Anti-derivatives

Introduction to Anti-differentiation

• Anti-differentiation is the reverse process of differentiation. It involves finding a function whose derivative is a given function.
• The anti-derivative is also known as the indefinite integral.
• Notation: If \(\frac{d}{dx}F(x) = f(x)\), then \(\int f(x) \, dx = F(x) + c\), where \(c\) is the constant of integration.

KEY TAKEAWAY: Anti-differentiation reverses the process of differentiation. Always include the constant of integration, c, for indefinite integrals.

Anti-derivatives of Polynomial Functions

Basic Power Rule

• If \(f(x) = x^n\), where \(n \in \mathbb{Q}\) and \(n \neq -1\), then
  \$\(\int x^n \, dx = \frac{x^{n+1}}{n+1} + c\)\$
• This rule applies to all rational powers of \(x\), except for \(n = -1\).

Examples

1. \(\int x^2 \, dx = \frac{x^{2+1}}{2+1} + c = \frac{x^3}{3} + c\)
2. \(\int \sqrt{x} \, dx = \int x^{\frac{1}{2}} \, dx = \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} + c = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + c = \frac{2}{3}x^{\frac{3}{2}} + c\)
3. \(\int \frac{1}{x^3} \, dx = \int x^{-3} \, dx = \frac{x^{-3+1}}{-3+1} + c = \frac{x^{-2}}{-2} + c = -\frac{1}{2x^2} + c\)

Linear Combinations

• The anti-derivative of a linear combination of functions is the linear combination of their anti-derivatives.
  \$\(\int [af(x) + bg(x)] \, dx = a\int f(x) \, dx + b\int g(x) \, dx\)\$
  where \(a\) and \(b\) are constants.

Examples

1. \(\int (3x^2 + 2x - 1) \, dx = 3\int x^2 \, dx + 2\int x \, dx - \int 1 \, dx = 3\left(\frac{x^3}{3}\right) + 2\left(\frac{x^2}{2}\right) - x + c = x^3 + x^2 - x + c\)
2. \(\int (4x^{\frac{1}{2}} - 6x^{-2}) \, dx = 4\int x^{\frac{1}{2}} \, dx - 6\int x^{-2} \, dx = 4\left(\frac{2}{3}x^{\frac{3}{2}}\right) - 6\left(\frac{x^{-1}}{-1}\right) + c = \frac{8}{3}x^{\frac{3}{2}} + \frac{6}{x} + c\)

EXAM TIP: Remember to simplify your answer after applying the power rule and combining terms.

Anti-derivatives of \(f(ax + b)\)

General Rule

• If \(\int f(x) \, dx = F(x) + c\), then \(\int f(ax + b) \, dx = \frac{1}{a}F(ax + b) + c\)

Anti-derivatives of \(x^n\) where \(f(x) = (ax+b)^n\)

• \[\int (ax + b)^n \, dx = \frac{1}{a} \cdot \frac{(ax + b)^{n+1}}{n+1} + c, \text{ for } n \neq -1\]

Anti-derivatives of Exponential Functions \(e^x\)

• \(\int e^x \, dx = e^x + c\)
• \(\int e^{ax + b} \, dx = \frac{1}{a}e^{ax + b} + c\)

Anti-derivatives of Trigonometric Functions \(\sin(x)\) and \(\cos(x)\)

• \(\int \sin(x) \, dx = -\cos(x) + c\)
• \(\int \cos(x) \, dx = \sin(x) + c\)
• \(\int \sin(ax + b) \, dx = -\frac{1}{a}\cos(ax + b) + c\)
• \(\int \cos(ax + b) \, dx = \frac{1}{a}\sin(ax + b) + c\)

Summary Table

Examples

1. \(\int (2x + 3)^4 \, dx = \frac{1}{2} \cdot \frac{(2x + 3)^{4+1}}{4+1} + c = \frac{(2x + 3)^5}{10} + c\)
2. \(\int e^{3x - 1} \, dx = \frac{1}{3}e^{3x - 1} + c\)
3. \(\int \sin(4x + 2) \, dx = -\frac{1}{4}\cos(4x + 2) + c\)
4. \(\int \cos(\frac{1}{2}x - 5) \, dx = 2\sin(\frac{1}{2}x - 5) + c\)
5. \(\int (2e^{-x} + 3\cos(2x)) \, dx = -2e^{-x} + \frac{3}{2}\sin(2x) + c\)

COMMON MISTAKE: Forgetting to divide by the coefficient ‘a’ when anti-differentiating \(f(ax+b)\).

Determining the Constant of Integration

Initial Conditions

• To find the specific anti-derivative, you need an initial condition, which is a point \((x_0, y_0)\) on the curve \(y = F(x)\). This allows you to solve for the constant of integration, c.

Example

• Find \(f(x)\) if \(f'(x) = 2x + 1\) and \(f(1) = 4\).
  1. Find the general anti-derivative: \(f(x) = \int (2x + 1) \, dx = x^2 + x + c\)
  2. Use the initial condition \(f(1) = 4\): \(4 = (1)^2 + (1) + c \Rightarrow 4 = 2 + c \Rightarrow c = 2\)
  3. The specific anti-derivative is \(f(x) = x^2 + x + 2\)

STUDY HINT: Practice a variety of anti-differentiation problems, including those with initial conditions, to master the techniques.

Applications of Anti-derivatives

Finding Displacement from Velocity

• If \(v(t)\) is the velocity function, then the displacement function \(s(t)\) is given by \(s(t) = \int v(t) \, dt\).

Finding Velocity from Acceleration

• If \(a(t)\) is the acceleration function, then the velocity function \(v(t)\) is given by \(v(t) = \int a(t) \, dt\).

Example

• A particle moves in a straight line with acceleration \(a(t) = 6t\). If its initial velocity is \(v(0) = 5\) and its initial displacement is \(s(0) = 0\), find the displacement function \(s(t)\).
  1. Find the velocity function: \(v(t) = \int 6t \, dt = 3t^2 + c_1\). Using \(v(0) = 5\), we get \(5 = 3(0)^2 + c_1 \Rightarrow c_1 = 5\). Thus, \(v(t) = 3t^2 + 5\).
  2. Find the displacement function: \(s(t) = \int (3t^2 + 5) \, dt = t^3 + 5t + c_2\). Using \(s(0) = 0\), we get \(0 = (0)^3 + 5(0) + c_2 \Rightarrow c_2 = 0\). Thus, \(s(t) = t^3 + 5t\).

APPLICATION: Anti-derivatives are fundamental in physics for calculating displacement, velocity, and other motion-related quantities.

Common Integrals to Remember

REMEMBER: Integrate means “to add the areas”. The constant c represents an infinite number of vertical shifts of the function.