Introduction to Antiderivatives

Understanding the reverse process of differentiation

🔄 Introduction to Antiderivatives

What is an Antiderivative?

An antiderivative of a function f(x)f(x) is a function F(x)F(x) whose derivative is f(x)f(x).

💡 Key Idea: If F(x)=f(x)F'(x) = f(x), then F(x)F(x) is an antiderivative of f(x)f(x).

Antidifferentiation is the reverse of differentiation!


The Basic Question

Differentiation asks: Given F(x)F(x), what is F(x)F'(x)?

Antidifferentiation asks: Given f(x)f(x), what is F(x)F(x) such that F(x)=f(x)F'(x) = f(x)?

Example

Differentiation: If F(x)=x2F(x) = x^2, then F(x)=2xF'(x) = 2x

Antidifferentiation: If f(x)=2xf(x) = 2x, then F(x)=x2F(x) = x^2 is an antiderivative


Notation

If F(x)=f(x)F'(x) = f(x), we write:

  • "F(x)F(x) is an antiderivative of f(x)f(x)"
  • Or: "F(x)F(x) is the integral of f(x)f(x)"

We'll learn the integral notation \int soon!


Simple Examples

Example 1: Power Function

Question: Find an antiderivative of f(x)=x3f(x) = x^3

Think: What function has derivative x3x^3?

Try F(x)=x4F(x) = x^4: F(x)=4x3x3F'(x) = 4x^3 \neq x^3

Try F(x)=14x4F(x) = \frac{1}{4}x^4: F(x)=14(4x3)=x3F'(x) = \frac{1}{4}(4x^3) = x^3

Answer: F(x)=14x4F(x) = \frac{1}{4}x^4 is an antiderivative of f(x)=x3f(x) = x^3


Example 2: Trigonometric Function

Question: Find an antiderivative of f(x)=cosxf(x) = \cos x

Think: What function has derivative cosx\cos x?

We know: ddx[sinx]=cosx\frac{d}{dx}[\sin x] = \cos x

Answer: F(x)=sinxF(x) = \sin x is an antiderivative of f(x)=cosxf(x) = \cos x


Example 3: Exponential Function

Question: Find an antiderivative of f(x)=exf(x) = e^x

Think: What function has derivative exe^x?

We know: ddx[ex]=ex\frac{d}{dx}[e^x] = e^x

Answer: F(x)=exF(x) = e^x is an antiderivative of f(x)=exf(x) = e^x


The "+C" Problem

Here's something important: antiderivatives are not unique!

Why Not?

If F(x)=x2F(x) = x^2, then F(x)=2xF'(x) = 2x

But also:

  • If G(x)=x2+5G(x) = x^2 + 5, then G(x)=2xG'(x) = 2x
  • If H(x)=x27H(x) = x^2 - 7, then H(x)=2xH'(x) = 2x
  • If K(x)=x2+100K(x) = x^2 + 100, then K(x)=2xK'(x) = 2x

All of these are antiderivatives of f(x)=2xf(x) = 2x!

The General Solution

The most general antiderivative of f(x)=2xf(x) = 2x is: F(x)=x2+CF(x) = x^2 + C

where CC is any constant (called the constant of integration).


Why Does +C Work?

Theorem: If F(x)F(x) is an antiderivative of f(x)f(x), then the most general antiderivative is: F(x)+CF(x) + C

where CC is any constant.

Reason: The derivative of any constant is zero! ddx[F(x)+C]=F(x)+0=f(x)\frac{d}{dx}[F(x) + C] = F'(x) + 0 = f(x)


Different Functions, Same Derivative

Two functions with the same derivative differ by a constant.

Theorem: If F(x)=G(x)F'(x) = G'(x) for all xx, then F(x)=G(x)+CF(x) = G(x) + C for some constant CC.

This is why we need the +C+C!


Finding Antiderivatives: The Reverse Rules

To find antiderivatives, we reverse our derivative rules!

Power Rule (Reversed)

Derivative: ddx[xn]=nxn1\frac{d}{dx}[x^n] = nx^{n-1}

Antiderivative: If f(x)=xnf(x) = x^n (where n1n \neq -1), then: F(x)=xn+1n+1+CF(x) = \frac{x^{n+1}}{n+1} + C

Remember:

  • Add 1 to the exponent
  • Divide by the new exponent
  • Don't forget +C!

