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Integration Made Simple: From Indefinite to Definite Integrals

2026.04.14

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Integration concept
Integration concept

Integration is the reverse of differentiation. It’s one of the two pillars of calculus and a must-know for AP Calculus, A-levels, and beyond.

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Key Point: Integration appears in 2–4 questions on the AP Calculus AB exam every year. Mastering the basics here sets you up for area, volume, and accumulation problems.

What Is Integration?

Definition

Integration finds a function F(x)F(x) whose derivative equals f(x)f(x).

Fβ€²(x)=f(x)β‡’βˆ«f(x) dx=F(x)+CF'(x) = f(x) \quad \Rightarrow \quad \int f(x)\,dx = F(x) + C

The CC is the constant of integration. Since differentiation erases constants, we add it back when integrating.


Indefinite vs Definite Integrals

TypeIndefiniteDefinite
Notation∫f(x) dx\int f(x)\,dx∫abf(x) dx\int_a^b f(x)\,dx
ResultFunction + CNumber (exact value)
MeaningFind antiderivativeCalculate area

Essential Integration Formulas

FunctionIntegral
xnx^n (nβ‰ βˆ’1n \neq -1)xn+1n+1+C\frac{x^{n+1}}{n+1} + C
1x\frac{1}{x}ln⁑∣x∣+C\ln|x| + C
exe^xex+Ce^x + C
cos⁑x\cos xsin⁑x+C\sin x + C
sin⁑x\sin xβˆ’cos⁑x+C-\cos x + C
sec⁑2x\sec^2 xtan⁑x+C\tan x + C

How to Integrate

Polynomial Integration

Integrate each term separately using the power rule.

∫(3x2βˆ’4x+1) dx=x3βˆ’2x2+x+C\int (3x^2 - 4x + 1)\,dx = x^3 - 2x^2 + x + C

U-Substitution

For composite functions, substitute the inner function as uu.

∫2x(x2+1)3 dx\int 2x(x^2+1)^3\,dx

Let u=x2+1u = x^2 + 1, du=2x dxdu = 2x\,dx:

∫u3 du=u44+C=(x2+1)44+C\int u^3\,du = \frac{u^4}{4} + C = \frac{(x^2+1)^4}{4} + C

Evaluating Definite Integrals

Find the antiderivative, then compute upper limit minus lower limit:

∫132x dx=[x2]13=9βˆ’1=8\int_1^3 2x\,dx = \left[x^2\right]_1^3 = 9 - 1 = 8
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Watch Out: For indefinite integrals, never forget the constant CC. It’s a guaranteed point deduction on exams. Definite integrals don’t need CC because it cancels out.

Applications of Integration

Area Under a Curve

The area between y=f(x)y = f(x) and the x-axis from aa to bb:

A=∫ab∣f(x)βˆ£β€‰dxA = \int_a^b |f(x)|\,dx

Use absolute value because regions below the x-axis give negative values.

Area Between Two Curves

A=∫ab∣f(x)βˆ’g(x)βˆ£β€‰dxA = \int_a^b |f(x) - g(x)|\,dx

Worked Examples

Example 1: Indefinite Integral

Find ∫(4x3βˆ’6x+2) dx\int (4x^3 - 6x + 2)\,dx.

Show Solution
∫(4x3βˆ’6x+2) dx=x4βˆ’3x2+2x+C\int (4x^3 - 6x + 2)\,dx = x^4 - 3x^2 + 2x + C

Example 2: Definite Integral

Evaluate ∫02(3x2+1) dx\int_0^2 (3x^2 + 1)\,dx.

Show Solution
[x3+x]02=(8+2)βˆ’(0+0)=10\left[x^3 + x\right]_0^2 = (8 + 2) - (0 + 0) = 10

Example 3: Area Problem

Find the area between y=x2y = x^2 and the x-axis from x=0x = 0 to x=3x = 3.

Show Solution

Since x2β‰₯0x^2 \geq 0 on [0,3][0, 3], no absolute value needed:

A=∫03x2 dx=[x33]03=9A = \int_0^3 x^2\,dx = \left[\frac{x^3}{3}\right]_0^3 = 9
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Pro Tip: Always sketch the graph before solving area problems. It helps you identify regions above and below the x-axis, so you know where to split the integral.

Top 3 Common Mistakes

  1. Forgetting +C in indefinite integrals β€” automatic point loss on every exam
  2. Ignoring absolute value for area β€” regions below the x-axis give negative integrals, which cancel positive areas
  3. Forgetting to back-substitute after u-sub β€” your final answer must be in terms of xx, not uu

Related Topics

  • What Is Differentiation?
  • U-Substitution Techniques
  • Integration by Parts
  • AP Calculus Exam Strategies