∫ Integral Calculus
Integration reverses differentiation — and separately, it calculates the exact area under a curve. Together these two ideas recover displacement from velocity, and total quantity from a rate.
2Topics
8Worked Examples
4Eng. Cases
12Practice Qs
∫Indefinite IntegrationReversing differentiation
📐Definite IntegralsArea under a curve
⚙️ApplicationsReal engineering
🎮Quick-Fire GameTest your speed
📝PracticeGraded questions
🎯What you will learn
Apply the power rule for integration to find indefinite integrals, always including the constant of integration; evaluate definite integrals using limits; and use definite integrals to calculate the exact area under a curve.
💡 Every indefinite integral needs "+C" — differentiating any constant gives zero, so integration can never recover exactly which constant was there originally, without extra information (like a boundary condition).
🧭How to use this resource
1. Read the rules banner at the top of each tab
2. Experiment with the interactive tool
3. Check the key vocabulary panel
4. Study the visual diagram, then the worked examples
∫
Indefinite Integration
Reversing differentiation — finding a family of functions that all share the same derivative.
🔧Power Rule IntegratorEnter a coefficient and power for axⁿ
Enter a and n for the term axⁿ, then press Integrate to see the power rule applied step by step…
📖 Key Vocabulary
Indefinite integral — the general antiderivative of a function, including the arbitrary constant C.
Constant of integration (C) — accounts for the fact that many functions share the same derivative.
Antiderivative — a function whose derivative equals the original function being integrated.
Boundary condition — extra information (like a known point) used to find the specific value of C.
👁️ Visual aid: all three curves have exactly the same derivative — they only differ by a vertical shift, which is precisely what "+C" represents.
✏️Worked Example 1 — Power rule on a polynomialType: Power rule
Question Integrate ∫(6x² − 4x + 3) dx.
Apply term by term
∫6x²dx=2x³, ∫−4x dx=−2x², ∫3 dx=3xAnswer = 2x³ − 2x² + 3x + C
📐Worked Example 2 — Finding C using a boundary conditionType: Boundary condition
Question Given dy/dx = 3x², and y=5 when x=1, find y.
Integrate
y = x³ + CSubstitute the boundary condition
5 = 1³ + C → C = 4Answer y = x³ + 4
〰️Worked Example 3 — Integrating a trigonometric functionType: Trig integral
Question Integrate ∫(4cosx − sinx) dx.
Apply rules
∫4cosx dx=4sinx, ∫−sinx dx=cosxAnswer = 4sinx + cosx + C
⚙️Worked Example 4 — Engineering context (velocity to displacement)Type: Engineering
Question A particle's velocity is v = 6t − 4 (m/s). If s=2m when t=0, find the displacement function.
Integrate
s = 3t² − 4t + CApply boundary condition
2 = 3(0)² − 4(0) + C → C = 2Answer s = 3t² − 4t + 2 (m) — this exact equation now predicts the particle's position at any future time.
🧠 Common slip: forgetting the "+C" entirely, or forgetting to solve for its actual value when a boundary condition is given — an indefinite integral without a stated C is technically incomplete.
📐
Definite Integrals & Area
Evaluating between two limits gives a specific number — often the exact area under a curve.
🔧Definite Integral CalculatorEvaluate ∫ₐᵇ (px²+qx+r) dx
Enter p, q, r, a and b, then press Evaluate for the full definite integral calculation…
📖 Key Vocabulary
Definite integral — an integral evaluated between two specific limits, giving a single number.
Limits of integration — the lower (a) and upper (b) x-values bounding the region.
[F(x)]ₐᵇ notation — shorthand meaning "evaluate F(x) at b, then subtract F(x) at a."
Area between two curves — found by integrating (upper curve − lower curve) between their intersection points.
👁️ Visual aid: the shaded region's exact area is what the definite integral calculates — no approximation needed, unlike counting grid squares.
✏️Worked Example 1 — Evaluating a definite integralType: Direct evaluation
Question Evaluate ∫₁³ (2x+1) dx.
Integrate
∫(2x+1)dx = x²+xApply limits
[x²+x]₁³ = (9+3) − (1+1) = 12−2Answer = 10
📊Worked Example 2 — Finding an area under a curveType: Area calculation
Question Find the area under y=x² between x=0 and x=4.
Integrate
∫x² dx = x³/3Apply limits
[x³/3]₀⁴ = 64/3 − 0Answer Area = 64/3 ≈ 21.33 square units
⚡Worked Example 3 — Displacement from velocity (definite integral)Type: Physical application
Question A particle has velocity v=3t²−2t (m/s). Find the displacement between t=1s and t=3s.