Examples Using Power Rule

Example 1: Find an antiderivative of f(x)=x5f(x) = x^5

F(x)=x5+15+1+C=x66+CF(x) = \frac{x^{5+1}}{5+1} + C = \frac{x^6}{6} + C

Check: F(x)=6x56=x5F'(x) = \frac{6x^5}{6} = x^5


Example 2: Find an antiderivative of f(x)=xf(x) = x

F(x)=x1+11+1+C=x22+CF(x) = \frac{x^{1+1}}{1+1} + C = \frac{x^2}{2} + C

Check: F(x)=2x2=xF'(x) = \frac{2x}{2} = x


Example 3: Find an antiderivative of f(x)=1=x0f(x) = 1 = x^0

F(x)=x0+10+1+C=x11+C=x+CF(x) = \frac{x^{0+1}}{0+1} + C = \frac{x^1}{1} + C = x + C

Check: F(x)=1F'(x) = 1


Example 4: Find an antiderivative of f(x)=1x2=x2f(x) = \frac{1}{x^2} = x^{-2}

F(x)=x2+12+1+C=x11+C=1x+CF(x) = \frac{x^{-2+1}}{-2+1} + C = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C

Check: F(x)=(1)x2=1x2F'(x) = -(-1)x^{-2} = \frac{1}{x^2}


Common Antiderivatives to Know

Basic Functions

| f(x)f(x) | F(x)F(x) (antiderivative) | Check | |--------|------------------------|-------| | kk (constant) | kx+Ckx + C | (kx)=k(kx)' = k | | xnx^n (n1n \neq -1) | xn+1n+1+C\frac{x^{n+1}}{n+1} + C | Power rule | | 1x\frac{1}{x} | lnx+C\ln|x| + C | (lnx)=1x(\ln|x|)' = \frac{1}{x} | | exe^x | ex+Ce^x + C | (ex)=ex(e^x)' = e^x |

Trigonometric Functions

| f(x)f(x) | F(x)F(x) | Check | |--------|--------|-------| | sinx\sin x | cosx+C-\cos x + C | (cosx)=sinx(-\cos x)' = \sin x | | cosx\cos x | sinx+C\sin x + C | (sinx)=cosx(\sin x)' = \cos x | | sec2x\sec^2 x | tanx+C\tan x + C | (tanx)=sec2x(\tan x)' = \sec^2 x | | secxtanx\sec x \tan x | secx+C\sec x + C | (secx)=secxtanx(\sec x)' = \sec x \tan x |

Note: These will be covered in more detail in later lessons!


Linearity Properties

Antiderivatives follow the same linearity rules as derivatives:

Constant Multiple Rule

If F(x)F(x) is an antiderivative of f(x)f(x), then: kF(x) is an antiderivative of kf(x)k \cdot F(x) \text{ is an antiderivative of } k \cdot f(x)

Example: Antiderivative of 5x25x^2: F(x)=5x33+C=5x33+CF(x) = 5 \cdot \frac{x^3}{3} + C = \frac{5x^3}{3} + C


Sum/Difference Rule

If F(x)F(x) is an antiderivative of f(x)f(x) and G(x)G(x) is an antiderivative of g(x)g(x), then: F(x)±G(x) is an antiderivative of f(x)±g(x)F(x) \pm G(x) \text{ is an antiderivative of } f(x) \pm g(x)

Example: Antiderivative of x2+xx^2 + x: F(x)=x33+x22+CF(x) = \frac{x^3}{3} + \frac{x^2}{2} + C


Complete Example: Polynomial

Find the most general antiderivative of f(x)=3x42x3+5x7f(x) = 3x^4 - 2x^3 + 5x - 7.

