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.