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high school Applications of Definite Integrals

Applications of Definite Integrals

The definite integral is not only a tool for measuring area. The same idea of slicing thin and adding up gives the area between two curves, the volume of a solid of revolution, and the distance traveled from a velocity graph. Problems that look completely different all trace back to one definite integral, the heart of this unit. Drag the lower a and upper b sliders and the shaded region between the two curves, and its area value, change.

Area Between Two Curves

📐 Area-Between Idea
①Area between f(x) and g(x) = ∫|f(x) − g(x)| dx
②Integrate (upper function) − (lower function)
③Find intersections first to set the limits!
-1.5
1.5
Area Between Curves
S = ∫ab |f(x) − g(x)| dx
Subtract the lower from the upper, then integrate (mind the absolute value)

Strategy for Area

Area Between Curve and x-axis
S = ∫ab |f(x)| dx
Split the interval if the curve crosses the x-axis
Area Against the y-axis (x = g(y))
S = ∫cd |g(y)| dy
Sometimes it is easier to swap the role of x and y
💡 Steps
①Find intersections: f(x) = g(x) or f(x) = 0
②Identify upper vs lower
③Split where they switch
④Integrate each piece and add

Volume of a Solid of Revolution

2
Volume Around x-axis (Disk)
V = π ∫ab [f(x)]² dx
Cross-sections are disks; radius = f(x), area = π[f(x)]²
Volume Around y-axis
V = π ∫cd [g(y)]² dy
For y-axis rotation, write x = g(y) and integrate w.r.t. y
🔑 Key Idea
①Identify which axis the figure rotates around
②x-axis: radius = |f(x)|, integrate w.r.t. x
③y-axis: radius = |g(y)|, integrate w.r.t. y
④This solid-of-revolution step is a Calculus preview

Velocity & Distance

2 s
Distance
distance = ∫ab |v(t)| dt
Integrate the absolute value of velocity for actual distance
💡 Displacement vs Distance
①Displacement = ∫v(t)dt (signed)
②Distance = ∫|v(t)|dt (always non-negative)
③Where v(t) ≥ 0, displacement = distance

Wrap-up

Area Between Curves
∫|f − g| dx
Split via intersections
Volume of Revolution
π∫[f(x)]² dx
Disk method
🎯 Exam Points
①Area: intersections → upper/lower → integrate per region → sum
②Absolute value: split above/below x-axis
③Volume: identify axis & radius
④Velocity → distance: ∫|v(t)|dt; displacement: ∫v(t)dt
⑤Sometimes integrating w.r.t. y is easier. Deeper area-between-curves and solids of revolution are on Calculus

Worked Examples & Past Exam

Example 1
Find the area enclosed by the curve y = x² and the line y = x.
1
Find the intersections (x = 0, 1); on 0 ≤ x ≤ 1 subtract the lower (x²) from the upper (x).
S = ∫01 (x - x2) dx
2
Integrate and compute.
[x22 - x33]01 = 12 - 13 = 16
▸ 1/6
The area between two curves integrates (upper − lower) between the intersection points.
Example 2
Find the volume of the solid formed by rotating y = √x (0 ≤ x ≤ 4) about the x-axis.
1
Use the disk method V = π∫ab [f(x)]² dx. [√x]² = x.
V = π ∫04 (√x)2 dx = π ∫04 x dx
2
Integrate and compute.
π [x22]04 = π × 8 = 8π
▸ 8π
The volume of a solid of revolution about the x-axis integrates the disk area π[f(x)]².
exam-style
What is the area enclosed by y = x², the x-axis, and the line x = 2?
①8/3
②4
③2
④8
⑤4/3
▸ ① 8/3
1
The area between the x-axis and the curve is S = ∫02 x² dx.
S = ∫02 x2 dx
2
Integrate and compute.
[x33]02 = 83
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