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🎯⭐ INTERACTIVE LESSON

Integration Applications

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Integration Applications - Complete Interactive Lesson

Part 1: Area Between Curves (Advanced)

Integration Applications

Part 1 of 7 — Area Between Curves

In This Topic

PartTopic
1Area Between Curves
2Cross-Sectional Volumes
3Disk & Washer Methods
4Riemann Sums & Trapezoidal Rule
5Rate Problems & Net Change
6Problem-Solving Workshop
7Comprehensive Assessment

Area Between Two Curves

A=∫ab[f(x)−g(x)] dxwhere f(x)≥g(x)\boxed{A = \int_a^b [f(x) - g(x)]\,dx \quad \text{where } f(x) \ge g(x)}

StepAction
1Sketch and identify which curve is on top
2Find intersection points (set f=gf = g)
3Set up ∫[top−bottom] dx\int [\text{top} - \text{bottom}]\,dx
4If curves cross, split at the crossing points

Key Fact: Always subtract bottom from top: ∫(top−bottom) dx\int (\text{top} - \text{bottom})\,dx. If you get a negative area, the curves were swapped.

Worked Examples

Example 1: Area between y=x2y = x^2 and y=xy = x on [0,1][0,1].

StepWork
Intersectionsx2=x⇒x=0,1x^2 = x \Rightarrow x=0,1
Top curvex≥x2x \ge x^2 on [0,1][0,1]
Integral∫01(x−x2)dx=[x22−x33]01\int_0^1 (x-x^2)dx = [\frac{x^2}{2}-\frac{x^3}{3}]_0^1
Answer12−13=16\frac{1}{2}-\frac{1}{3} = \frac{1}{6}

Example 2: Area enclosed by y=x2y = x^2 and y=2xy = 2x.

StepWork
Intersectionsx2=2x⇒x=0,2x^2 = 2x \Rightarrow x=0,2
Top curve2x≥x22x \ge x^2 on [0,2][0,2]
Integral∫02(2x−x2)dx=[x2−x33]02\int_0^2(2x-x^2)dx = [x^2-\frac{x^3}{3}]_0^2
Answer4−83=434-\frac{8}{3} = \frac{4}{3}

Integrating with Respect to yy

A=∫cd[right−left] dy\boxed{A = \int_c^d [\text{right} - \text{left}]\,dy}

Use when curves are easier to express as x=f(y)x = f(y).

AP Tip: If the region is bounded by x=y2x = y^2 and x=4x = 4, integrate with respect to yy. It avoids splitting the integral.

Area Between Curves 🎯

Set up the integral. 🔍

Compute the area. ✍️

Key Takeaways — Part 1

ConceptFormula
Area (horizontal)∫ab[top−bottom] dx\int_a^b [\text{top}-\text{bottom}]\,dx
Area (vertical)∫cd[right−left] dy\int_c^d [\text{right}-\text{left}]\,dy
Crossing curvesSplit at intersection points

Up Next: Part 2 — Cross-Sectional Volumes.

Part 2: Cross-Sectional Volumes

Integration Applications

Part 2 of 7 — Cross-Sectional Volumes

Volume with Known Cross Sections

V=∫abA(x) dx\boxed{V = \int_a^b A(x)\,dx}

where A(x)A(x) is the area of the cross section at position xx.

Cross-Section Area Formulas

If the side length (or diameter) of each cross section is s(x)=f(x)−g(x)s(x) = f(x) - g(x):

ShapeArea A(x)A(x)Common AP form
Squares2s^2(f(x)−g(x))2(f(x)-g(x))^2
Semicircleπ8s2\frac{\pi}{8}s^2π8(f(x)−g(x))2\frac{\pi}{8}(f(x)-g(x))^2
Equilateral triangle34s2\frac{\sqrt{3}}{4}s^234(f(x)−g(x))2\frac{\sqrt{3}}{4}(f(x)-g(x))^2
Isosceles right triangle (leg)12s2\frac{1}{2}s^212(f(x)−g(x))2\frac{1}{2}(f(x)-g(x))^2
Rectangle (h=2sh = 2s)2s22s^22(f(x)−g(x))22(f(x)-g(x))^2

Key Fact: The shape of the cross section only affects the constant multiplier. The integral setup is always ∫ab(constant)⋅s2 dx\int_a^b (\text{constant}) \cdot s^2\,dx.

Worked Example

Base region: Between y=xy = \sqrt{x} and y=0y = 0 from x=0x=0 to x=4x=4.

Cross sections (perpendicular to xx-axis) are squares.

