The Central Idea
Suppose
\[ F'(x)=f(x). \]Then \(F\) is called an antiderivative of \(f\).
Symbolically,
\[ \boxed{F'(x)=f(x)} \]means that \(F\) is one function whose derivative is \(f\).
For example, because
\[ \frac{d}{dx}(x^2)=2x, \]the function
\[ F(x)=x^2 \]is an antiderivative of
\[ f(x)=2x. \]Antidifferentiation Reverses Differentiation
Differentiation moves from a function to its rate of change:
\[ x^3 \longrightarrow 3x^2. \]Antidifferentiation moves in the reverse direction:
\[ 3x^2 \longrightarrow x^3. \]This reversal is the guiding idea:
\[ \boxed{\text{differentiate forward, antidifferentiate backward}.} \]The process is not always a perfect one-to-one reversal, however, because differentiation loses constant terms. That observation leads to one of the most important features of antiderivatives.
Why Antiderivatives Come in Families
Consider the functions
\[ x^2,\qquad x^2+3,\qquad x^2-10,\qquad x^2+\pi. \]Every one of them has derivative
\[ 2x. \]The reason is that the derivative of a constant is zero:
\[ \frac{d}{dx}(C)=0. \]Therefore,
\[ \frac{d}{dx}(x^2+C)=2x. \]So \(2x\) does not have just one antiderivative. It has an entire family:
\[ \boxed{x^2+C}. \]The Constant of Integration
The symbol \(C\) represents an arbitrary real constant and is called the constant of integration.
If
\[ F'(x)=f(x), \]then the complete family of antiderivatives is
\[ \boxed{F(x)+C}. \]The constant is not an optional decoration. It records the fact that infinitely many functions can have the same derivative.
The Indefinite Integral
The standard notation for the family of antiderivatives of \(f(x)\) is
\[ \boxed{\int f(x)\,dx=F(x)+C}, \]provided
\[ F'(x)=f(x). \]This expression is called an indefinite integral.
At this stage, the integral sign can be read as:
find all functions whose derivative is this integrand.
Parts of the Indefinite Integral
In
\[ \int f(x)\,dx, \]- \(\int\) is the integral sign;
- \(f(x)\) is the integrand;
- \(dx\) identifies \(x\) as the variable of integration.
The notation
\[ \int f(x)\,dx \]asks for an antiderivative with respect to \(x\).
Example: A Basic Antiderivative
Find
\[ \int 6x\,dx. \]We ask which function differentiates to \(6x\).
Since
\[ \frac{d}{dx}(3x^2)=6x, \]we have
\[ \boxed{\int 6x\,dx=3x^2+C}. \]Verification by Differentiation
The most reliable way to check an antiderivative is to differentiate the proposed answer.
Suppose
\[ \int 6x\,dx=3x^2+C. \]Differentiate:
\[ \frac{d}{dx}(3x^2+C)=6x. \]The original integrand returns, so the antiderivative is correct.
This gives a general checking rule:
\[ \boxed{\text{Differentiate your answer. The integrand should return.}} \]Antiderivative of a Constant
Because
\[ \frac{d}{dx}(kx)=k, \]we have
\[ \boxed{\int k\,dx=kx+C}. \]For example,
\[ \int 7\,dx=7x+C. \]Checking:
\[ \frac{d}{dx}(7x+C)=7. \]The Power Rule for Antiderivatives
Recall the derivative power rule:
\[ \frac{d}{dx}\left(x^{n+1}\right)=(n+1)x^n. \]Solving backward gives the antiderivative power rule:
\[ \boxed{\int x^n\,dx=\frac{x^{n+1}}{n+1}+C}, \]provided
\[ n\neq -1. \]Why the Power Rule Adds One to the Exponent
Suppose we want an antiderivative of
\[ x^4. \]Differentiation reduces an exponent by one. To reverse that effect, begin with exponent \(5\):
\[ \frac{d}{dx}(x^5)=5x^4. \]The coefficient \(5\) is too large, so divide by \(5\):
\[ \frac{d}{dx}\left(\frac{x^5}{5}\right)=x^4. \]Therefore,
\[ \boxed{\int x^4\,dx=\frac{x^5}{5}+C}. \]A Practical Power-Rule Shortcut
For a power \(x^n\), the routine is:
- Add \(1\) to the exponent.
- Divide by the new exponent.
