The definite integral gives the exact area between a curve and the x-axis. The shaded region below has area = F(b) − F(a) where F is the anti-derivative.
Area = F(b) − F(a)Always positive above x-axisSplit if curve goes below axis
📏 The Rules
Area = F(b) − F(a)
The definite integral gives the exact area, when the curve is above the x-axis.
always positive above the axis
When f(x) ≥ 0 between a and b, the integral is automatically a positive area.
split if curve dips below
If the curve crosses the x-axis, integrate each section separately and take the modulus.
📖 Key Vocabulary
Anti-derivative — the function F(x) you get by integrating f(x).
Definite integral — an integral evaluated between two limits, giving a numerical answer.
Bounded region — the shaded area enclosed between a curve, the x-axis, and the limits a and b.
Area = ∫ba f(x) dx = F(b) − F(a) when f(x) ≥ 0 between a and b
How to find the area
1Integrate f(x) to get F(x)
2Calculate F(b) — upper limit
3Calculate F(a) — lower limit
4Area = F(b) − F(a)
⚠ If curve dips below x-axis, the integral is negative. Use |integral| for area, or split at the root.
Worked: area under y = x²−2x+3, x=1 to 3
Integrate:F(x) = x3/3 − x2 + 3x
F(3):27/3 − 9 + 9 = 9
F(1):1/3 − 1 + 3 = 7/3
Area:9 − 7/3 = 20/3 ≈ 6.67 ✓
🎯 Live Area Explorer — drag a and b to update the area
ℹ️ How to use: Drag the slider to explore how the area changes live.
1.0
ℹ️ How to use: Drag the slider to explore how the area changes live.
3.0
∫[1.0→3.0] (x²−2x+3) dx = 6.6667
🔀
Area Between Two Curves
Integrate (top curve minus bottom curve) between the intersection points.
📏 The Rules
top curve − bottom curve
Always subtract the lower function from the upper one, over the shared interval.
find the intersection points first
Set f(x) = g(x) to find the limits a and b — where the curves cross.
check which curve is on top
Sketch or test a value between a and b to confirm which function is greater.
📖 Key Vocabulary
Intersection point — where two curves meet, found by solving f(x) = g(x).
Enclosed region — the area trapped between two curves.
Area = ∫ba [f(x) − g(x)] dx where f(x) is the top curve and g(x) is the bottom
4 steps
1Find intersections — set f(x)=g(x), solve for x
2Identify top curve — which is larger between the roots?
3Integrate (top − bottom) between intersection x values
4Always positive — top minus bottom ensures this
Worked: y=4x−x² and y=x²
Intersect:4x−x²=x² → 4x=2x² → x=0 or x=2
Integrate:∫20 [(4x−x²)−x²] dx = ∫20 (4x−2x²) dx
F(x):2x2 − 2x3/3
F(2)−F(0):8 − 16/3 = 8/3
Area:8/3 ≈ 2.67 ✓
🔀 Between Curves Explorer — drag k to change y=kx−x²
ℹ️ How to use: Drag the slider to explore how the area changes live.
4.0
Area between y=4x−x² and y=x² = 8/3
🚗
Motion — Integration in Kinematics
Integrate acceleration to get velocity; integrate velocity to get displacement. Use initial conditions to find each constant.
📏 The Rules
v = ∫a dt + C
Integrate acceleration to get velocity.
s = ∫v dt + C
Integrate velocity to get displacement.
use initial conditions for C
Substitute a known starting value (e.g. t=0) to find the exact constant each time.
📖 Key Vocabulary
Displacement (s) — the position of an object relative to a starting point.
Velocity (v) — the rate of change of displacement.
Initial condition — a known value at a specific time, usually t = 0, used to find C.
v = ∫ a dt + Cs = ∫ v dt + C Use initial conditions to find C
Worked: a = 6t − 4, v₀=5, s₀=0
Find s at t = 3 s
∫ a dt:v = 3t2 − 4t + C
v(0)=5:C = 5 → v = 3t2 − 4t + 5
∫ v dt:s = t3 − 2t2 + 5t + D
s(0)=0:D = 0
s(3):27 − 18 + 15 = 24 m ✓
Distance vs Displacement
Displacement = ∫ v dt (signed)
Distance = ∫ |v| dt (always positive)
💡 Split at times when v=0. Take |integral| of each part.
v = t²−4. Distance 0→3 s.
v=0 at:t=2
∫20 (t²−4)dt:−16/3
∫32 (t²−4)dt:7/3
Distance:16/3 + 7/3 = 23/3 ✓
🚗 Motion Animator — drag or play to see the particle move
ℹ️ How to use: Drag the slider to explore how the area changes live.