StepWork
Side lengths(x)=x−0=xs(x) = \sqrt{x} - 0 = \sqrt{x}
Cross-section areaA(x)=(x)2=xA(x) = (\sqrt{x})^2 = x
Volume integralV=∫04x dxV = \int_0^4 x\,dx
Evaluate[x22]04=8[\frac{x^2}{2}]_0^4 = 8

Same base, semicircular cross sections:

StepWork
Diameterd=xd = \sqrt{x}, radius =x/2= \sqrt{x}/2
Areaπ2(x/2)2=πx8\frac{\pi}{2}(\sqrt{x}/2)^2 = \frac{\pi x}{8}
Volume∫04πx8dx=π8⋅8=π\int_0^4 \frac{\pi x}{8}dx = \frac{\pi}{8} \cdot 8 = \pi

AP Tip: Cross-section volumes are a favorite AP FRQ topic. Make sure you can set up the integral for any shape.

Cross-Sectional Volumes 🎯

Set up the volume integral. 🔍

Compute the volume. ✍️

Key Takeaways — Part 2

ShapeArea multiplier
Square1⋅s21 \cdot s^2
Semicircleπ8s2\frac{\pi}{8}s^2
Equilateral △\triangle34s2\frac{\sqrt{3}}{4}s^2
Isosceles right △\triangle12s2\frac{1}{2}s^2

Up Next: Part 3 — Disk & Washer Methods.

Part 3: Volumes: Disk and Washer Methods

Integration Applications

Part 3 of 7 — Disk & Washer Methods

Disk Method (Solid with No Hole)

V=π∫ab[R(x)]2 dx\boxed{V = \pi\int_a^b [R(x)]^2\,dx}

Use when: The region is rotated around an axis and there is no gap between the region and the axis.

Washer Method (Solid with a Hole)

V=π∫ab([R(x)]2−[r(x)]2)dx\boxed{V = \pi\int_a^b \left([R(x)]^2 - [r(x)]^2\right)dx}

VariableMeaning
R(x)R(x)Outer radius (farther curve from axis)
r(x)r(x)Inner radius (closer curve to axis)

Key Fact: NEVER subtract the radii first. It’s R2−r2R^2 - r^2, NOT (R−r)2(R-r)^2.

Rotation About Different Lines

Axis of RotationRadius Setup
xx-axis (y=0y=0)R=f(x)R = f(x)
yy-axis (x=0x=0)R=f(y)R = f(y), integrate dydy
y=ky = k (horizontal)$R =
x=hx = h (vertical)$R =

Worked Example

Rotate y=x2y = x^2 about the xx-axis from x=0x=0 to x=2x=2:

StepWork
RadiusR=x2R = x^2
IntegralV=π∫02(x2)2dx=π∫02x4 dxV = \pi\int_0^2 (x^2)^2 dx = \pi\int_0^2 x^4\,dx
Evaluateπ[x55]02=32π5\pi[\frac{x^5}{5}]_0^2 = \frac{32\pi}{5}

Washer Example

Rotate region between y=xy=x and y=x2y=x^2 about xx-axis on [0,1][0,1]:

R=xR = x (outer), r=x2r = x^2 (inner).

V=π∫01(x2−x4)dx=π[x33−x55]01=π(13−15)=2π15V = \pi\int_0^1(x^2-x^4)dx = \pi[\frac{x^3}{3}-\frac{x^5}{5}]_0^1 = \pi(\frac{1}{3}-\frac{1}{5}) = \frac{2\pi}{15}

Disk & Washer 🎯

Identify the setup. 🔍

Compute the volume. ✍️

Key Takeaways — Part 3

MethodWhen to UseFormula
DiskNo gap from axisπ∫R2 dx\pi\int R^2\,dx
WasherGap creates holeπ∫(R2−r2) dx\pi\int (R^2-r^2)\,dx

Up Next: Part 4 — Riemann Sums & Trapezoidal Rule.

Part 4: Riemann Sums and Trapezoidal Rule

Integration Applications

Part 4 of 7 — Riemann Sums & Trapezoidal Rule

Riemann Sum Formulas

Ln=∑i=0n−1f(xi)ΔxRn=∑i=1nf(xi)ΔxMn=∑i=1nf(xˉi)Δx\boxed{L_n = \sum_{i=0}^{n-1} f(x_i)\Delta x \qquad R_n = \sum_{i=1}^{n} f(x_i)\Delta x \qquad M_n = \sum_{i=1}^{n} f(\bar{x}_i)\Delta x}

Trapezoidal Rule

Tn=Δx2[f(x0)+2f(x1)+2f(x2)+⋯+2f(xn−1)+f(xn)]\boxed{T_n = \frac{\Delta x}{2}[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)]}

Over/Underestimate Guide

Methodff Increasingff Decreasing
LeftUnderOver
RightOverUnder
MidpointDepends on concavityDepends on concavity
Methodff Concave Upff Concave Down
TrapezoidalOverUnder
MidpointUnderOver

Key Fact: Trapezoidal = average of Left and Right: Tn=Ln+Rn2T_n = \frac{L_n + R_n}{2}.

