🎯⭐ INTERACTIVE LESSON

Antiderivatives & Indefinite Integrals

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Antiderivatives & Indefinite Integrals - Complete Interactive Lesson

Part 1: Antiderivative Basics

Antiderivatives & Indefinite Integrals

Part 1 of 7 — What is an Antiderivative?

Definition

An antiderivative of f(x)f(x) is a function F(x)F(x) such that F(x)=f(x)F'(x) = f(x).

The indefinite integral represents the family of all antiderivatives:

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

The "+C+ C" is essential! Since the derivative of a constant is 0, there are infinitely many antiderivatives.

Power Rule for Integration

xndx=xn+1n+1+C(n1)\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)

This reverses the power rule for differentiation.

Worked Examples

f(x)f(x)f(x)dx\int f(x)\,dx
x4x^4x55+C\frac{x^5}{5} + C
x3x^{-3}x22+C=12x2+C\frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C
x=x1/2\sqrt{x} = x^{1/2}x3/23/2+C=23x3/2+C\frac{x^{3/2}}{3/2} + C = \frac{2}{3}x^{3/2} + C
1x4=x4\frac{1}{x^4} = x^{-4}x33+C=13x3+C\frac{x^{-3}}{-3} + C = -\frac{1}{3x^3} + C

Power Rule for Integration 🎯

Key Takeaways — Part 1

  1. An antiderivative reverses differentiation: if F=fF' = f, then f=F+C\int f = F + C
  2. Power Rule: add 1 to the exponent, divide by the new exponent
  3. Always include +C+C for indefinite integrals
  4. The special case n=1n = -1: x1dx=lnx+C\int x^{-1}\,dx = \ln|x| + C (next part)

Part 2: Power Rule for Integration

Antiderivatives

Part 2 of 7 — Essential Antiderivative Formulas

Complete Table of Basic Antiderivatives

FunctionAntiderivative
xnx^n (n1)(n \neq -1)xn+1n+1+C\frac{x^{n+1}}{n+1} + C
1x\frac{1}{x}$\ln
exe^xex+Ce^x + C
axa^xaxlna+C\frac{a^x}{\ln a} + C
sinx\sin xcosx+C-\cos x + C
cosx\cos xsinx+C\sin x + C
sec2x\sec^2 xtanx+C\tan x + C
csc2x\csc^2 xcotx+C-\cot x + C
secxtanx\sec x \tan xsecx+C\sec x + C
cscxcotx\csc x \cot xcscx+C-\csc x + C
11x2\frac{1}{\sqrt{1-x^2}}arcsinx+C\arcsin x + C
11+x2\frac{1}{1+x^2}arctanx+C\arctan x + C

Memorize this table! These are the building blocks of all integration.

Essential Antiderivatives 🎯

Key Takeaways — Part 2

  1. Memorize the complete table of basic antiderivatives
  2. 1xdx=lnx+C\int \frac{1}{x}\,dx = \ln|x| + C (absolute value matters!)
  3. axdx=axlna+C\int a^x\,dx = \frac{a^x}{\ln a} + C (not the power rule!)
  4. Linearity: [af+bg]dx=afdx+bgdx\int [af + bg]\,dx = a\int f\,dx + b\int g\,dx

Part 3: Trig Antiderivatives

Antiderivatives

Part 3 of 7 — Initial Value Problems (IVPs)

Finding Specific Antiderivatives

An initial condition pins down the value of CC:

Given: f(x)=3x24x+1f'(x) = 3x^2 - 4x + 1 and f(0)=5f(0) = 5. Find f(x)f(x).