- Add \(C\).
In compact form:
\[ x^n \longrightarrow \frac{x^{n+1}}{n+1}+C. \]This works for every real exponent for which the expression is being considered on a suitable domain, except \(n=-1\).
Example: Power Rule
Find
\[ \int x^7\,dx. \]Add one to the exponent:
\[ 7+1=8. \]Divide by the new exponent:
\[ \boxed{\int x^7\,dx=\frac{x^8}{8}+C}. \]Check:
\[ \frac{d}{dx}\left(\frac{x^8}{8}+C\right)=x^7. \]Constant Multiples
Constants can be carried through antidifferentiation:
\[ \boxed{\int kf(x)\,dx=k\int f(x)\,dx}. \]For example,
\[ \int 5x^3\,dx = 5\int x^3\,dx = 5\left(\frac{x^4}{4}\right)+C. \]Therefore,
\[ \boxed{\int 5x^3\,dx=\frac{5}{4}x^4+C}. \]Sums and Differences
Antidifferentiation distributes across sums and differences:
\[ \boxed{ \int [f(x)+g(x)]\,dx = \int f(x)\,dx+\int g(x)\,dx }. \]Likewise,
\[ \boxed{ \int [f(x)-g(x)]\,dx = \int f(x)\,dx-\int g(x)\,dx }. \]This allows polynomials to be integrated term by term.
Example: Antiderivative of a Polynomial
Find
\[ \int (6x^2-4x+9)\,dx. \]Integrate term by term:
\[ \int 6x^2\,dx - \int 4x\,dx + \int 9\,dx. \]This gives
\[ 6\left(\frac{x^3}{3}\right) - 4\left(\frac{x^2}{2}\right) + 9x+C. \]Simplify:
\[ \boxed{2x^3-2x^2+9x+C}. \]Why We Usually Write Only One \(C\)
If we integrate several terms separately, we might temporarily imagine
\[ F(x)+C_1+G(x)+C_2. \]But the sum of two arbitrary constants is still an arbitrary constant:
\[ C_1+C_2=C. \]Therefore, after integrating a sum, we normally write only one constant of integration at the end.
Negative Powers
The power rule also works for negative exponents other than \(-1\).
For example,
\[ \int x^{-3}\,dx = \frac{x^{-2}}{-2}+C. \]So
\[ \boxed{\int x^{-3}\,dx=-\frac{1}{2x^2}+C}. \]Differentiate to check:
\[ \frac{d}{dx}\left(-\frac12x^{-2}\right)=x^{-3}. \]Fractional Powers
Radical expressions can often be rewritten using rational exponents before antidifferentiating.
For example,
\[ \sqrt{x}=x^{1/2}. \]Therefore,
\[ \int \sqrt{x}\,dx = \int x^{1/2}\,dx = \frac{x^{3/2}}{3/2}+C. \]Dividing by \(3/2\) is the same as multiplying by \(2/3\):
\[ \boxed{\int \sqrt{x}\,dx=\frac{2}{3}x^{3/2}+C}. \]Example: Rewrite Before Integrating
Find
\[ \int \frac{1}{\sqrt{x}}\,dx. \]Rewrite:
\[ \frac{1}{\sqrt{x}}=x^{-1/2}. \]Apply the power rule:
\[ \int x^{-1/2}\,dx = \frac{x^{1/2}}{1/2}+C. \]Hence,
\[ \boxed{\int \frac{1}{\sqrt{x}}\,dx=2\sqrt{x}+C}. \]The Important Exception: \(n=-1\)
The power rule formula
\[ \frac{x^{n+1}}{n+1} \]cannot be used when
\[ n=-1, \]because then the denominator would be zero.
The exceptional integrand is
\[ x^{-1}=\frac{1}{x}. \]Its antiderivative comes from the logarithmic derivative:
\[ \frac{d}{dx}\ln|x|=\frac{1}{x}. \]Therefore,
\[ \boxed{\int \frac{1}{x}\,dx=\ln|x|+C}. \]Why the Logarithm Uses Absolute Value
The derivative
\[ \frac{d}{dx}\ln|x|=\frac{1}{x} \]holds wherever \(x\neq 0\).
The absolute value allows one formula to represent the logarithmic antiderivative on both the positive and negative parts of the domain.
Thus the standard form is
\[ \boxed{\ln|x|+C}, \]not merely \(\ln x+C\) unless the domain has already been restricted to \(x>0\).
