Applications of integration

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1. Motion with integrals: position, displacement, distance

Warm-up

Three quick questions from earlier lessons.

Discover

A particle moves right for 2 s at 3 m/s, then left for 1 s at 3 m/s. Its displacement is 6−3=36 - 3 = 3 m. How far did it travel in total?

  1. 9 m
  2. 3 m

Key idea

Integrals of velocity give position changes; integrals of speed give distance.

s(b)=s(a)+∫abv(t) dts(b) = s(a) + \displaystyle\int_a^b v(t)\,dt. Displacement: ∫abv(t) dt\displaystyle\int_a^b v(t)\,dt. Total distance: ∫ab∣v(t)∣ dt\displaystyle\int_a^b |v(t)|\,dt, found by splitting where vv changes sign.

Worked example

v(t)=t−2v(t) = t - 2 on [0,5][0, 5]. Displacement and distance.

Displacement =12.5−10=2.5= 12.5 - 10 = 2.5
∫(t − 2) dt = t²/2 − 2t from 0 to 5.
∫02∣v∣=2\displaystyle\int_0^2 |v| = 2, ∫25∣v∣=4.5\displaystyle\int_2^5 |v| = 4.5
v < 0 on [0, 2]: that piece has area 2.
Distance =6.5= 6.5
Add the pieces.

Your turn, guided

Your turn: v(t)=3t2v(t) = 3t^2, s(0)=4s(0) = 4. Find s(2)s(2).

s(2)=12s(2) = 12
4 + 8.

Practice ★

  1. A particle has velocity v(t)=2t+7v(t) = 2t + 7 and starts at s(0)=4s(0) = 4. Find its position at t=5t = 5.
  2. A particle has velocity v(t)=2t+2v(t) = 2t + 2 and starts at s(0)=8s(0) = 8. Find its position at t=5t = 5.
  3. A particle has velocity v(t)=2t+9v(t) = 2t + 9 and starts at s(0)=1s(0) = 1. Find its position at t=5t = 5.

Watch out!

Practice ★★

  1. A particle's velocity is v(t)=t−4v(t) = t - 4 m/s. Find the total distance it travels from t=0t = 0 to t=6t = 6, in meters.
  2. A particle's velocity is v(t)=t−4v(t) = t - 4 m/s. Find the total distance it travels from t=0t = 0 to t=9t = 9, in meters.
  1. v(t)v(t) is a velocity in m/s. What does ∫09∣v(t)∣ dt\displaystyle\int_0^{ 9 } |v(t)|\,dt give?
    1. The total distance traveled from t=0t = 0 to t=9t = 9
    2. The displacement from t=0t = 0 to t=9t = 9
    3. The average speed on [0,9][0, 9]

Exit ticket

Three questions to check you have got it.

  1. A particle has velocity v(t)=2t+7v(t) = 2t + 7 and starts at s(0)=4s(0) = 4. Find its position at t=5t = 5.
  2. A particle has velocity v(t)=2t+2v(t) = 2t + 2 and starts at s(0)=8s(0) = 8. Find its position at t=5t = 5.
  3. A particle has velocity v(t)=2t+9v(t) = 2t + 9 and starts at s(0)=1s(0) = 1. Find its position at t=5t = 5.

Summary card

Motion with integrals

  • Position: start + ∫v\int v.
  • Displacement ∫v\int v; distance ∫∣v∣\int |v|.

Watch out: split where vv changes sign.

2. Area between curves

Warm-up

Three quick questions from earlier lessons.

Discover

The violet curve is y=x2y = x^2, the blue line y=kxy = kx. They enclose a region between x=0x = 0 and x=kx = k. Move kk.

The line y = kx and the parabola y = x². They meet at x = 0 and x = k; between them the line is on top.

On [0,k][0, k], which curve is on top?

  1. The line y=kxy = kx
  2. The parabola y=x2y = x^2

Key idea

Area between curves: integrate top minus bottom.

If f(x)≥g(x)f(x) \ge g(x) on [a,b][a, b], the area between them is ∫ab(f(x)−g(x)) dx\displaystyle\int_a^b \big(f(x) - g(x)\big)\,dx. The limits are usually where the curves meet; if they cross inside, split there.

Worked example

Area between y=2xy = 2x and y=x2y = x^2.

Limits 00 and 22
2x = x² at 0 and 2.
∫02(2x−x2) dx\displaystyle\int_0^2 (2x - x^2)\,dx
Top minus bottom.
=4−83=43= 4 - \tfrac{8}{3} = \tfrac{4}{3}
x² − x³/3 from 0 to 2.

Your turn, guided

Your turn: area between y=3xy = 3x and y=x2y = x^2.

92\tfrac{9}{2}
3³/6.

