Find The Particular Antiderivative That Satisfies The Following Conditions

Diving into the realm of calculus, we often encounter the task of finding antiderivatives, the reverse operation of differentiation. But what happens when we need to pinpoint a specific antiderivative that aligns with certain predefined conditions? This is where the concept of finding a particular antiderivative comes into play, a fundamental skill vital for solving various problems in physics, engineering, and economics.

Understanding Antiderivatives: The Basics

At its core, finding the antiderivative, also known as indefinite integration, means determining a function whose derivative is a given function. If F(x) is an antiderivative of f(x), then F'(x) = f(x). The challenge lies in the fact that antiderivatives are not unique. Adding any constant, C, to an antiderivative will still result in a valid antiderivative, as the derivative of a constant is always zero. This gives rise to the general antiderivative, expressed as F(x) + C.

The Constant of Integration: "C"

The "+ C" isn't just a formality; it represents an infinite family of functions that differ only by a constant. Each of these functions has the same derivative, f(x). To nail down a particular antiderivative, we need additional information, a specific condition that singles out one function from this infinite family.

What Are "Conditions" in This Context?

The "conditions" we're referring to are usually presented as an initial value or a boundary condition. This is a known value of the antiderivative at a specific point. For example:

F(0) = 5: This means that when x is 0, the value of the antiderivative F(x) is 5.
F(2) = -1: This means that when x is 2, the value of the antiderivative F(x) is -1.
y(1) = 3: This means that when x is 1, the value of the antiderivative y(x) is 3. (Note the use of y(x) as an alternative notation for the antiderivative).

These conditions provide a fixed point through which the particular antiderivative must pass. This allows us to solve for the constant of integration, C, and thus determine the unique antiderivative that satisfies the given condition.

Steps to Find the Particular Antiderivative

Here's a systematic approach to finding the particular antiderivative that satisfies given conditions:

Find the General Antiderivative: Calculate the indefinite integral of the given function f(x). This will result in F(x) + C, where F(x) is any antiderivative of f(x) and C is the constant of integration. This step uses integration rules and techniques.
Apply the Given Condition: Substitute the given x value and the corresponding F(x) value (from the initial or boundary condition) into the general antiderivative equation F(x) + C. This creates an equation with C as the only unknown.
Solve for C: Solve the equation from step 2 for the value of C. This value represents the specific constant of integration for this particular antiderivative.
Write the Particular Antiderivative: Substitute the value of C you just found back into the general antiderivative equation F(x) + C. This results in the particular antiderivative, F(x) with the specific constant, satisfying the given condition.

Example Problems: Putting the Steps into Action

Let's illustrate this process with several examples:

Example 1:

Find the particular antiderivative of f(x) = 2x that satisfies F(1) = 4.

Step 1: Find the General Antiderivative

The antiderivative of 2x is x² + C.

∫ 2x dx = x² + C
Step 2: Apply the Given Condition

We know that F(1) = 4. So, substitute x = 1 and F(1) = 4 into the general antiderivative:

4 = (1)² + C
Step 3: Solve for C

Solve the equation for C:

4 = 1 + C C = 3
Step 4: Write the Particular Antiderivative

Substitute C = 3 back into the general antiderivative:

F(x) = x² + 3

Therefore, the particular antiderivative of f(x) = 2x that satisfies F(1) = 4 is F(x) = x² + 3.

Example 2:

Find the particular antiderivative of f(x) = cos(x) that satisfies F(π) = 0.

Step 1: Find the General Antiderivative

The antiderivative of cos(x) is sin(x) + C.

∫ cos(x) dx = sin(x) + C
Step 2: Apply the Given Condition

We know that F(π) = 0. So, substitute x = π and F(π) = 0 into the general antiderivative:

0 = sin(π) + C
Step 3: Solve for C

Solve the equation for C:

0 = 0 + C C = 0
Step 4: Write the Particular Antiderivative

Substitute C = 0 back into the general antiderivative:

F(x) = sin(x) + 0 F(x) = sin(x)

Therefore, the particular antiderivative of f(x) = cos(x) that satisfies F(π) = 0 is F(x) = sin(x).

Example 3:

Find the particular antiderivative of f(x) = e^x that satisfies F(0) = 7.

Step 1: Find the General Antiderivative

The antiderivative of e^x is e^x + C.

∫ e^x dx = e^x + C
Step 2: Apply the Given Condition

We know that F(0) = 7. So, substitute x = 0 and F(0) = 7 into the general antiderivative:

7 = e^0 + C
Step 3: Solve for C

Solve the equation for C:

7 = 1 + C C = 6
Step 4: Write the Particular Antiderivative

Substitute C = 6 back into the general antiderivative:

F(x) = e^x + 6

Therefore, the particular antiderivative of f(x) = e^x that satisfies F(0) = 7 is F(x) = e^x + 6.

Example 4:

Find the particular antiderivative of f(x) = 3x² - 2x + 1 that satisfies F(2) = 5.

Step 1: Find the General Antiderivative

The antiderivative of 3x² - 2x + 1 is x³ - x² + x + C.

∫ (3x² - 2x + 1) dx = x³ - x² + x + C
Step 2: Apply the Given Condition

We know that F(2) = 5. So, substitute x = 2 and F(2) = 5 into the general antiderivative:

5 = (2)³ - (2)² + (2) + C
Step 3: Solve for C

Solve the equation for C:

5 = 8 - 4 + 2 + C 5 = 6 + C C = -1
Step 4: Write the Particular Antiderivative

Substitute C = -1 back into the general antiderivative:

F(x) = x³ - x² + x - 1

Therefore, the particular antiderivative of f(x) = 3x² - 2x + 1 that satisfies F(2) = 5 is F(x) = x³ - x² + x - 1.