Antiderivative of \(e^x\)
The exponential function \(e^x\) is special because it is its own derivative:
\[ \frac{d}{dx}e^x=e^x. \]Therefore,
\[ \boxed{\int e^x\,dx=e^x+C}. \]Antiderivative of \(a^x\)
Recall
\[ \frac{d}{dx}a^x=a^x\ln a \]for \(a>0\) and \(a\neq 1\).
Therefore,
\[ \boxed{\int a^x\,dx=\frac{a^x}{\ln a}+C}. \]The factor \(1/\ln a\) compensates for the \(\ln a\) introduced by differentiation.
Basic Trigonometric Antiderivatives
Because
\[ \frac{d}{dx}(\sin x)=\cos x, \]we have
\[ \boxed{\int \cos x\,dx=\sin x+C}. \]And because
\[ \frac{d}{dx}(\cos x)=-\sin x, \]we obtain
\[ \boxed{\int \sin x\,dx=-\cos x+C}. \]More Basic Trigonometric Antiderivatives
Several derivative formulas can be reversed directly:
\[ \boxed{\int \sec^2 x\,dx=\tan x+C}, \] \[ \boxed{\int \csc^2 x\,dx=-\cot x+C}, \] \[ \boxed{\int \sec x\tan x\,dx=\sec x+C}, \] \[ \boxed{\int \csc x\cot x\,dx=-\csc x+C}. \]These are best understood as reverse derivative patterns rather than as isolated formulas.
Core Antiderivative Table
| Integrand | Antiderivative |
|---|---|
| \(k\) | \(kx+C\) |
| \(x^n,\ n\neq -1\) | \(\dfrac{x^{n+1}}{n+1}+C\) |
| \(\dfrac{1}{x}\) | \(\ln|x|+C\) |
| \(e^x\) | \(e^x+C\) |
| \(a^x\) | \(\dfrac{a^x}{\ln a}+C\) |
| \(\cos x\) | \(\sin x+C\) |
| \(\sin x\) | \(-\cos x+C\) |
| \(\sec^2 x\) | \(\tan x+C\) |
| \(\csc^2 x\) | \(-\cot x+C\) |
| \(\sec x\tan x\) | \(\sec x+C\) |
| \(\csc x\cot x\) | \(-\csc x+C\) |
Antidifferentiation Is Pattern Recognition
Derivative rules tell us what patterns to look for in reverse.
If you recognize
\[ 4x^3, \]think of the derivative of \(x^4\).
If you recognize
\[ \cos x, \]think of the derivative of \(\sin x\).
If you recognize
\[ \frac{1}{x}, \]think of the derivative of \(\ln|x|\).
This pattern-based viewpoint becomes increasingly important as integration methods grow more sophisticated.
From a Family to One Particular Antiderivative
An indefinite integral produces a family
\[ F(x)+C. \]Additional information can determine the constant \(C\).
For example, suppose
\[ F'(x)=6x \]and
\[ F(1)=5. \]First antidifferentiate:
\[ F(x)=3x^2+C. \]Then use the condition \(F(1)=5\):
\[ 5=3(1)^2+C. \]Therefore,
\[ C=2. \]The particular function is
\[ \boxed{F(x)=3x^2+2}. \]Initial Value Problems
A problem that gives a derivative together with a function value is often called an initial value problem.
Its basic structure is
\[ F'(x)=f(x), \]together with a condition such as
\[ F(a)=b. \]The workflow is:
- Find the general antiderivative \(F(x)+C\).
- Substitute the given point.
- Solve for \(C\).
- Write the particular function.
- Differentiate to verify the derivative condition.
Antiderivatives in Motion
Antiderivatives naturally connect quantities related by rates of change.
If position is \(s(t)\), then velocity is
\[ v(t)=s'(t). \]Therefore, position can be recovered from velocity by antidifferentiation:
\[ s(t)=\int v(t)\,dt. \]Likewise, if
\[ a(t)=v'(t), \]then velocity can be recovered from acceleration:
\[ v(t)=\int a(t)\,dt. \]Each reversal introduces a constant that must be determined from additional information.