Step 1: Find the general antiderivative. f(x)=x32x2+x+Cf(x) = x^3 - 2x^2 + x + C

Step 2: Use the initial condition to find CC. f(0)=00+0+C=5    C=5f(0) = 0 - 0 + 0 + C = 5 \implies C = 5

Answer: f(x)=x32x2+x+5f(x) = x^3 - 2x^2 + x + 5

Position-Velocity-Acceleration

If a(t)=a(t) = acceleration, then:

  • v(t)=a(t)dtv(t) = \int a(t)\,dt (velocity)
  • s(t)=v(t)dts(t) = \int v(t)\,dt (position)

Each integration introduces a constant determined by initial conditions.

Initial Value Problems 🎯

Key Takeaways — Part 3

  1. An IVP consists of a derivative equation plus initial condition(s)
  2. Find the general antiderivative, then plug in the initial condition to find CC
  3. For particle motion: integrate a(t)a(t) to get v(t)v(t), integrate v(t)v(t) to get s(t)s(t)

Part 4: Initial Value Problems

Antiderivatives

Part 4 of 7 — Rewriting Before Integrating

Algebraic Manipulation

Many integrals require rewriting before applying basic rules.

Expand Products

(x+1)(x3)dx=(x22x3)dx=x33x23x+C\int (x+1)(x-3)\,dx = \int (x^2 - 2x - 3)\,dx = \frac{x^3}{3} - x^2 - 3x + C

Split Fractions

x3+2xx2dx=(x+2x)dx=x22+2lnx+C\int \frac{x^3 + 2x}{x^2}\,dx = \int \left(x + \frac{2}{x}\right)\,dx = \frac{x^2}{2} + 2\ln|x| + C

Rewrite Radicals

3xdx=3x1/2dx=3x1/21/2+C=6x+C\int \frac{3}{\sqrt{x}}\,dx = \int 3x^{-1/2}\,dx = 3 \cdot \frac{x^{1/2}}{1/2} + C = 6\sqrt{x} + C

Simplify Then Integrate 🎯

Key Takeaways — Part 4

  1. Expand products before integrating
  2. Split fractions into separate terms when possible
  3. Rewrite radicals using fractional exponents
  4. These techniques reduce complex integrals to sums of power rule applications

Part 5: Motion Applications

Antiderivatives

Part 5 of 7 — Inverse Trig Antiderivatives

Three Key Formulas

1a2x2dx=arcsin(xa)+C\int \frac{1}{\sqrt{a^2 - x^2}}\,dx = \arcsin\left(\frac{x}{a}\right) + C

1a2+x2dx=1aarctan(xa)+C\int \frac{1}{a^2 + x^2}\,dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C

1xx2a2dx=1aarcsec(xa)+C\int \frac{1}{x\sqrt{x^2 - a^2}}\,dx = \frac{1}{a}\text{arcsec}\left(\frac{|x|}{a}\right) + C

Recognition is Key

The AP Exam tests whether you can recognize these forms:

  • Square root of a2x2a^2 - x^2 in denominator → arcsin\arcsin
  • Sum of squares a2+x2a^2 + x^2 in denominator → arctan\arctan

Inverse Trig Integrals 🎯

Key Takeaways — Part 5

  1. a2x2\sqrt{a^2 - x^2} in denominator → arcsin(x/a)\arcsin(x/a)
  2. a2+x2a^2 + x^2 in denominator → 1aarctan(x/a)\frac{1}{a}\arctan(x/a)
  3. Often need u-sub to get into the standard form first

Part 6: Problem-Solving Workshop

Antiderivatives

Part 6 of 7 — Mixed Practice

Time to combine all antiderivative techniques.

Mixed Antiderivative Problems 🎯

Workshop Complete!

Practice makes perfect with antiderivatives. Always verify by differentiating your answer.

Part 7: Review & Applications

Antiderivatives — Review

Part 7 of 7 — Comprehensive Assessment

Final Assessment 🎯

Antiderivatives — Complete! ✅

You have mastered:

  • ✅ Power Rule for integration
  • ✅ All basic antiderivative formulas
  • ✅ Initial Value Problems
  • ✅ Rewriting before integrating
  • ✅ Inverse trig antiderivatives