Beyond Basic Integration: Handling More Complex Functions

The fundamental steps remain the same, but the complexity of finding the general antiderivative can significantly increase. You might need to employ various integration techniques:

Substitution (u-substitution): Useful when the integrand contains a composite function and its derivative (or a multiple thereof).
Integration by Parts: Applicable when the integrand is a product of two functions, often when one function simplifies upon differentiation and the other simplifies upon integration. The formula is: ∫ u dv = uv - ∫ v du.
Trigonometric Integrals: Requires using trigonometric identities to simplify the integrand before integration.
Partial Fraction Decomposition: Used to integrate rational functions (polynomials divided by polynomials) by breaking them down into simpler fractions.

Example 5: Using u-Substitution

Find the particular antiderivative of f(x) = 2x * cos(x²) that satisfies F(√(π/2)) = 1.

Step 1: Find the General Antiderivative

This requires u-substitution. Let u = x². Then, du = 2x dx. The integral becomes:

∫ cos(u) du = sin(u) + C

Substituting back for u:

∫ 2x cos(x²) dx = sin(x²) + C
Step 2: Apply the Given Condition

We know that F(√(π/2)) = 1. Substitute x = √(π/2) into the general antiderivative:

1 = sin((√(π/2))²) + C 1 = sin(π/2) + C
Step 3: Solve for C

Solve the equation for C:

1 = 1 + C C = 0
Step 4: Write the Particular Antiderivative

Substitute C = 0 back into the general antiderivative:

F(x) = sin(x²)

Therefore, the particular antiderivative of f(x) = 2x * cos(x²) that satisfies F(√(π/2)) = 1 is F(x) = sin(x²).

Example 6: Using Integration by Parts

Find the particular antiderivative of f(x) = x * e^x that satisfies F(0) = 3.

Step 1: Find the General Antiderivative

This requires integration by parts. Let u = x and dv = e^x dx. Then, du = dx and v = e^x. Using the integration by parts formula:

∫ x e^x dx = x e^x - ∫ e^x dx = x e^x - e^x + C
Step 2: Apply the Given Condition

We know that F(0) = 3. Substitute x = 0 into the general antiderivative:

3 = (0)e^0 - e^0 + C 3 = 0 - 1 + C
Step 3: Solve for C

Solve the equation for C:

3 = -1 + C C = 4
Step 4: Write the Particular Antiderivative

Substitute C = 4 back into the general antiderivative:

F(x) = x e^x - e^x + 4

Therefore, the particular antiderivative of f(x) = x * e^x that satisfies F(0) = 3 is F(x) = x e^x - e^x + 4.

Applications of Particular Antiderivatives

Finding particular antiderivatives isn't just an abstract mathematical exercise. It has profound applications in various fields:

Physics: Determining the position of an object given its velocity (velocity is the derivative of position). The initial condition might be the object's starting position.
Engineering: Calculating the deflection of a beam under load. Boundary conditions might include fixed points or supports.
Economics: Modeling the growth of a population or the accumulation of capital. Initial conditions could represent the starting population size or initial capital investment.
Differential Equations: Finding particular solutions to differential equations, where the conditions are typically initial values.

Common Mistakes to Avoid

Forgetting the Constant of Integration: Always include "+ C" when finding the general antiderivative. This is the most common mistake.
Incorrectly Applying Integration Rules: Ensure you are using the correct integration rules and techniques. Double-check your work, especially when using substitution or integration by parts.
Algebra Errors: Careless algebra mistakes when solving for C can lead to an incorrect particular antiderivative.
Misinterpreting the Initial Condition: Make sure you correctly identify the x value and the corresponding F(x) value from the given condition.

Advanced Scenarios: Multiple Conditions and Higher-Order Derivatives

Sometimes, you might be given multiple conditions or information about higher-order derivatives. For example:

Given f''(x) (the second derivative), and conditions like f(0) = 2 and f'(0) = -1.

In these cases, you need to perform the integration multiple times, applying the conditions at each step.

Example 7: Higher-Order Derivative

Find f(x) given that f''(x) = 6x + 4, f(0) = 3, and f'(0) = -2.

Step 1: Find f'(x)

Integrate f''(x) to find f'(x):

f'(x) = ∫ (6x + 4) dx = 3x² + 4x + C₁

Apply the condition f'(0) = -2:

-2 = 3(0)² + 4(0) + C₁ C₁ = -2

So, f'(x) = 3x² + 4x - 2
Step 2: Find f(x)

Integrate f'(x) to find f(x):

f(x) = ∫ (3x² + 4x - 2) dx = x³ + 2x² - 2x + C₂

Apply the condition f(0) = 3:

3 = (0)³ + 2(0)² - 2(0) + C₂ C₂ = 3

So, f(x) = x³ + 2x² - 2x + 3

Therefore, the function is f(x) = x³ + 2x² - 2x + 3.

Tips for Success

Practice Regularly: The more you practice, the more comfortable you will become with different integration techniques and types of problems.
Master Basic Integration Rules: Know the fundamental integration rules inside and out.
Show Your Work: Write out each step clearly and carefully. This helps prevent errors and makes it easier to identify any mistakes.
Check Your Answer: Differentiate your final answer to see if you get back the original function. Also, verify that your answer satisfies the given conditions.
Use Online Resources: There are many excellent online resources available, such as calculators, tutorials, and practice problems.

Conclusion: Mastering the Art of the Particular

Finding the particular antiderivative that satisfies specific conditions is a core skill in calculus with widespread practical applications. By understanding the concept of the constant of integration, following the systematic steps outlined above, mastering various integration techniques, and avoiding common mistakes, you can confidently tackle these types of problems. Remember that practice is key to developing proficiency and solidifying your understanding. So, keep practicing, keep exploring, and keep pushing your calculus skills to new heights!