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Grade 11-12 (age 16-18)

high school Applications of Definite Integrals

A definite integral does more than find the area under a curve. The same idea of slicing finely and adding up gives the area between two curves, the volume of a solid of revolution, and the distance travelled from a velocity graph. The problems look different, but each one fixes the amount in a small slice and then gathers the slices with a definite integral. The first picture shows the area between two curves as a and b move, the second the volume of a solid of revolution as b changes, and the third the distance travelled up to time t.

The area between two curves

📐 Finding the area between curves
①The area between f(x) and g(x) is the integral of |f(x) − g(x)|
②Subtract the lower function from the upper one
③Find the intersections first to fix the interval
④Moving b past 1.56 in the picture swaps which graph is on top, and the area still adds up the absolute difference
-1.5
1.5
Area Between Curves
S = ∫ab |f(x) − g(x)| dx
Integrate the upper function minus the lower one; where they swap, use the absolute value

How to set up an area

Area between a curve and the x-axis
S = ∫ab |f(x)| dx
If the curve dips below the x-axis, split the interval
Area between a curve and the y-axis (x = g(y))
S = ∫cd |g(y)| dy
Used when it is easier to swap the roles of x and y and integrate with respect to y
💡 Steps for an area
①Find the intersections: f(x) = g(x) or f(x) = 0
②Decide which function is on top
③Split the interval where the order changes
④Integrate on each piece and add

Volumes of solids of revolution

2
Volume of rotation about the x-axis
V = π ∫ab [f(x)]² dx
Each cross-section is a circle of radius f(x), so its area is π[f(x)]²
Volume of rotation about the y-axis
V = π ∫cd [g(y)]² dy
When rotating about the y-axis, write x = g(y) and integrate with respect to y
🔑 Finding volumes of revolution
①First check which axis the region turns about
②About the x-axis: radius |f(x)|, integrate in x
③About the y-axis: radius |g(y)|, integrate in y
④Volumes of revolution are treated fully later in calculus; this is a preview

Velocity and distance travelled

2 s
Distance travelled
distance = ∫ab |v(t)| dt
Integrating the absolute value of velocity gives the actual distance travelled
💡 Change in position versus distance
①The change in position integrates v(t) as it is, so travel in the opposite direction subtracts
②The distance travelled integrates |v(t)|, so it is never negative
③Where v(t) ≥ 0, the two are equal

Summary

Area between curves
∫|f − g| dx
Split the interval at the intersections
Volume of revolution
π∫[f(x)]² dx
Integrate the area of the disk cross-sections
🎯 Exam Points
①Area problems: intersections → which is on top → integrate on each piece → add
②Handle absolute values by splitting above and below the x-axis
③Solids of revolution: decide the axis and the radius first
④Distance is the integral of |v(t)|; change in position is the integral of v(t)
⑤Rewriting as x = g(y) and integrating in y often makes things easier

Worked Examples & Past Exam

Example 1
Find the area enclosed by the curve y = x² and the line y = x.
1
x² = x gives the intersections x = 0 and 1, and between them the line y = x lies above the curve y = x².
S = ∫01 (x - x2) dx
2
Integrate the upper function minus the lower one, x − x², from 0 to 1.
[x22 - x33]01 = 12 - 13 = 16
▸ 1/6
Reversing the order and integrating x² − x gives −1/6. For an area, always subtract the lower function from the upper one.
Example 2
Find the volume of the solid formed by rotating y = √x (0 ≤ x ≤ 4) about the x-axis.
1
Rotating about the x-axis, the cross-section at x is a circle of radius √x, with area π(√x)² = πx.
V = π ∫04 (√x)2 dx = π ∫04 x dx
2
Integrate the cross-sectional area πx from 0 to 4.
π [x22]04 = π × 8 = 8π
▸ 8π
The second picture cannot reach b = 4, but putting b = 4 into its formula πb²/2 also gives 8π.
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
1
On 0 ≤ x ≤ 2, y = x² lies above the x-axis, so the area is the integral of x² from 0 to 2.
S = ∫02 x2 dx
2
Put 2 and 0 into x³/3 and subtract.
[x33]02 = 83
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