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 watch 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!

Lower a-1.5

Upper b1.5

Area Between Curves

S = ∫_a^b |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 = ∫_a^b |f(x)| dx

Split the interval if the curve crosses the x-axis

Area Against the y-axis (x = g(y))

S = ∫_c^d |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

Upper b2

Volume Around x-axis (Disk)

V = π ∫_a^b [f(x)]² dx

Cross-sections are disks; radius = f(x), area = π[f(x)]²

Volume Around y-axis

V = π ∫_c^d [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

Velocity & Distance

Time t2 s

Distance

distance = ∫_a^b |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

Worked Examples & Past Exam

Example 1

Find the area enclosed by the curve y = x² and the line y = x.

Find the intersections (x = 0, 1); on 0 ≤ x ≤ 1 subtract the lower (x²) from the upper (x).

S = ∫₀¹ (x - x²) dx

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.

Use the disk method V = π∫_a^b [f(x)]² dx. [√x]² = x.

V = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx

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.

2023 CSAT Math type, adapted

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

The area between the x-axis and the curve is S = ∫₀² x² dx.

S = ∫₀² x² dx

Integrate and compute.

[x³3]₀² = 83

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Definite Integral

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