Worked Example from Table

xx00112233
f(x)f(x)11332255

Δx=1\Delta x = 1, n=3n = 3.

MethodComputationValue
Left sumf(0)+f(1)+f(2)=1+3+2f(0)+f(1)+f(2) = 1+3+266
Right sumf(1)+f(2)+f(3)=3+2+5f(1)+f(2)+f(3) = 3+2+51010
Trapezoidal12(1+2(3)+2(2)+5)=162\frac{1}{2}(1+2(3)+2(2)+5) = \frac{16}{2}88

Check: T=L+R2=6+102=8T = \frac{L+R}{2} = \frac{6+10}{2} = 8 ✔

Unequal Subintervals

When Δx\Delta x varies, use individual trapezoids:

T=Δx12(f0+f1)+Δx22(f1+f2)+⋯T = \frac{\Delta x_1}{2}(f_0+f_1) + \frac{\Delta x_2}{2}(f_1+f_2) + \cdots

AP Tip: Table problems with unequal spacing appear frequently. Apply the trapezoidal formula to each subinterval separately.

Riemann Sums 🎯

Over or under? 🔍

Compute from a table. ✍️

Key Takeaways — Part 4

MethodFormula
Left∑f(xi)Δx\sum f(x_i)\Delta x (skip last)
Right∑f(xi)Δx\sum f(x_i)\Delta x (skip first)
TrapezoidalΔx2[f0+2f1+⋯+2fn−1+fn]\frac{\Delta x}{2}[f_0+2f_1+\cdots+2f_{n-1}+f_n]
Key relationT=(L+R)/2T = (L+R)/2

Up Next: Part 5 — Rate Problems & Net Change.

Part 5: Rate Problems & Net Change

Integration Applications

Part 5 of 7 — Rate Problems & Net Change

The Net Change Theorem

∫abf′(x) dx=f(b)−f(a)\boxed{\int_a^b f'(x)\,dx = f(b) - f(a)}

The integral of a rate of change gives the net change in the quantity.

Common AP Contexts

Rate FunctionWhat ∫\int GivesUnits
v(t)v(t) (velocity)Displacementdistance
$v(t)$ (speed)
R(t)R(t) (flow rate)Net volume changevolume
P′(t)P'(t) (pop. rate)Net population changecount
C′(x)C'(x) (marginal cost)Total cost changedollars

Key Fact: "Rate" in the problem statement means you’re given f′f', and integration gives f(b)−f(a)f(b)-f(a).

Rate In / Rate Out

Q(t)=Q0+∫0t[Rin(s)−Rout(s)] ds\boxed{Q(t) = Q_0 + \int_0^t [R_{in}(s) - R_{out}(s)]\,ds}

AP FRQ Pattern

PartTypical Question
(a)Compute ∫abR(t) dt\int_a^b R(t)\,dt and interpret
(b)Is quantity increasing or decreasing at t=ct = c?
(c)Find absolute min/max on interval
(d)Average rate over interval

Interpretation sentence template:

"∫abR(t) dt=N\int_a^b R(t)\,dt = N means that NN [units] of [quantity] [entered/left] from t=at=a to t=bt=b."

AP Tip: The AP exam almost always asks you to interpret the integral in context. Include units and the time interval.

Rate Problems 🎯

Interpret the integral. 🔍

Apply net change. ✍️

Key Takeaways — Part 5

ConceptFormula
Net change∫abf′(t) dt=f(b)−f(a)\int_a^b f'(t)\,dt = f(b)-f(a)
Rate in/outQ0+∫0t(Rin−Rout)dsQ_0 + \int_0^t(R_{in}-R_{out})ds
InterpretationInclude units and time interval

Up Next: Part 6 — Problem-Solving Workshop.

