Integral Symbol
\[ \int \]Indicates integration.
Calculus Guide
Learn how integrals represent accumulation, reverse differentiation, and measure quantities such as signed area over an interval.
Accumulation and Antiderivatives
Integration is one of the two central operations of calculus.
While derivatives measure instantaneous change, integrals measure accumulation.
Integrals also reverse the process of differentiation by finding antiderivatives.
Integral
Integration asks for a function whose derivative is \(f(x)\), or measures accumulated change over an interval.
Big Idea
If
\[ F'(x)=f(x), \]then
\[ \int f(x)\,dx = F(x)+C. \]The function \(F\) is an antiderivative of \(f\).
Reverse Differentiation
An antiderivative of \(f(x)\) is a function whose derivative equals \(f(x)\).
Definition
\[ \boxed{ F'(x)=f(x) } \]means \(F(x)\) is an antiderivative of \(f(x)\).
Example
We need a function whose derivative is \(6x\).
Since
\[ \frac{d}{dx}(3x^2)=6x, \]one antiderivative is
\[ 3x^2. \]But so are
\[ 3x^2+1, \qquad 3x^2-5, \qquad 3x^2+100. \]All of these differ only by a constant.
Families of Antiderivatives
Integral Symbol
\[ \int \]Indicates integration.
Integrand
\[ f(x) \]The expression being integrated.
Differential
\[ dx \]Indicates the variable of integration.
Constant of Integration
\[ C \]Represents all possible constant shifts of the antiderivative.
Do Not Forget \(+C\)
Function
\[ F(x)=x^2 \] \[ F'(x)=2x \]Shifted Function
\[ G(x)=x^2+7 \] \[ G'(x)=2x \]Why \(+C\)?
Every function of the form
\[ x^2+C \]has derivative
\[ 2x. \]Therefore,
\[ \int 2x\,dx = x^2+C. \]Core Rules
Constant Rule
\[ \int c\,dx = cx+C \]Power Rule
\[ \int x^n\,dx = \frac{x^{n+1}}{n+1}+C \]for \(n\neq -1\)
Constant Multiple
\[ \int c f(x)\,dx = c \int f(x)\,dx \]Sum Rule
\[ \int \left[ f(x)+g(x) \right] dx = \int f(x)\,dx + \int g(x)\,dx \]Difference Rule
\[ \int \left[ f(x)-g(x) \right] dx = \int f(x)\,dx - \int g(x)\,dx \]The Main Algebraic Rule
provided \(n\neq -1\).
Example
\[ \int x^4\,dx \] \[ = \frac{x^5}{5}+C \]Example
\[ \int x^{-3}\,dx \] \[ = \frac{x^{-2}}{-2}+C \] \[ = -\frac{1}{2x^2}+C \]Example
\[ \int \sqrt{x}\,dx \]Rewrite:
\[ \sqrt{x} = x^{1/2} \] \[ = \frac{x^{3/2}}{3/2}+C \] \[ = \frac{2}{3}x^{3/2}+C \]Integrate Term by Term
Example
Integrate each term:
\[ \int 4x^3\,dx = x^4 \] \[ \int -6x^2\,dx = -2x^3 \] \[ \int 5x\,dx = \frac{5}{2}x^2 \] \[ \int -8\,dx = -8x. \]The Exception to the Power Rule
Power Rule Exception
Since \(x^{-1}=\frac1x\), using the power rule would require dividing by zero.
Accumulation Over an Interval
A definite integral gives a number rather than a family of functions.
Lower Limit
\[ a \]Starting \(x\)-value.
Upper Limit
\[ b \]Ending \(x\)-value.
Integrand
\[ f(x) \]Quantity being accumulated.
Geometric Meaning
Above the \(x\)-Axis
\[ f(x)>0 \]Contributes positive area.
Below the \(x\)-Axis
\[ f(x)<0 \]Contributes negative signed area.
Important Distinction
A definite integral gives net signed area. Total geometric area may require splitting the interval where the function crosses the \(x\)-axis.
The Central Connection
The Fundamental Theorem connects differentiation and integration.
Evaluating a Definite Integral
\[ \boxed{ \int_a^b f(x)\,dx = F(b)-F(a) } \]where \(F'(x)=f(x)\).
Example
An antiderivative of \(2x\) is
\[ F(x)=x^2. \]Apply the Fundamental Theorem:
\[ \left[ x^2 \right]_1^3 = 3^2-1^2. \] \[ = 9-1. \]Accumulation Functions
Example
By the Fundamental Theorem,
Dummy Variable
The \(t\) inside the integral is a temporary variable. The result is expressed in terms of the upper limit \(x\).
