Applications of Integration

Study Sheet

Applications of Integration

Everything you need for AP Calculus BC, Topic 8

Area Between Two Curves

Concept
Area formulas

If f(x)≥g(x)f(x)\ge g(x) on [a,b][a,b], the area of the region between the curves is

A=∫ab[ f(x)⏟top−g(x)⏟bottom ] dx.A=\int_a^b\big[\,\underbrace{f(x)}_{\text{top}}-\underbrace{g(x)}_{\text{bottom}}\,\big]\,dx.

If instead the region is bounded left/right by x=v(y)x=v(y) (right) and x=u(y)x=u(y) (left) for c≤y≤dc\le y\le d, integrate in yy:

A=∫cd[ v(y)⏟right−u(y)⏟left ] dy.A=\int_c^d\big[\,\underbrace{v(y)}_{\text{right}}-\underbrace{u(y)}_{\text{left}}\,\big]\,dy.

Choosing dxdx vs. dydy: integrate in xx when the region has a clear top curve and bottom curve; integrate in yy when it has a clear right curve and left curve (this often avoids splitting into two integrals). Always find the intersection points first --- they are your limits.

Example
Integrating in xx

Find the area between y=2xy=2x and y=x2y=x^2. Intersections: 2x=x2⇒x=0,22x=x^2\Rightarrow x=0,2. Top is 2x2x:

A=∫02(2x−x2) dx=[x2−x33]02=4−83=43.A=\int_0^2(2x-x^2)\,dx=\Big[x^2-\tfrac{x^3}{3}\Big]_0^2=4-\tfrac{8}{3}=\tfrac{4}{3}.
Example
Integrating in yy (avoids splitting)

Find the area bounded by x=y2x=y^2 and x=y+2x=y+2. Intersections: y2=y+2⇒y2−y−2=0⇒y=−1,2y^2=y+2\Rightarrow y^2-y-2=0\Rightarrow y=-1,2. Right curve is x=y+2x=y+2:

A=∫−12[(y+2)−y2] dy=[y22+2y−y33]−12=103−(−76)=92.A=\int_{-1}^{2}\big[(y+2)-y^2\big]\,dy=\Big[\tfrac{y^2}{2}+2y-\tfrac{y^3}{3}\Big]_{-1}^{2}=\tfrac{10}{3}-\Big(-\tfrac{7}{6}\Big)=\tfrac{9}{2}.

Doing this in xx would require two integrals because the bottom boundary changes.

Tip

Tip: If curves cross inside [a,b][a,b], the region flips top/bottom --- split at each crossing and integrate ∣ top−bottom ∣|\,\text{top}-\text{bottom}\,| piece by piece.

Average Value, Accumulation & Net Change

Concept
Average value of a function

The average value of ff on [a,b][a,b] is

favg=1b−a∫abf(x) dx.f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx.

By the Mean Value Theorem for Integrals, if ff is continuous there is at least one c∈(a,b)c\in(a,b) with f(c)=favgf(c)=f_{\text{avg}}.

Reminder — The Mean Value Theorem:  f′(c)=f(b)−f(a)b−afor some c∈(a,b)f'(c)=\frac{f(b)-f(a)}{b-a} \quad\text{for some } c \in (a,b)
Example
Average value

Average value of f(x)=x2f(x)=x^2 on [0,3][0,3]:

favg=13−0∫03x2 dx=13[x33]03=13⋅9=3.f_{\text{avg}}=\frac{1}{3-0}\int_0^3 x^2\,dx=\frac{1}{3}\Big[\tfrac{x^3}{3}\Big]_0^3=\frac{1}{3}\cdot 9=3.
Concept
Accumulation & net change

If F′(t)=r(t)F'(t)=r(t) is a rate of change, the net change of FF over [a,b][a,b] is

F(b)−F(a)=∫abr(t) dt.F(b)-F(a)=\int_a^b r(t)\,dt.