Practice ★

  1. Find the area of the region between y=5xy = 5x and y=x2y = x^2.
  2. Find the area of the region between y=4xy = 4x and y=x2y = x^2.
  3. Find the area of the region between y=1xy = 1x and y=x2y = x^2.

Watch out!

Practice ★★

  1. Find the area of the region between y=x2y = x^2 and y=4x−x2y = 4x - x^2.
  2. Find the area of the region between y=x2y = x^2 and y=2x−x2y = 2x - x^2.
  1. Find the area between y=1cos⁡xy = 1\cos x and y=1sin⁡xy = 1\sin x for 0≤x≤π40 \le x \le \dfrac{\pi}{4}. Give an exact value.

Exit ticket

Three questions to check you have got it.

  1. Find the area of the region between y=5xy = 5x and y=x2y = x^2.
  2. Find the area of the region between y=4xy = 4x and y=x2y = x^2.
  3. Find the area of the region between y=1xy = 1x and y=x2y = x^2.

Summary card

Area between curves

  1. Find where they meet.
  2. Decide which is on top.
  3. ∫(top−bottom) dx\int (\text{top} - \text{bottom})\,dx.

Watch out: a negative result means the order is reversed.

3. Area between curves with respect to y

Warm-up

Three quick questions from earlier lessons.

Discover

The region between x=y2x = y^2 and x=4x = 4 is easiest to slice horizontally. A horizontal strip at height yy runs from the parabola to the line. How long is it?

  1. 4−y24 - y^2
  2. y2−4y^2 - 4

Key idea

When curves are given as x=g(y)x = g(y), integrate in yy.

If f(y)≥g(y)f(y) \ge g(y) for c≤y≤dc \le y \le d, the area is ∫cd(f(y)−g(y)) dy\displaystyle\int_c^d \big(f(y) - g(y)\big)\,dy: right minus left, with yy-limits.

Worked example

Area between x=y2x = y^2 and x=4x = 4.

yy from −2-2 to 22
y² = 4: y = ±2.
∫−22(4−y2) dy\displaystyle\int_{-2}^{2} (4 - y^2)\,dy
Right minus left.
=323= \tfrac{32}{3}
4y − y³/3.

Your turn, guided

Your turn: between x=y2x = y^2 and x=1x = 1.

43\tfrac{4}{3}
∫ from −1 to 1 of (1 − y²) dy.

Practice ★

  1. Find the area of the region between x=y2x = y^2 and x=9x = 9, integrating with respect to yy.
  2. Find the area of the region between x=y2x = y^2 and x=16x = 16, integrating with respect to yy.
  1. Which integral gives the area between x=y2x = y^2 and x=16x = 16?
    1. ∫−44(16−y2) dy\displaystyle\int_{-4}^{ 4 } (16 - y^2)\,dy
    2. ∫016(16−y2) dy\displaystyle\int_0^{ 16 } (16 - y^2)\,dy
    3. ∫−44(y2−16) dy\displaystyle\int_{-4}^{ 4 } (y^2 - 16)\,dy

Watch out!

Practice ★★

  1. Find the area of the region between x=y2x = y^2 and x=2y+8x = 2y + 8.
  2. Find the area of the region between x=y2x = y^2 and x=3y+4x = 3y + 4.

Exit ticket

Three questions to check you have got it.

  1. Find the area of the region between x=y2x = y^2 and x=9x = 9, integrating with respect to yy.
  2. Find the area of the region between x=y2x = y^2 and x=16x = 16, integrating with respect to yy.
  3. Find the area of the region between x=y2x = y^2 and x=4x = 4, integrating with respect to yy.

Summary card

Area in yy

  • Right minus left, dydy.
  • Limits: the yy-values where the curves meet.

Watch out: xx-values are not yy-limits.

4. Volumes with known cross-sections

Warm-up

Three quick questions from earlier lessons.

Discover

A loaf of bread is sliced into thin slices. If each slice at position xx has area A(x)A(x) and thickness dxdx, what adds up to the loaf's volume?

  1. ∫A(x) dx\displaystyle\int A(x)\,dx
  2. The area of the base

Key idea

Volume is the integral of cross-sectional area.

V=∫abA(x) dxV = \displaystyle\int_a^b A(x)\,dx, where A(x)A(x) is the area of the cross-section at xx. Square of side ss: s2s^2. Semicircle of diameter ss: πs28\tfrac{\pi s^2}{8}. Equilateral triangle of side ss: 34s2\tfrac{\sqrt{3}}{4}s^2.

Worked example

Base: under y=2−xy = 2 - x for 0≤x≤20 \le x \le 2; square cross-sections.

s=2−xs = 2 - x
The side at x is the height of the base.
A(x)=(2−x)2A(x) = (2 - x)^2
Area of the square.
V=83V = \tfrac{8}{3}
−(2 − x)³/3 from 0 to 2.