Integrate
∫v dt = t³−t²Apply limits
[t³−t²]₁³ = (27−9) − (1−1) = 18−0Answer Displacement = 18m over that time interval
⚙️Worked Example 4 — Engineering context (work done by a variable force)Type: Engineering
Question A variable force F=(20−2x) N acts on an object as it moves from x=0 to x=8m. Find the total work done.
Recall Work = ∫F dx
Integrate
∫(20−2x)dx = 20x−x²Apply limits
[20x−x²]₀⁸ = (160−64) − 0Answer Work done = 96 J — the definite integral correctly accounts for the force weakening as x increases, unlike simply multiplying a single force value by distance.
🧠 Common slip: forgetting to subtract the value at the lower limit — a definite integral always requires F(b) minus F(a), not F(b) alone.
⚙️
Engineering Applications
Four live scenarios — each with its own calculator. Enter your own numbers.
🚗 Displacement from Velocity
Integrating a velocity function recovers the displacement over a specified time interval.
s = ∫v dt
📟 Try it — Displacement Calculator (v=At²+Bt+C)
Enter A, B, C, t₁ and t₂, then press Calculate…
💪 Work Done by a Variable Force
When force varies with position, the definite integral gives the exact total work — not just an approximation.
W = ∫ₐᵇ F dx
📟 Try it — Variable Force Work Calculator (F=px+q)
Enter p, q, x₁ and x₂, then press Calculate…
📊 Area Under a Load Diagram
Integrating a distributed load function gives the total force it represents — vital in structural analysis.
Total load = ∫ₐᵇ w(x) dx
📟 Try it — Total Load Calculator (w=px²+qx+r)
Enter p, q, r, x₁ and x₂, then press Calculate…
🔋 Charge from Current
Integrating a time-varying current gives the total electric charge that has flowed.
Q = ∫ I dt
📟 Try it — Charge Calculator (I=At+B)
Enter A, B, t₁ and t₂, then press Calculate…
🎮
Quick-Fire Integration Game
12 questions on indefinite and definite integrals. How many can you get?
⚡ Integration Challenge
Score: 0Question: 1/12Streak: 0🔥
Press Start to begin!
📝
Practice Questions
Graded from warm-up to challenge. Attempt fully before revealing the working.
Question 1 🥉2 marks
Integrate ∫4x³ dx.
x⁴ + C
Question 2 🥉2 marks
Integrate ∫(6x − 2) dx.
3x² − 2x + C
Question 3 🥉2 marks
Evaluate ∫₀² 3 dx.
[3x]₀² = 6−0 = 6
Question 4 🥉2 marks
Integrate ∫cosx dx.
sinx + C
Question 5 🥈3 marks
Evaluate ∫₁⁴ x² dx.
[x³/3]₁⁴ = 64/3 − 1/3 = 63/3
= 21
= 21
Question 6 🥈4 marks
Given dy/dx=4x−1 and y=3 when x=0, find y.
y=2x²−x+C
3=0−0+C → C=3
y=2x²−x+3
3=0−0+C → C=3
y=2x²−x+3
Question 7 🥈3 marks
Find the area under y=3x² between x=0 and x=2.
[x³]₀² = 8−0 = 8 square units
Question 8 🥈4 marks
A particle has v=2t+3 (m/s). Find its displacement between t=0 and t=4.
[t²+3t]₀⁴=(16+12)−0
=28m
=28m
Question 9 🥇4 marks
Evaluate ∫₀^π sinx dx.
[−cosx]₀^π = (−cosπ)−(−cos0) = (1)−(−1)
= 2
= 2
Question 10 🥇5 marks
A variable force F=(30−3x) N acts from x=0 to x=6m. Find the total work done.
∫(30−3x)dx=30x−1.5x²
[30x−1.5x²]₀⁶=(180−54)−0
=126 J
[30x−1.5x²]₀⁶=(180−54)−0
=126 J
Question 11 🥇5 marks
A current I=(6−0.8t) A flows for 5 seconds. Find the total charge that flows.
Q=∫₀⁵(6−0.8t)dt=[6t−0.4t²]₀⁵=(30−10)−0
=20 C
=20 C
Question 12 🥇5 marks
Given acceleration a=6t−4 (m/s²), with v=2 m/s at t=0 and s=5m at t=0, find the displacement function s(t).
v=∫a dt=3t²−4t+C₁; v=2 at t=0 → C₁=2, so v=3t²−4t+2
s=∫v dt=t³−2t²+2t+C₂; s=5 at t=0 → C₂=5
s = t³ − 2t² + 2t + 5
s=∫v dt=t³−2t²+2t+C₂; s=5 at t=0 → C₂=5
s = t³ − 2t² + 2t + 5