Step 1: Break it down

Use sum rule and constant multiple rule:

  • Antiderivative of 3x43x^4 is 3x55=3x553 \cdot \frac{x^5}{5} = \frac{3x^5}{5}
  • Antiderivative of 2x3-2x^3 is 2x44=x42-2 \cdot \frac{x^4}{4} = -\frac{x^4}{2}
  • Antiderivative of 5x5x is 5x22=5x225 \cdot \frac{x^2}{2} = \frac{5x^2}{2}
  • Antiderivative of 7-7 is 7x-7x

Step 2: Combine

F(x)=3x55x42+5x227x+CF(x) = \frac{3x^5}{5} - \frac{x^4}{2} + \frac{5x^2}{2} - 7x + C


Step 3: Check (optional but recommended!)

F(x)=35x454x32+52x27F'(x) = \frac{3 \cdot 5x^4}{5} - \frac{4x^3}{2} + \frac{5 \cdot 2x}{2} - 7

=3x42x3+5x7= 3x^4 - 2x^3 + 5x - 7


The Special Case: 1x\frac{1}{x}

Why can't we use the power rule?

For f(x)=1x=x1f(x) = \frac{1}{x} = x^{-1}, the power rule would give: F(x)=x1+11+1=x00=10F(x) = \frac{x^{-1+1}}{-1+1} = \frac{x^0}{0} = \frac{1}{0}

Division by zero! ❌

The answer: 1xdx=lnx+C\int \frac{1}{x}\,dx = \ln|x| + C

This is a special case you must memorize!


⚠️ Common Mistakes

Mistake 1: Forgetting +C

WRONG: The antiderivative of x2x^2 is x33\frac{x^3}{3}

RIGHT: The antiderivative of x2x^2 is x33+C\frac{x^3}{3} + C

Always include the constant of integration!

Mistake 2: Wrong Power Rule

WRONG: Antiderivative of x3x^3 is x33\frac{x^3}{3} (forgot to add 1 to exponent!)

RIGHT: Antiderivative of x3x^3 is x44+C\frac{x^4}{4} + C

Mistake 3: Treating Product Like Sum

WRONG: Antiderivative of xx2=x3x \cdot x^2 = x^3 is... (treating separately)

RIGHT: Simplify first! xx2=x3x \cdot x^2 = x^3, so antiderivative is x44+C\frac{x^4}{4} + C

Mistake 4: Sine and Cosine Signs

Remember:

  • Antiderivative of sinx\sin x is cosx+C-\cos x + C (negative!)
  • Antiderivative of cosx\cos x is sinx+C\sin x + C (positive!)

Checking Your Work

Always check by differentiating your answer!

If you found F(x)F(x) as an antiderivative of f(x)f(x), verify: F(x)=f(x)F'(x) = f(x)

This is a foolproof way to catch errors!


Initial Value Problems

Sometimes we can find the specific value of CC using an initial condition.

Example

Find F(x)F(x) such that F(x)=2xF'(x) = 2x and F(1)=5F(1) = 5.

Step 1: Find general antiderivative F(x)=x2+CF(x) = x^2 + C

Step 2: Use initial condition F(1)=5F(1) = 5 F(1)=12+C=5F(1) = 1^2 + C = 5 1+C=51 + C = 5 C=4C = 4

Step 3: Write specific solution F(x)=x2+4F(x) = x^2 + 4


Physical Interpretations

Position, Velocity, Acceleration

If a(t)a(t) is acceleration:

  • v(t)v(t) is velocity (antiderivative of aa)
  • s(t)s(t) is position (antiderivative of vv)

Example: If a(t)=32a(t) = -32 ft/s² (gravity), then: v(t)=32t+C1v(t) = -32t + C_1 s(t)=16t2+C1t+C2s(t) = -16t^2 + C_1 t + C_2

The constants are determined by initial conditions!


📝 Practice Strategy

  1. Memorize basic antiderivatives (powers, trig, exponential)
  2. Always include +C in your answer
  3. Use linearity to break complex problems into simpler parts
  4. Check your work by differentiating
  5. For power rule: Add 1 to exponent, divide by new exponent
  6. Watch for special cases like 1x\frac{1}{x}
  7. For initial value problems: Find general solution first, then solve for CC

📚 Practice Problems

1Problem 1easy

Question:

Find the most general antiderivative of f(x)=6x24x+3f(x) = 6x^2 - 4x + 3.