Example: Recovering Velocity from Acceleration
Suppose
\[ a(t)=6t-4 \]and the initial velocity is
\[ v(0)=3. \]Since \(v'(t)=a(t)\),
\[ v(t)=\int (6t-4)\,dt. \]Antidifferentiate:
\[ v(t)=3t^2-4t+C. \]Use \(v(0)=3\):
\[ 3=C. \]Therefore,
\[ \boxed{v(t)=3t^2-4t+3}. \]Antiderivatives of Rates in Applications
The same idea appears whenever one quantity is given as the derivative of another.
Examples include:
- velocity as the derivative of position;
- acceleration as the derivative of velocity;
- marginal cost as the derivative of total cost;
- marginal revenue as the derivative of total revenue;
- population growth rate as the derivative of population;
- flow rate as the derivative of accumulated volume.
In each case, antidifferentiation recovers a quantity from its rate of change, up to an additive constant.
Geometric Meaning of the Constant \(C\)
If \(F(x)\) is one antiderivative of \(f(x)\), then every function
\[ F(x)+C \]has the same derivative.
Graphically, changing \(C\) shifts the graph of \(F\) vertically without changing its slope at any \(x\)-value.
Thus all members of an antiderivative family have the same shape and differ only by vertical translation.
Why Any Two Antiderivatives Differ by a Constant
Suppose \(F\) and \(G\) are antiderivatives of the same function \(f\):
\[ F'(x)=f(x) \]and
\[ G'(x)=f(x). \]Then
\[ (F-G)'=F'-G'=f-f=0. \]On an interval, a differentiable function with derivative zero is constant. Therefore,
\[ F(x)-G(x)=C, \]or
\[ \boxed{F(x)=G(x)+C}. \]Domain Awareness
Antiderivative formulas must be interpreted on intervals where the integrand is defined.
For example,
\[ \frac{1}{x} \]is not defined at \(x=0\). Its domain is split into
\[ (-\infty,0) \]and
\[ (0,\infty). \]The formula \(\ln|x|+C\) works on either interval, but the constants may be chosen independently when treating disconnected pieces of the domain.
Rewrite into a Familiar Form First
Many elementary antiderivative problems become easier after algebraic rewriting.
For example,
\[ \int \frac{x^3+2x}{x}\,dx \]can first be simplified:
\[ \frac{x^3+2x}{x}=x^2+2 \]for \(x\neq 0\).
Then
\[ \int (x^2+2)\,dx = \frac{x^3}{3}+2x+C. \]So, on intervals not crossing \(x=0\),
\[ \boxed{\int \frac{x^3+2x}{x}\,dx=\frac{x^3}{3}+2x+C}. \]Sometimes Expanding Helps
If an integrand contains a simple product of polynomials, expansion may reveal terms that can be integrated directly.
For example,
\[ \int x(x+3)\,dx \]becomes
\[ \int (x^2+3x)\,dx. \]Therefore,
\[ \boxed{ \int x(x+3)\,dx = \frac{x^3}{3}+\frac{3x^2}{2}+C }. \]Not Every Product Can Be Integrated Factor by Factor
In general,
\[ \int f(x)g(x)\,dx \]is not equal to
\[ \left(\int f(x)\,dx\right) \left(\int g(x)\,dx\right). \]Likewise, there is no general rule that lets us integrate quotients by integrating the numerator and denominator separately.
Products and quotients often require algebraic simplification or later techniques such as substitution, integration by parts, or partial fractions.
Common Error: Forgetting \(+C\)
Writing
\[ \int 4x^3\,dx=x^4 \]identifies one antiderivative, but it does not describe the complete family.
The indefinite integral should be written
\[ \boxed{\int 4x^3\,dx=x^4+C}. \]Common Error: Using the Derivative Power Rule
Differentiation and antidifferentiation move exponents in opposite directions.
For differentiation,
\[ x^n \longrightarrow nx^{n-1}. \]For antidifferentiation,
\[ x^n \longrightarrow \frac{x^{n+1}}{n+1}+C. \]Do not multiply by the old exponent and subtract one when the task is to antidifferentiate.
Common Error: Dividing by the Old Exponent
For
\[ \int x^5\,dx, \]the new exponent is
\[ 5+1=6. \]Therefore, divide by \(6\), not by \(5\):
\[ \boxed{\int x^5\,dx=\frac{x^6}{6}+C}. \]Common Error: Applying the Power Rule to \(x^{-1}\)
If \(n=-1\), then
\[ n+1=0, \]so the power-rule denominator would be zero.
