Integrals: Formulas, Definite vs Indefinite & Examples

Integrals reverse differentiation. Where a derivative gives the rate of change, an integral recovers the quantity from its rate, or measures the area under a curve. The skill that matters most is reading the integrand first: does it match a basic formula, hide a chain-rule pattern for substitution, or split into a product needing integration by parts? Get the structure right and the computation follows in a few lines. This page covers the core formulas, indefinite vs definite integrals, worked examples, and common mistakes.

Indefinite vs Definite Integrals

An indefinite integral asks for a function: the family of all antiderivatives of $f(x)$. Since the derivative of a constant is zero, the answer carries an arbitrary constant $C$.

A definite integral asks for a number: the signed area between the curve and the $x$-axis from $x=a$ to $x=b$. No $+C$ appears because it cancels when you subtract.

This link between antiderivatives and area is the Fundamental Theorem of Calculus, why one set of formulas serves both jobs.

The Core Integration Formulas

These are the building blocks: each is a derivative rule read backwards.

Two linearity rules break any sum apart before you apply the table:

Why The Power Rule Adds One To The Exponent

The power rule looks arbitrary until you differentiate it back. If $F(x) = \dfrac{x^{n+1}}{n+1}$, then $F'(x) = \dfrac{(n+1)x^{n}}{n+1} = x^n$: the $n+1$ on top and bottom cancel. This is also why $n=-1$ breaks the rule, forcing division by $n+1 = 0$. That gap is filled by the logarithm, since $\dfrac{d}{dx}\ln\lvert x\rvert = \dfrac{1}{x}$. Every formula verifies the same way: differentiate and check you recover the integrand.

Reading The Integrand: Which Method?

Classifying the structure first saves more time than any formula.

• Direct formula — matches the table after splitting a sum or factoring out a constant.
• u-substitution — an inner function and its derivative both appear; the integral form of the Chain Rule. Look for $g'(x)$ next to $g(x)$.
• Integration by parts — a product of two unrelated pieces, like $x e^x$ or $x\ln x$.

Worked Example 1: Substitution

Evaluate $\displaystyle\int 2x\,(x^2 + 1)^3\,dx$. The cube $(x^2+1)^3$ has an inner function $x^2 + 1$ whose derivative $2x$ sits right in front. That is the substitution signal. Let $u = x^2 + 1$, so $du = 2x\,dx$. The $2x\,dx$ becomes exactly $du$, collapsing the integral to a power rule in $u$:

Substitute back $u = x^2 + 1$:

Differentiating with the Chain Rule returns $2x(x^2+1)^3$, confirming the result.

Worked Example 2: A Definite Integral

Evaluate $\displaystyle\int_1^3 (2x + 1)\,dx$. An antiderivative is $F(x) = x^2 + x$. Apply the Fundamental Theorem by subtracting at the bounds; $\big[\,F(x)\,\big]_1^3$ means $F(3) - F(1)$:

The area under $y = 2x+1$ on $[1,3]$ is $10$ square units. The $+C$ never appeared: it cancels in the subtraction, which is why definite integrals omit it.

Practice Problems

1. $\displaystyle \int (3x^2 - 4x + 5)\,dx$
2. $\displaystyle \int_0^2 x^2\,dx$
3. $\displaystyle \int 6x\,(x^2 - 7)^5\,dx$

Solutions. (1) Term by term: $x^3 - 2x^2 + 5x + C$. (2) Antiderivative $\tfrac{x^3}{3}$, so $\tfrac{8}{3} - 0 = \tfrac{8}{3}$. (3) Let $u = x^2 - 7$, $du = 2x\,dx$, so $6x\,dx = 3\,du$: $\int 3u^5\,du = \tfrac{(x^2-7)^6}{2} + C$.

For more problems with full step-by-step solutions, try the GPAI Solver on any integral you are stuck on.

Common Mistakes

Treat every answer as testable: differentiate it, and the integrand should reappear. That habit catches sign slips, missing constants, and botched substitutions before the grader does. Once reading the integrand becomes automatic, integration turns from guesswork into a short, checkable routine.