0.0 s
t=0.0s | a=−4.0 m/s² | v=5.0 m/s | s=0.0 m
⚙️
Applications
Where this topic is used in engineering, manufacturing, maintenance and daily life. Try the calculators in the Engineering Applications and Worked Examples tabs to explore these with real numbers.
📖 Key Vocabulary
Charge — the total electric charge, found by integrating current over time.
Accumulated quantity — a total built up over time or distance by integration, e.g. charge, work, heat.
⚡Electrical:Area under i–t graph = charge to battery.
🔧Mechanical:Area under F–x = work. Area between curves = net work.
🚗Kinematics:Area under v–t = displacement of vehicle.
🏭Manufacturing:Non-standard profile area by numerical integration.
🔨
Engineering Applications
Integration accumulates quantities — charge from current, work from force, heat from power.
📏 The Rules
q = ∫i dt
Charge is current integrated over time.
W = ∫F dx
Work done is force integrated over distance.
integration = accumulation
Anywhere a total builds up from a varying rate, integration finds it.
📖 Key Vocabulary
Charge (q) — the total electric charge, measured in coulombs.
Work done (W) — energy transferred when a force moves an object.
Heat (Q) — thermal energy, found by integrating power over time.
🔧 Accumulated Quantity CalculatorCharge, Work, Heat — all by integration
ℹ️ How to use: Pick a scenario from the dropdown. Type the function coefficients and time/distance limits. The solver integrates and evaluates the definite integral, explaining why integration applies in each case.
Select a scenario and enter values above.
👆
Try: i=4t+2, t=0 to 5 ·
F=20x+500, x=0 to 4 ·
P=3t²+10, t=0 to 6
⚡ Charge q = ∫ i dt
Current is rate of change of charge. Integrate to find charge.
i = 4t + 2 A, t = 0 to 5 s
q =∫₀⁵(4t+2)dt = [2t²+2t]₀⁵
Answer:50+10 = 60 Coulombs ✓
🔧 Work W = ∫ F dx
When force varies with distance, work = area under F–x graph.
F = 500+20x N, x = 0 to 4 m
W =[500x+10x²]₀⁴
Answer:2000+160 = 2160 J ✓
🌡️ Heat Q = ∫ P dt
P = 3t² + 10 W, t = 0 to 6 s
Q =[t³+10t]₀⁶
Answer:216+60 = 276 J ✓
📡 RMS Current
For i = I₀ sin(ωt): Irms = I₀/√2
I²rms = (1/T)∫₀ᵗI₀²sin²(ωt)dt = I₀²/2
💡 Mains: I₀=14.14 A → Irms = 10 A
✏️
Worked Examples — Area Problems
📏 The Rules
solve for the intersections
Set the two functions equal to find the limits before integrating.
line vs curve
Sketch or test a mid-point to confirm which function is the "top" one over that interval.
📖 Key Vocabulary
Gradient — the slope of a straight line, the m in y = mx + k.
Intercept — where a line crosses the y-axis, the k in y = mx + k.
✏️ Area Between Two Curves Calculatory = mx + k vs y = x²
ℹ️ How to use: Type the gradient and intercept of a straight line, and the limits. The solver finds where the line and y=x² intersect (or uses your limits directly), checks which curve is on top, and calculates the area between them with full working.
Enter the line and limits above.
👆
Try: y=x+2 vs y=x², -1 to 2 ·
y=6x vs y=x² (intersections!)
Example 1 — Area between y=x+2 and y=x², x=−1 to x=2
Top curve:x+2 > x² for −1<x<2 ✓
Integrate:∫2-1 [(x+2)−x²] dx
F(x):x2/2 + 2x − x3/3
F(2):2+4−8/3 = 10/3
F(−1):1/2−2+1/3 = −7/6
Area:10/3−(−7/6) = 27/6 = 4.5 ✓
Example 2 — y=6x−x² and y=x² (find intersections first)
Intersect:6x−x²=x² → x=0 or x=3
Integrate:∫30 (6x−2x²) dx
F(x):3x2−2x3/3
F(3)−F(0):27−18=9
Area:9 sq units ✓
🎮
Quick Fire Quiz
10 questions, instant feedback.
⚡ Quick Blitz
ℹ️ How to play: An area/integration question appears. Click the correct answer. Score and streak update after each — click Next to continue.
Score: 0Streak: 0 🔥Q 1/10
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Score: 0/10
📝
Practice — Area & Applications
🟢 Tier 1
ℹ️ How to use: Work through each question and write down your answer. When ready, click Show Answer for the full step-by-step solution.
Q1
Find the area enclosed between y = x² and the x-axis from x = 0 to x = 3.
[3]
[x³/3]₀³ = 27/3 − 0 = 9 sq units ✓
🟡 Tier 2
Q2
Find the area between y = 4x − x² and y = x² from x = 0 to x = 2.