For a particle with velocity v(t)v(t):

displacement=∫abv(t) dt,total distance=∫ab∣v(t)∣ dt.\text{\bf displacement}=\int_a^b v(t)\,dt,\qquad \text{\bf total distance}=\int_a^b |v(t)|\,dt.
Example
Displacement vs. total distance

A particle has velocity v(t)=t2−4v(t)=t^2-4 (m/s) on [0,3][0,3]. Since v(t)=0v(t)=0 at t=2t=2, with v<0v<0 on [0,2][0,2] and v>0v>0 on [2,3][2,3]:

displacement=∫03(t2−4) dt=[t33−4t]03=9−12=−3 m.\text{displacement}=\int_0^3(t^2-4)\,dt=\Big[\tfrac{t^3}{3}-4t\Big]_0^3=9-12=-3\ \text{m}.
total distance=∫02 ⁣ ⁣−(t2−4) dt+∫23(t2−4) dt=163+73=233 m.\text{total distance}=\int_0^2\!\!-(t^2-4)\,dt+\int_2^3(t^2-4)\,dt=\tfrac{16}{3}+\tfrac{7}{3}=\tfrac{23}{3}\ \text{m}.
Tip

Tip: Position at time bb is s(b)=s(a)+∫abv(t) dts(b)=s(a)+\displaystyle\int_a^b v(t)\,dt. Speed is ∣v(t)∣|v(t)|; total distance integrates speed, so you must break the integral at every sign change of vv.

Volumes of Revolution: Disks & Washers

Concept
Disk and washer method (slices ⊥\perp to the axis)

Revolving a region about a horizontal axis and slicing perpendicular to it gives circular cross-sections.

Disk: V=π∫ab[R(x)]2 dx,Washer: V=π∫ab([R(x)]2−[r(x)]2) dx,\textbf{Disk: } V=\pi\int_a^b\big[R(x)\big]^2\,dx,\qquad \textbf{Washer: } V=\pi\int_a^b\Big(\big[R(x)\big]^2-\big[r(x)\big]^2\Big)\,dx,

where RR is the outer radius (far edge from axis) and rr the inner radius (near edge). For revolution about a vertical axis, integrate in yy with radii measured horizontally.

A representative disk of radius R=xR=\sqrt{x} generated by revolving the shaded region about the xx-axis.

Example
Disk

Region under y=xy=\sqrt{x} from x=0x=0 to 44, revolved about the xx-axis. Radius R=xR=\sqrt{x}:

V=π∫04(x)2 dx=π∫04x dx=π[x22]04=8π.V=\pi\int_0^4(\sqrt{x})^2\,dx=\pi\int_0^4 x\,dx=\pi\Big[\tfrac{x^2}{2}\Big]_0^4=8\pi.
Example
Washer

Region between y=2xy=2x (top) and y=x2y=x^2 (bottom) on [0,2][0,2], revolved about the xx-axis. Outer R=2xR=2x, inner r=x2r=x^2:

V=π∫02 ⁣[(2x)2−(x2)2]dx=π∫02(4x2−x4) dx=π[4x33−x55]02=64π15.V=\pi\int_0^2\!\big[(2x)^2-(x^2)^2\big]dx=\pi\int_0^2(4x^2-x^4)\,dx=\pi\Big[\tfrac{4x^3}{3}-\tfrac{x^5}{5}\Big]_0^2=\tfrac{64\pi}{15}.
Tip

Tip: When revolving about a line y=ky=k (not the xx-axis), each radius is a distance: R=∣ curve−k ∣R=|\,\text{curve}-k\,|. Square the distance, not the raw function value.

Volumes of Revolution: Cylindrical Shells

Concept
Shell method (slices ∥\parallel to the axis)

Slicing parallel to the axis of revolution produces thin cylindrical shells:

V=2π∫ab(radius)(height) dx.V=2\pi\int_a^b (\text{radius})(\text{height})\,dx.