💡 Show Solution

Step 1: Apply power rule to each term

For 6x26x^2: Antiderivative=6x2+12+1=6x33=2x3\text{Antiderivative} = 6 \cdot \frac{x^{2+1}}{2+1} = 6 \cdot \frac{x^3}{3} = 2x^3

For 4x=4x1-4x = -4x^1: Antiderivative=4x1+11+1=4x22=2x2\text{Antiderivative} = -4 \cdot \frac{x^{1+1}}{1+1} = -4 \cdot \frac{x^2}{2} = -2x^2

For 3=3x03 = 3x^0: Antiderivative=3x\text{Antiderivative} = 3x


Step 2: Combine and add +C

F(x)=2x32x2+3x+CF(x) = 2x^3 - 2x^2 + 3x + C


Step 3: Check by differentiating

F(x)=2(3x2)2(2x)+3=6x24x+3F'(x) = 2(3x^2) - 2(2x) + 3 = 6x^2 - 4x + 3

Answer: F(x)=2x32x2+3x+CF(x) = 2x^3 - 2x^2 + 3x + C

2Problem 2medium

Question:

Find f(x)f(x) if f(x)=4x36xf'(x) = 4x^3 - 6x and f(2)=10f(2) = 10.

💡 Show Solution

Step 1: Find the general antiderivative

f(x)=(4x36x)dxf(x) = \int (4x^3 - 6x)\,dx

For 4x34x^3: 4x44=x4\frac{4x^4}{4} = x^4

For 6x-6x: 6x22=3x2-6 \cdot \frac{x^2}{2} = -3x^2

General solution: f(x)=x43x2+Cf(x) = x^4 - 3x^2 + C


Step 2: Use initial condition f(2)=10f(2) = 10

f(2)=(2)43(2)2+C=10f(2) = (2)^4 - 3(2)^2 + C = 10

163(4)+C=1016 - 3(4) + C = 10

1612+C=1016 - 12 + C = 10

4+C=104 + C = 10

C=6C = 6


Step 3: Write the specific solution

f(x)=x43x2+6f(x) = x^4 - 3x^2 + 6


Check:

  • f(x)=4x36xf'(x) = 4x^3 - 6x
  • f(2)=1612+6=10f(2) = 16 - 12 + 6 = 10

Answer: f(x)=x43x2+6f(x) = x^4 - 3x^2 + 6

3Problem 3hard

Question:

A particle moves along a line with acceleration a(t)=12t6a(t) = 12t - 6 m/s². At t=0t=0, the velocity is v(0)=2v(0) = 2 m/s and position is s(0)=5s(0) = 5 m. Find the position function s(t)s(t).

💡 Show Solution

Step 1: Find velocity from acceleration

v(t)=a(t)dt=(12t6)dtv(t) = \int a(t)\,dt = \int (12t - 6)\,dt

v(t)=12t226t+C1=6t26t+C1v(t) = \frac{12t^2}{2} - 6t + C_1 = 6t^2 - 6t + C_1


Step 2: Use initial velocity v(0)=2v(0) = 2

v(0)=6(0)26(0)+C1=2v(0) = 6(0)^2 - 6(0) + C_1 = 2

C1=2C_1 = 2

So: v(t)=6t26t+2v(t) = 6t^2 - 6t + 2


Step 3: Find position from velocity

s(t)=v(t)dt=(6t26t+2)dts(t) = \int v(t)\,dt = \int (6t^2 - 6t + 2)\,dt

s(t)=6t336t22+2t+C2s(t) = \frac{6t^3}{3} - \frac{6t^2}{2} + 2t + C_2

s(t)=2t33t2+2t+C2s(t) = 2t^3 - 3t^2 + 2t + C_2


Step 4: Use initial position s(0)=5s(0) = 5

s(0)=2(0)33(0)2+2(0)+C2=5s(0) = 2(0)^3 - 3(0)^2 + 2(0) + C_2 = 5

C2=5C_2 = 5


Answer: s(t)=2t33t2+2t+5s(t) = 2t^3 - 3t^2 + 2t + 5 meters


Summary:

  • Acceleration: a(t)=12t6a(t) = 12t - 6
  • Velocity: v(t)=6t26t+2v(t) = 6t^2 - 6t + 2
  • Position: s(t)=2t33t2+2t+5s(t) = 2t^3 - 3t^2 + 2t + 5