Part 6: Practice Workshop

Integration Applications

Part 6 of 7 — Problem-Solving Workshop

Integration Application Decision Guide

GivenMethod
Two curves, find area∫[top−bottom]dx\int [\text{top}-\text{bottom}]dx
Region + cross-section shape∫A(x) dx\int A(x)\,dx with shape formula
Rotation about axis, one boundaryDisk: π∫R2 dx\pi\int R^2\,dx
Rotation about axis, two boundariesWasher: π∫(R2−r2)dx\pi\int(R^2-r^2)dx
Table of valuesRiemann sum or trapezoidal rule
Rate functionNet change: ∫abf′(t) dt\int_a^b f'(t)\,dt
Rate in / rate outQ0+∫(Rin−Rout)dtQ_0 + \int(R_{in}-R_{out})dt

Key Fact: The first step is always identifying the problem type. The correct setup determines 90% of your score.

AP-Style Worked Problem

Region RR is bounded by y=4−x2y = 4-x^2 and y=0y = 0.

(a) Find the area of RR.

A=∫−22(4−x2)dx=[4x−x33]−22=323A = \int_{-2}^{2}(4-x^2)dx = [4x-\frac{x^3}{3}]_{-2}^{2} = \frac{32}{3}

(b) Cross sections perpendicular to xx-axis are squares. Find volume.

s=4−x2s = 4-x^2. V=∫−22(4−x2)2dxV = \int_{-2}^{2}(4-x^2)^2 dx

=∫−22(16−8x2+x4)dx=2[16x−8x33+x55]02= \int_{-2}^{2}(16-8x^2+x^4)dx = 2[16x-\frac{8x^3}{3}+\frac{x^5}{5}]_0^2

=2(32−643+325)=2⋅480−320+9615=51215= 2(32-\frac{64}{3}+\frac{32}{5}) = 2 \cdot \frac{480-320+96}{15} = \frac{512}{15}

(c) Rotate RR about the xx-axis. Find volume.

V=π∫−22(4−x2)2dx=512π15V = \pi\int_{-2}^{2}(4-x^2)^2 dx = \frac{512\pi}{15}

AP Tip: Parts (b) and (c) have the same integral! Cross sections → no π\pi. Revolution → multiply by π\pi.

Mixed Applications 🎯

Choose the right setup. 🔍

Mixed problem. ✍️

Key Takeaways — Part 6

Problem TypeKey Setup
Area∫(top−bottom)\int(\text{top}-\text{bottom})
Cross-section∫A(x) dx\int A(x)\,dx
Revolutionπ∫R2\pi\int R^2 or π∫(R2−r2)\pi\int(R^2-r^2)
Table dataRiemann sums or trapezoidal
Rates∫f′(t) dt=f(b)−f(a)\int f'(t)\,dt = f(b)-f(a)

Up Next: Part 7 — Comprehensive Assessment.

Part 7: Final Assessment

Integration Applications

Part 7 of 7 — Comprehensive Assessment

Complete Formula Reference

ApplicationFormula
Area (horizontal)∫ab[f(x)−g(x)]dx\int_a^b[f(x)-g(x)]dx
Area (vertical)∫cd[right−left]dy\int_c^d[\text{right}-\text{left}]dy
Cross-section volume∫abA(x) dx\int_a^b A(x)\,dx
Disk methodπ∫abR2 dx\pi\int_a^b R^2\,dx
Washer methodπ∫ab(R2−r2)dx\pi\int_a^b(R^2-r^2)dx
Trapezoidal ruleΔx2[f0+2f1+⋯+fn]\frac{\Delta x}{2}[f_0+2f_1+\cdots+f_n]
Net change∫abf′(t) dt=f(b)−f(a)\int_a^b f'(t)\,dt = f(b)-f(a)
Average value1b−a∫abf(x) dx\frac{1}{b-a}\int_a^b f(x)\,dx

Top AP Mistakes

MistakeCorrection
Subtracting radii: (R−r)2(R-r)^2Use R2−r2R^2-r^2 in washer method
Forgetting π\pi in revolutionCross-section: no π\pi. Revolution: include π\pi.
Wrong over/underestimateCheck increasing/decreasing AND concavity
Missing intersection pointsAlways find where curves cross
Wrong axis of rotationAdjust radii: distance = $
Not interpreting with unitsAlways state quantity, units, and time interval

Quiz — Area & Volume 🎯

Quiz — Numerical & Rates 🎯

Final classification. 🔍

Final Challenge ✍️

Integration Applications — Complete!

You’ve mastered:

PartTopic
1Area between curves
2Cross-sectional volumes
3Disk & washer methods
4Riemann sums & trapezoidal rule
5Rate problems & net change
6Problem-solving workshop
7Comprehensive assessment

You’re ready for AP-level integration application problems!