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
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
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Area Between Two Curves
Integrate (top curve minus bottom curve) between the intersection points.
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
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Motion — Integration in Kinematics
Integrate acceleration to get velocity; integrate velocity to get displacement. Use initial conditions to find each constant.
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
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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.
ā”Electrical:Area under iāt graph = charge to battery.
š§Mechanical:Area under Fāx = work. Area between curves = net work.
šKinematics:Area under vāt = displacement of vehicle.
šManufacturing:Non-standard profile area by numerical integration.
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Engineering Applications
Integration accumulates quantities — charge from current, work from force, heat from power.
🔧 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
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Worked Examples — Area Problems
✏️ 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
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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.