For revolution about the yy-axis with a region described by xx: radius =x=x, height == top −- bottom. For revolution about the xx-axis, integrate in yy: radius =y=y, height == right −- left.

Example
Shells about the yy-axis

Region under y=xy=\sqrt{x} on [0,4][0,4], revolved about the yy-axis. Shell radius =x=x, height =x=\sqrt{x}:

V=2π∫04xx dx=2π∫04x3/2 dx=2π[25x5/2]04=2π⋅645=128π5.V=2\pi\int_0^4 x\sqrt{x}\,dx=2\pi\int_0^4 x^{3/2}\,dx=2\pi\Big[\tfrac{2}{5}x^{5/2}\Big]_0^4=2\pi\cdot\tfrac{64}{5}=\tfrac{128\pi}{5}.
Tip

Tip: Disks/washers slice perpendicular to the axis; shells slice parallel to it. Use shells when perpendicular slices would force you to solve y=f(x)y=f(x) for xx or would create a washer with an awkward inner radius.

Volumes by Known Cross-Sections

Concept
Cross-section method

If every slice perpendicular to the xx-axis has area A(x)A(x), then

V=∫abA(x) dx.V=\int_a^b A(x)\,dx.

Common cross-sections built on a segment of length ss (usually s=top−bottoms=\text{top}-\text{bottom} of the base region):

Square: A=s2,Semicircle (diameter s): A=π8s2,Equilateral △: A=34s2.\text{Square: } A=s^2,\quad \text{Semicircle (diameter }s\text{): } A=\tfrac{\pi}{8}s^2,\quad \text{Equilateral }\triangle:\ A=\tfrac{\sqrt3}{4}s^2.

Base region under y=xy=\sqrt{x}; a square cross-section (drawn in perspective) stands on the slice of length s=xs=\sqrt{x}.

Example
Square cross-sections

The base is the region under y=xy=\sqrt{x} on [0,4][0,4]. Cross-sections ⊥\perp to the xx-axis are squares with side s=xs=\sqrt{x}:

V=∫04(x)2 dx=∫04x dx=[x22]04=8.V=\int_0^4(\sqrt{x})^2\,dx=\int_0^4 x\,dx=\Big[\tfrac{x^2}{2}\Big]_0^4=8.
Example
Semicircular cross-sections

The base is bounded by y=xy=x, x=2x=2, and the xx-axis. Cross-sections ⊥\perp to the xx-axis are semicircles with diameter s=xs=x, so A=π8x2A=\tfrac{\pi}{8}x^2:

V=∫02π8x2 dx=π8[x33]02=π8⋅83=π3.V=\int_0^2\tfrac{\pi}{8}x^2\,dx=\tfrac{\pi}{8}\Big[\tfrac{x^3}{3}\Big]_0^2=\tfrac{\pi}{8}\cdot\tfrac{8}{3}=\tfrac{\pi}{3}.
Tip

Tip: No π\pi appears for square or triangular cross-sections --- only circular/semicircular ones carry a π\pi. Read whether the base segment is the diameter or the radius of a circular slice; they differ by a factor of 4 in area.

Arc Length of y=f(x)y=f(x)

Concept
Arc length formula

The length of the smooth curve y=f(x)y=f(x) from x=ax=a to x=bx=b is

L=∫ab1+[f′(x)]2 dx.L=\int_a^b\sqrt{1+\big[f'(x)\big]^2}\,dx.

It comes from summing hypotenuses (dx)2+(dy)2=1+(dy/dx)2 dx\sqrt{(dx)^2+(dy)^2}=\sqrt{1+(dy/dx)^2}\,dx of tiny right triangles along the curve.