Instead remember the logarithmic pattern:
\[ \boxed{\int x^{-1}\,dx=\ln|x|+C}. \]Common Error: Sign Mistakes with Sine
Because
\[ \frac{d}{dx}(\cos x)=-\sin x, \]the antiderivative of \(\sin x\) is
\[ \boxed{-\cos x+C}. \]A quick derivative check catches the sign:
\[ \frac{d}{dx}(-\cos x)=\sin x. \]Common Error: Dropping \(C\) Before Using an Initial Condition
Suppose
\[ F'(x)=2x \]and
\[ F(3)=12. \]First write
\[ F(x)=x^2+C. \]Then use the condition:
\[ 12=9+C. \]Thus
\[ C=3 \]and
\[ \boxed{F(x)=x^2+3}. \]A Reliable Antiderivative Procedure
- Identify the variable of integration.
- Simplify or rewrite the integrand when useful.
- Separate sums and differences into recognizable pieces.
- Pull out constant factors when convenient.
- Match each term to a known reverse-derivative pattern.
- Apply the appropriate antiderivative rule.
- Write one constant of integration, \(+C\).
- Simplify the result.
- Differentiate the answer to verify it.
- If an initial condition is given, use it to determine \(C\).
Example: Full Polynomial Workflow
Find
\[ \int \left(3x^4-8x^3+5x^{-2}+6\right)\,dx. \]Integrate term by term:
\[ 3\frac{x^5}{5} - 8\frac{x^4}{4} + 5\frac{x^{-1}}{-1} + 6x + C. \]Simplify:
\[ \boxed{ \frac{3}{5}x^5 - 2x^4 - \frac{5}{x} + 6x + C }. \]Differentiate:
\[ 3x^4-8x^3+5x^{-2}+6. \]The integrand returns.
Example: Mixed Elementary Functions
Find
\[ \int \left(4e^x+3\cos x-\frac{2}{x}\right)\,dx. \]Use the basic patterns:
\[ \int 4e^x\,dx=4e^x, \] \[ \int 3\cos x\,dx=3\sin x, \]and
\[ \int -\frac{2}{x}\,dx=-2\ln|x|. \]Therefore,
\[ \boxed{ 4e^x+3\sin x-2\ln|x|+C }. \]Example: Initial Value Problem
Find \(F\) if
\[ F'(x)=3x^2-4 \]and
\[ F(2)=7. \]First antidifferentiate:
\[ F(x)=x^3-4x+C. \]Apply \(F(2)=7\):
\[ 7=8-8+C. \]Hence
\[ C=7. \]Therefore,
\[ \boxed{F(x)=x^3-4x+7}. \]Recovering a Function from a Second Derivative
Sometimes we are given
\[ F''(x) \]and must antidifferentiate twice.
Suppose
\[ F''(x)=6x. \]Integrate once:
\[ F'(x)=3x^2+C_1. \]Integrate again:
\[ F(x)=x^3+C_1x+C_2. \]Two integrations introduce two independent constants:
\[ \boxed{F(x)=x^3+C_1x+C_2}. \]Using Two Conditions After Two Antiderivatives
Suppose
\[ F''(x)=6x, \] \[ F'(0)=2, \]and
\[ F(0)=5. \]From the previous section,
\[ F'(x)=3x^2+C_1. \]Using \(F'(0)=2\) gives
\[ C_1=2. \]Then
\[ F(x)=x^3+2x+C_2. \]Using \(F(0)=5\) gives
\[ C_2=5. \]Therefore,
\[ \boxed{F(x)=x^3+2x+5}. \]Antiderivatives and Differential Equations
An equation involving an unknown function and its derivatives is called a differential equation.
The simple equation
\[ y'=f(x) \]can be solved by antidifferentiation:
\[ \boxed{y=\int f(x)\,dx}. \]Thus antiderivatives are the first and simplest examples of solving differential equations.
From Antiderivatives to Accumulation
So far, the integral sign has represented a family of antiderivatives. Integral calculus also studies accumulation: the total effect produced by continuously adding small contributions.
That second viewpoint will lead to Riemann sums and definite integrals.
The major structural result of integral calculus is that these two ideas— antiderivatives and accumulation—are deeply connected. That connection is formalized by the Fundamental Theorem of Calculus.
