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What is an Antiderivative?

An antiderivative is the reverse process of differentiation. If we have a function $ f(x) $ , its antiderivative, often denoted as $ F(x) $ , satisfies the equation:

$ F’(x) = f(x) $

In simpler terms, the derivative of the antiderivative is the original function. Note that antiderivatives are not unique. If $ F(x) $ is an antiderivative of $ f(x) $ , then so is $ F(x) + C $ , where $ C $ is any constant. This is because the derivative of a constant is always zero.

Finding Antiderivatives

Finding antiderivatives often involves reversing the rules of differentiation. Here are some basic examples:

• Power Rule: If $ f(x) = x^n $ , then $ F(x) = \frac{x^{n+1}}{n+1} + C $ for $ n \neq -1 $ .

• Constant Multiple Rule: If $ F(x) $ is an antiderivative of $ f(x) $ , then $ kF(x) $ is an antiderivative of $ kf(x) $ , where $ k $ is a constant.

• Sum/Difference Rule: If $ F(x) $ is an antiderivative of $ f(x) $ and $ G(x) $ is an antiderivative of $ g(x) $ , then $ F(x) + G(x) $ is an antiderivative of $ f(x) + g(x) $ , and $ F(x) - G(x) $ is an antiderivative of $ f(x) - g(x) $ .

• Exponential Functions: If $ f(x) = e^x $ , then $ F(x) = e^x + C $ . More generally, if $ f(x) = e^{kx} $ , then $ F(x) = \frac{1}{k}e^{kx} + C $ .

• Trigonometric Functions:

  • If $ f(x) = \sin(x) $ , then $ F(x) = -\cos(x) + C $ .
  • If $ f(x) = \cos(x) $ , then $ F(x) = \sin(x) + C $ .

The Indefinite Integral

The indefinite integral is a notation used to represent the antiderivative. It’s written as:

$ \int f(x) , dx = F(x) + C $

where:

• $ \int $ is the integral symbol.
• $ f(x) $ is the integrand (the function being integrated).
• $ dx $ indicates that the integration is with respect to $ x $ .
• $ F(x) $ is an antiderivative of $ f(x) $ .
• $ C $ is the constant of integration.

Initial Value Problems

An initial value problem provides an initial condition, which allows us to determine the specific value of the constant of integration, $ C $ . For example, if we are given that $ F(0) = 5 $ and $ F’(x) = 2x $ , then we can find the specific antiderivative.

Since $ \int 2x , dx = x^2 + C $ , we can use the initial condition:

$ F(0) = 0^2 + C = 5 $ , so $ C = 5 $ . Therefore, the specific antiderivative is $ F(x) = x^2 + 5 $ .

Examples

Example 1: Find the antiderivative of $ f(x) = 3x^2 + 2x - 1 $ .

$ \int (3x^2 + 2x - 1) , dx = x^3 + x^2 - x + C $

Example 2: Find the antiderivative of $ f(x) = e^{2x} + \cos(x) $ .

$ \int (e^{2x} + \cos(x)) , dx = \frac{1}{2}e^{2x} + \sin(x) + C $

Example 3: Solve the initial value problem: $ F’(x) = 4x^3 - 6x $ and $ F(1) = 2 $ .

$ \int (4x^3 - 6x) , dx = x^4 - 3x^2 + C $

Using the initial condition: $ F(1) = 1^4 - 3(1)^2 + C = 2 $ , which gives $ C = 4 $ .

Therefore, $ F(x) = x^4 - 3x^2