Example
Arc length that simplifies to a perfect square

Find the length of y=13(x2+2)3/2y=\tfrac{1}{3}(x^2+2)^{3/2} on [0,3][0,3]. Here y′=xx2+2y'=x\sqrt{x^2+2}, so

1+(y′)2=1+x2(x2+2)=x4+2x2+1=(x2+1)2.1+(y')^2=1+x^2(x^2+2)=x^4+2x^2+1=(x^2+1)^2.
L=∫03(x2+1)2 dx=∫03(x2+1) dx=[x33+x]03=9+3=12.L=\int_0^3\sqrt{(x^2+1)^2}\,dx=\int_0^3(x^2+1)\,dx=\Big[\tfrac{x^3}{3}+x\Big]_0^3=9+3=12.
Tip

Tip: Textbook arc-length problems are engineered so 1+(y′)21+(y')^2 becomes a perfect square. If it does not, you may only be asked to set up the integral --- or to evaluate it numerically on a calculator.

Going Deeper: Advanced Applications

Concept
Surface area of a solid of revolution

Revolving y=f(x)≥0y=f(x)\ge 0 on [a,b][a,b] about the xx-axis produces a surface of area

S=∫ab2π y 1+(y′)2 dx=2π∫abf(x)1+[f′(x)]2 dx.S=\int_a^b 2\pi\,y\,\sqrt{1+(y')^2}\,dx=2\pi\int_a^b f(x)\sqrt{1+[f'(x)]^2}\,dx.

About the yy-axis (region in the first quadrant), S=∫2π x 1+(y′)2 dxS=\displaystyle\int 2\pi\,x\,\sqrt{1+(y')^2}\,dx. Think “circumference 2πr2\pi r times arc-length element.”

Example
A surface-area setup

Revolve y=xy=\sqrt{x}, 1≤x≤41\le x\le 4, about the xx-axis. With y′=12xy'=\tfrac{1}{2\sqrt x},   1+(y′)2=1+14x=4x+14x\;1+(y')^2=1+\tfrac{1}{4x}=\tfrac{4x+1}{4x}, so

S=2π∫14x⋅4x+14x dx=2π∫14124x+1 dx=π[16(4x+1)3/2]14=π6(173/2−53/2).S=2\pi\int_1^4\sqrt{x}\cdot\sqrt{\tfrac{4x+1}{4x}}\,dx=2\pi\int_1^4\tfrac{1}{2}\sqrt{4x+1}\,dx=\pi\Big[\tfrac{1}{6}(4x+1)^{3/2}\Big]_1^4=\tfrac{\pi}{6}\big(17^{3/2}-5^{3/2}\big).
Concept
Work: variable force and Hooke's law

Work done by a variable force F(x)F(x) moving an object from x=ax=a to x=bx=b is

W=∫abF(x) dx.W=\int_a^b F(x)\,dx.

For a spring, Hooke's law gives F(x)=kxF(x)=kx, so W=∫0dkx dx=12kd2W=\int_0^d kx\,dx=\tfrac12 kd^2.

Concept
Work: pumping a fluid

To pump fluid out of a tank, slice the fluid into horizontal layers at height yy of area A(y)A(y) and thickness dydy. A layer has weight ρg A(y) dy\rho g\,A(y)\,dy and must be lifted a distance D(y)D(y):

W=∫cdρg A(y) D(y) dy,W=\int_c^d \rho g\,A(y)\,D(y)\,dy,

where ρg\rho g is the fluid's weight density (for water, ρg≈9800 N/m3\rho g\approx 9800\ \text{N/m}^3, or 62.4 lb/ft362.4\ \text{lb/ft}^3).

Example
Mass from a density function

A thin rod along [0,4][0,4] has linear density ρ(x)=3+2x\rho(x)=3+2x (kg/m). Its mass is the accumulation of density:

m=∫04(3+2x) dx=[3x+x2]04=12+16=28 kg.m=\int_0^4(3+2x)\,dx=\Big[3x+x^2\Big]_0^4=12+16=28\ \text{kg}.