Practice 1: Constant
Find the indefinite integral:
\[ \int 12\,dx. \]Practice 2: Power Rule
Find:
\[ \int x^9\,dx. \]Practice 3: Constant Multiple
Find:
\[ \int 7x^4\,dx. \]Practice 4: Polynomial
Find:
\[ \int (4x^3-6x^2+8)\,dx. \]Practice 5: Negative Power
Find:
\[ \int x^{-4}\,dx. \]Practice 6: Radical
Find:
\[ \int \sqrt[3]{x}\,dx. \]Practice 7: Logarithmic Pattern
Find:
\[ \int \frac{5}{x}\,dx. \]Practice 8: Trigonometric Pattern
Find:
\[ \int (3\cos x-2\sin x)\,dx. \]Practice 9: Initial Condition
Find \(F(x)\) if
\[ F'(x)=4x^3+2 \]and
\[ F(1)=6. \]Practice 10: Motion
A particle has acceleration
\[ a(t)=8t \]and initial velocity
\[ v(0)=-3. \]Find \(v(t)\).
Practice 11: Second Derivative
Find the general form of \(F(x)\) if
\[ F''(x)=12x^2. \]Practice 12: Identify the Error
A student writes
\[ \int x^6\,dx=\frac{x^7}{6}+C. \]Identify and correct the error.
Practice Answers
Practice 1
\[ \boxed{\int 12\,dx=12x+C}. \]Practice 2
\[ \boxed{\int x^9\,dx=\frac{x^{10}}{10}+C}. \]Practice 3
\[ \boxed{\int 7x^4\,dx=\frac{7}{5}x^5+C}. \]Practice 4
\[ \int (4x^3-6x^2+8)\,dx = x^4-2x^3+8x+C. \]Therefore,
\[ \boxed{x^4-2x^3+8x+C}. \]Practice 5
\[ \int x^{-4}\,dx = \frac{x^{-3}}{-3}+C. \]Thus,
\[ \boxed{-\frac{1}{3x^3}+C}. \]Practice 6
Rewrite
\[ \sqrt[3]{x}=x^{1/3}. \]Then
\[ \int x^{1/3}\,dx = \frac{x^{4/3}}{4/3}+C = \boxed{\frac34x^{4/3}+C}. \]Practice 7
\[ \boxed{\int \frac{5}{x}\,dx=5\ln|x|+C}. \]Practice 8
\[ \int (3\cos x-2\sin x)\,dx = 3\sin x+2\cos x+C. \]Therefore,
\[ \boxed{3\sin x+2\cos x+C}. \]Practice 9
First antidifferentiate:
\[ F(x)=x^4+2x+C. \]Apply \(F(1)=6\):
\[ 6=1+2+C. \]Hence \(C=3\), so
\[ \boxed{F(x)=x^4+2x+3}. \]Practice 10
Since
\[ v'(t)=a(t)=8t, \]we have
\[ v(t)=4t^2+C. \]Using \(v(0)=-3\) gives \(C=-3\). Therefore,
\[ \boxed{v(t)=4t^2-3}. \]Practice 11
Integrate once:
\[ F'(x)=4x^3+C_1. \]Integrate again:
\[ \boxed{F(x)=x^4+C_1x+C_2}. \]Practice 12
The exponent was increased correctly from \(6\) to \(7\), but the student divided by the old exponent instead of the new one.
The correct result is
\[ \boxed{\int x^6\,dx=\frac{x^7}{7}+C}. \]What to Remember
An antiderivative reverses differentiation:
\[ \boxed{F'(x)=f(x)} \]means that \(F\) is an antiderivative of \(f\).
The indefinite integral represents the complete family:
\[ \boxed{\int f(x)\,dx=F(x)+C}. \]The central power rule is
\[ \boxed{\int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\qquad n\neq -1}. \]The exceptional logarithmic pattern is
\[ \boxed{\int \frac{1}{x}\,dx=\ln|x|+C}. \]Most importantly:
- recognize antiderivatives as reverse derivatives;
- include the constant of integration;
- rewrite algebraically when that reveals familiar forms;
- use initial conditions to determine a particular constant;
- differentiate the result to check your work.
Where We Go Next
Antiderivatives recover functions from their rates of change. The next lesson introduces a different but complementary idea: approximating total accumulation by adding many small contributions.
In Riemann Sums, we will divide an interval into pieces, form rectangular approximations, and begin building the geometric and numerical meaning of the definite integral.