Your turn, guided

Your turn: base under y=xy = \sqrt{x}, 0≤x≤20 \le x \le 2, squares.

V=2V = 2
∫₀² x dx.

Practice ★

  1. A solid's base is the region under y=xy = \sqrt{x} for 0≤x≤30 \le x \le 3. Cross-sections perpendicular to the xx-axis are squares. Find its volume.
  2. A solid's base is the region under y=xy = \sqrt{x} for 0≤x≤40 \le x \le 4. Cross-sections perpendicular to the xx-axis are squares. Find its volume.
  1. A solid's base is the triangle under y=4xy = 4x for 0≤x≤40 \le x \le 4. Cross-sections perpendicular to the xx-axis are squares. Find its volume.
  2. A solid's base is the triangle under y=1xy = 1x for 0≤x≤40 \le x \le 4. Cross-sections perpendicular to the xx-axis are squares. Find its volume.

Watch out!

Practice ★★

  1. Same base (under y=xy = \sqrt{x}, 0≤x≤20 \le x \le 2), but the cross-sections are semicircles whose diameter lies in the base. Find the volume.
  2. Same base (under y=xy = \sqrt{x}, 0≤x≤40 \le x \le 4), but the cross-sections are semicircles whose diameter lies in the base. Find the volume.

Exit ticket

Three questions to check you have got it.

  1. A solid's base is the region under y=xy = \sqrt{x} for 0≤x≤30 \le x \le 3. Cross-sections perpendicular to the xx-axis are squares. Find its volume.
  2. A solid's base is the region under y=xy = \sqrt{x} for 0≤x≤40 \le x \le 4. Cross-sections perpendicular to the xx-axis are squares. Find its volume.
  3. A solid's base is the region under y=xy = \sqrt{x} for 0≤x≤80 \le x \le 8. Cross-sections perpendicular to the xx-axis are squares. Find its volume.

Summary card

Known cross-sections

  1. Side s(x)s(x) from the base.
  2. Area A(x)A(x) of the shape.
  3. V=∫A(x) dxV = \int A(x)\,dx.

Watch out: a diameter is twice the radius.

5. Volumes of revolution: the disc method

Warm-up

Three quick questions from earlier lessons.

Discover

Spin the region under y=xy = x, 0≤x≤30 \le x \le 3, about the xx-axis: a cone. The slice at xx is a disc. What is its radius?

  1. xx
  2. πx\pi x

Key idea

Revolving a region about an axis: slices are discs.

Disc method: revolving the region under y=f(x)≥0y = f(x) \ge 0, a≤x≤ba \le x \le b, about the xx-axis gives V=π∫abf(x)2 dxV = \pi\displaystyle\int_a^b f(x)^2\,dx.

Worked example

Revolve the region under y=xy = x, 0≤x≤30 \le x \le 3.

V=π∫03x2 dxV = \pi\displaystyle\int_0^3 x^2\,dx
Radius x.
=9π= 9\pi
x³/3.
Matches 13πr2h\tfrac{1}{3}\pi r^2 h
Check: a cone of radius 3 and height 3 has (1/3)π·9·3.

Your turn, guided

Your turn: the region under y=xy = \sqrt{x}, 0≤x≤40 \le x \le 4.

V=8πV = 8\pi
π∫₀⁴ x dx.

Practice ★

  1. The region under y=xy = \sqrt{x}, 0≤x≤10 \le x \le 1, is revolved about the xx-axis. Find the volume.
  2. The region under y=xy = \sqrt{x}, 0≤x≤30 \le x \le 3, is revolved about the xx-axis. Find the volume.
  1. The region under y=xy = x, 0≤x≤40 \le x \le 4, is revolved about the xx-axis (a cone). Find the volume.
  2. The region under y=xy = x, 0≤x≤70 \le x \le 7, is revolved about the xx-axis (a cone). Find the volume.

Watch out!

Practice ★★

  1. The region under y=exy = e^x, 0≤x≤10 \le x \le 1, is revolved about the xx-axis. Find the volume.
  2. The region under y=exy = e^x, 0≤x≤30 \le x \le 3, is revolved about the xx-axis. Find the volume.

Exit ticket

Three questions to check you have got it.

  1. The region under y=xy = \sqrt{x}, 0≤x≤10 \le x \le 1, is revolved about the xx-axis. Find the volume.
  2. The region under y=xy = \sqrt{x}, 0≤x≤30 \le x \le 3, is revolved about the xx-axis. Find the volume.
  3. The region under y=xy = \sqrt{x}, 0≤x≤40 \le x \le 4, is revolved about the xx-axis. Find the volume.