Variable Upper Bounds
Example
\[ G(x) = \int_0^{x^2} \cos t\,dt. \]Evaluate the integrand at the upper bound:
\[ \cos(x^2). \]Then multiply by the derivative of the upper bound:
\[ \frac{d}{dx}(x^2)=2x. \]Essential Formulas
Cosine
\[ \boxed{ \int \cos x\,dx = \sin x+C } \]Sine
\[ \boxed{ \int \sin x\,dx = -\cos x+C } \]Secant Squared
\[ \boxed{ \int \sec^2x\,dx = \tan x+C } \]Cosecant Squared
\[ \boxed{ \int \csc^2x\,dx = -\cot x+C } \]Secant-Tangent
\[ \boxed{ \int \sec x\tan x\,dx = \sec x+C } \]Cosecant-Cotangent
\[ \boxed{ \int \csc x\cot x\,dx = -\csc x+C } \]Reverse Chain Rule
\(u\)-substitution simplifies an integral when one part of the integrand is the derivative of another part.
Main Idea
If
\[ u=g(x), \]then
\[ du=g'(x)\,dx. \]Choose
Differentiate
Rewrite
Integrate
Substitute Back
Worked Example
Example
Choose
\[ u=x^2+1. \]Differentiate:
\[ du=2x\,dx. \]Rewrite the integral:
\[ \int u^4\,du. \]Integrate:
\[ \frac{u^5}{5}+C. \]Substitute back:
Missing a Constant?
Example
\[ \int x(x^2+4)^3\,dx. \]Let
\[ u=x^2+4. \]Then
\[ du=2x\,dx. \]Therefore,
\[ x\,dx = \frac12\,du. \]Rewrite:
\[ \frac12 \int u^3\,du. \] \[ = \frac12 \cdot \frac{u^4}{4} + C. \] \[ = \frac{u^4}{8}+C. \]Substitution with Bounds
Method 1
Integrate in \(u\), replace \(u\) with the original expression, then use the original \(x\)-bounds.
Method 2
Convert the original \(x\)-bounds into \(u\)-bounds and finish entirely in \(u\).
Example
\[ \int_0^1 2x(x^2+1)^2\,dx. \]Let
\[ u=x^2+1, \qquad du=2x\,dx. \]Change the bounds:
\[ x=0 \Rightarrow u=1 \] \[ x=1 \Rightarrow u=2. \]The integral becomes
\[ \int_1^2u^2\,du. \] \[ = \left[ \frac{u^3}{3} \right]_1^2. \] \[ = \frac83-\frac13. \]Useful Properties
Same Bounds
\[ \int_a^a f(x)\,dx = 0 \]Reverse Bounds
\[ \int_b^a f(x)\,dx = - \int_a^b f(x)\,dx \]Split an Interval
\[ \int_a^c f(x)\,dx + \int_c^b f(x)\,dx = \int_a^b f(x)\,dx \]Constant Multiple
\[ \int_a^b cf(x)\,dx = c \int_a^b f(x)\,dx \]Even and Odd Functions
Even Function
\[ f(-x)=f(x) \] \[ \boxed{ \int_{-a}^{a} f(x)\,dx = 2 \int_0^a f(x)\,dx } \]Odd Function
\[ f(-x)=-f(x) \] \[ \boxed{ \int_{-a}^{a} f(x)\,dx = 0 } \]A Powerful Check
Suppose you found
\[ \int \left( 3x^2+4 \right) dx = x^3+4x+C. \]Differentiate your answer:
\[ \frac{d}{dx} \left( x^3+4x+C \right) = 3x^2+4. \]Watch Out
Indefinite integrals represent a family of antiderivatives.
Integration adds one to the exponent and divides by the new exponent.
Remember: \(\int \frac1x\,dx=\ln|x|+C\).
A definite integral evaluates to a number, so the constants cancel.
The Fundamental Theorem uses upper minus lower: \(F(b)-F(a)\).
Area below the axis contributes negatively to a definite integral.
After substitution, the entire integral should be expressed in the new variable.
Either substitute back to \(x\), or convert the bounds to \(u\).
Problem-Solving Roadmap
Identify
Indefinite integrals need \(+C\). Definite integrals use bounds.
Simplify
Convert radicals and denominator powers before integrating.
Choose a Rule
Look first for a direct antiderivative rule, then consider \(u\)-substitution.
Integrate
Finish
Check
For indefinite integrals, differentiating your answer should reproduce the integrand.
Keep This Handy
Power Rule
\[ \int x^n\,dx = \frac{x^{n+1}}{n+1} + C \]\(n\neq -1\)
Reciprocal
\[ \int \frac1x\,dx = \ln|x|+C \]Cosine
\[ \int \cos x\,dx = \sin x+C \]Sine
\[ \int \sin x\,dx = -\cos x+C \]Secant Squared
\[ \int \sec^2x\,dx = \tan x+C \]Fundamental Theorem
\[ \int_a^b f(x)\,dx = F(b)-F(a) \]FTC Derivative Form
\[ \frac{d}{dx} \left[ \int_a^x f(t)\,dt \right] = f(x) \]\(u\)-Substitution
\[ u=g(x), \qquad du=g'(x)\,dx \]Reverse Bounds
\[ \int_b^a f(x)\,dx = - \int_a^b f(x)\,dx \]Split Interval
\[ \int_a^c f(x)\,dx + \int_c^b f(x)\,dx = \int_a^b f(x)\,dx \]Before You Practice
Your Turn
Practice antiderivatives, indefinite and definite integrals, the Fundamental Theorem of Calculus, \(u\)-substitution, and basic trigonometric integrals.
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