unique trajectory. This is why the term initial value problem is common: you have a differential equation (here, (F'(x)=f(x))) and an extra condition that selects the appropriate member of the family Took long enough..
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Mathematically, the condition can be expressed as
[ F(x_0) = y_0, ]
where ((x_0, y_0)) is a known point on the graph of the desired antiderivative. Solving for C gives
[ C = y_0 - F(x_0). ]
Thus, the specific antiderivative is
[ F_{\text{specific}}(x) = \int f(x),dx + \bigl[y_0 - F(x_0)\bigr]. ]
This formulation highlights that the specific antiderivative is simply the general antiderivative shifted vertically so that it passes through the prescribed point.
Example 1: Polynomial Function
Problem: Find the specific antiderivative (F(x)) of (f(x)=3x^{2}+2x) such that (F(1)=10).
Solution:
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Integrate:
[ \int (3x^{2}+2x),dx = 3\frac{x^{3}}{3}+2\frac{x^{2}}{2}+C = x^{3}+x^{2}+C. ]
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Apply condition:
[ F(1)=1^{3}+1^{2}+C = 2 + C = 10 ;\Longrightarrow; C = 8. ]
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Specific antiderivative:
[ F_{\text{specific}}(x)=x^{3}+x^{2}+8. ]
Verification: Differentiating gives (3x^{2}+2x), matching f(x), and (F_{\text{specific}}(1)=10) as required.
Example 2: Exponential with Initial Condition
Problem: Determine the specific antiderivative of (f(x)=e^{2x}) that satisfies (F(0)=5).
Solution:
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Integrate:
[ \int e^{2x},dx = \frac{1}{2}e^{2x}+C. ]
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Apply condition:
[ F(0)=\frac{1}{2}e^{0}+C = \frac{1}{2}+C = 5 ;\Longrightarrow; C = \frac{9}{2}. ]
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Specific antiderivative:
[ F_{\text{specific}}(x)=\frac{1}{2}e^{2x}+\frac{9}{2}. ]
Verification: Differentiation yields (e^{2x}), and (F_{\text{specific}}(0)=5) as needed.
Example 3: Trigonometric Function
Problem: Find the specific antiderivative of (f(x)=\sin(3x)) such that (F\bigl(\frac{\pi}{6}\bigr)=2).
Solution:
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Integrate:
[ \int \sin(3x),dx = -\frac{1}{3}\cos(3x)+C. ]
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Apply condition:
[ F!In practice, \left(3\cdot\frac{\pi}{6}\right)+C = -\frac{1}{3}\cos! Think about it: \left(\frac{\pi}{6}\right)= -\frac{1}{3}\cos! \left(\frac{\pi}{2}\right)+C = -\frac{1}{3}\cdot0 + C = C.
Hence (C = 2).
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Specific antiderivative:
[ F_{\text{specific}}(x)= -\frac{1}{3}\cos(3x)+
Example 3 (completed)
Problem: Find the specific antiderivative of (f(x)=\sin(3x)) such that (F!\bigl(\frac{\pi}{6}\