The same idea gives total population from a population-density function, or total rainfall from a rate.

Concept
Choosing disks/washers vs. shells wisely

Match the slice to the axis:

  • [leftmargin=*]
  • Slice perpendicular to the axis ⇒\Rightarrow disks/washers; the variable of integration matches the axis direction.
  • Slice parallel to the axis ⇒\Rightarrow shells.

Prefer whichever keeps your radii/heights as single functions and avoids solving the boundary curve for the other variable. Revolving a region under y=f(x)y=f(x) about the yy-axis is usually one clean shell integral 2π∫xf(x) dx2\pi\int x f(x)\,dx, whereas washers would require inverting ff.

Example
Cross-sections on a non-trivial base

The base is the region between y=xy=\sqrt{x} and y=x2y=\tfrac{x}{2} (they meet at x=0x=0 and x=4x=4). Cross-sections ⊥\perp to the xx-axis are squares whose side is the vertical gap s=x−x2s=\sqrt{x}-\tfrac{x}{2}:

V=∫04(x−x2)2dx=∫04(x−x3/2+x24)dx=[x22−25x5/2+x312]04=8−645+163=815.V=\int_0^4\Big(\sqrt{x}-\tfrac{x}{2}\Big)^2 dx=\int_0^4\Big(x-x^{3/2}+\tfrac{x^2}{4}\Big)dx=\Big[\tfrac{x^2}{2}-\tfrac{2}{5}x^{5/2}+\tfrac{x^3}{12}\Big]_0^4=8-\tfrac{64}{5}+\tfrac{16}{3}=\tfrac{8}{15}.

The side length is the difference of the two boundary curves, exactly as in an area problem.

Concept
Arc length of a parametric curve (BC preview)

For a curve given by x=x(t)x=x(t), y=y(t)y=y(t) on [α,β][\alpha,\beta], the arc length is

L=∫αβ(dxdt)2+(dydt)2 dt.L=\int_\alpha^\beta\sqrt{\Big(\tfrac{dx}{dt}\Big)^2+\Big(\tfrac{dy}{dt}\Big)^2}\,dt.

Example: for x=cos⁡tx=\cos t, y=sin⁡ty=\sin t,   (−sin⁡t)2+(cos⁡t)2=1\;\sqrt{(-\sin t)^2+(\cos t)^2}=1, so one full loop t∈[0,2π]t\in[0,2\pi] has length ∫02π1 dt=2π\int_0^{2\pi}1\,dt=2\pi --- the circumference of the unit circle, as expected. This unifies with y=f(x)y=f(x): taking t=xt=x recovers 1+(dy/dx)2\sqrt{1+(dy/dx)^2}.

Tip

Big picture: Every application here is the same move --- slice the quantity into thin pieces, write the piece as (something) ⋅ dx\,\cdot\,dx, and integrate. Area, volume, distance, work, mass, and length are all definite integrals of a rate or a cross-sectional measure.

Formulas, Proofs & Tips

Tip
Area between curves and volumes of revolution
A=∫ab(f−g) dx,V=π∫ab(R2−r2) dxA=\int_a^b\big(f-g\big)\,dx,\qquad V=\pi\int_a^b\big(R^{2}-r^{2}\big)\,dx

What it means. Integrate top minus bottom for area; sum circular slices for a solid of revolution.

Example. Between y=xy=x and y=x2y=x^2 on [0,1][0,1]: ∫01(x−x2) dx=16\int_0^1(x-x^2)\,dx=\tfrac16.

Why it works. A thin vertical strip has height f−gf-g and width dxdx, so its area is (f−g)dx(f-g)dx. Revolving that strip sweeps a washer of outer radius RR and inner radius rr, whose area is πR2−πr2\pi R^2-\pi r^2; multiplying by thickness dxdx and integrating stacks them.

Tip. Find the intersection points first — they are the limits, and they tell you which curve is on top.