Summary card

Disc method

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

Watch out: square the radius; keep the π\pi.

6. Volumes of revolution: the washer method

Warm-up

Three quick questions from earlier lessons.

Discover

Revolve the region between y=2y = 2 and y=1y = 1, 0≤x≤30 \le x \le 3, about the xx-axis: a thick pipe. Each slice is a washer with radii 2 and 1. What is its area?

  1. 3π3\pi
  2. π\pi

Key idea

A hole in the middle: subtract the inner disc.

Washer method: V=π∫ab(R(x)2−r(x)2) dxV = \pi\displaystyle\int_a^b \big(R(x)^2 - r(x)^2\big)\,dx, with RR the outer radius and rr the inner radius, both measured from the axis of rotation.

Worked example

Region between y=xy = x and y=x2y = x^2 revolved about the xx-axis.

R=xR = x, r=x2r = x^2
They meet at 0 and 1; y = x is farther from the axis.
V=π∫01(x2−x4) dxV = \pi\displaystyle\int_0^1 (x^2 - x^4)\,dx
R² − r².
=2π15= \tfrac{2\pi}{15}
1/3 − 1/5.

Your turn, guided

Your turn: revolve the region under y=1y = 1, 0≤x≤20 \le x \le 2, about y=−1y = -1.

V=6πV = 6\pi
R = 2, r = 1: π(4 − 1) × 2.

Practice ★

  1. The region between y=16y = 16 and y=x2y = x^2 is revolved about the xx-axis. Which integral is its volume?
    1. π∫−44(256−x4) dx\pi\displaystyle\int_{-4}^{ 4 } \big(256 - x^4\big)\,dx
    2. π∫−44(16−x2)2 dx\pi\displaystyle\int_{-4}^{ 4 } \big(16 - x^2\big)^2\,dx
    3. π∫−44(16−x2) dx\pi\displaystyle\int_{-4}^{ 4 } \big(16 - x^2\big)\,dx
  1. The region between y=3xy = 3x and y=x2y = x^2 is revolved about the xx-axis. Find the volume.
  2. The region between y=1xy = 1x and y=x2y = x^2 is revolved about the xx-axis. Find the volume.

Watch out!

Practice ★★

  1. The region under y=xy = \sqrt{x}, 0≤x≤90 \le x \le 9, is revolved about the line y=−1y = -1. Find the volume.
  2. The region under y=xy = \sqrt{x}, 0≤x≤10 \le x \le 1, is revolved about the line y=−1y = -1. Find the volume.
  1. Boss problem. The region RR between y=2xy = 2x and y=x2y = x^2.

Exit ticket

Three questions to check you have got it.

  1. The region between y=3xy = 3x and y=x2y = x^2 is revolved about the xx-axis. Find the volume.
  2. The region between y=1xy = 1x and y=x2y = x^2 is revolved about the xx-axis. Find the volume.
  3. The region between y=2xy = 2x and y=x2y = x^2 is revolved about the xx-axis. Find the volume.

Summary card

Washer method

V=π∫ab(R2−r2) dxV = \pi\displaystyle\int_a^b (R^2 - r^2)\,dx, radii from the axis.

Watch out: R2−r2R^2 - r^2, not (R−r)2(R - r)^2.

Answers: Applications of integration

1. Motion with integrals: position, displacement, distance

A

  1. 6464
  2. 4343
  3. 7171

B

  1. 1010
  2. 412\frac{41}{2}

C

  1. The total distance traveled from t=0t = 0 to t=9t = 9

D

  1. 6464
  2. 4343
  3. 7171

2. Area between curves

A

  1. 1256\frac{125}{6}
  2. 323\frac{32}{3}
  3. 16\frac{1}{6}

B

  1. 83\frac{8}{3}
  2. 13\frac{1}{3}

C

  1. −1+2-1 + \sqrt{2}

D

  1. 1256\frac{125}{6}
  2. 323\frac{32}{3}
  3. 16\frac{1}{6}

3. Area between curves with respect to y

A

  1. 3636
  2. 2563\frac{256}{3}

B

  1. ∫−44(16−y2) dy\displaystyle\int_{-4}^{ 4 } (16 - y^2)\,dy

C

  1. 3636
  2. 1256\frac{125}{6}

D

  1. 3636
  2. 2563\frac{256}{3}
  3. 323\frac{32}{3}

4. Volumes with known cross-sections

A

  1. 92\frac{9}{2}
  2. 88

B

  1. 10243\frac{1024}{3}
  2. 643\frac{64}{3}

C

  1. π4\frac{\pi}{4}
  2. π\pi

D

  1. 92\frac{9}{2}
  2. 88
  3. 3232

5. Volumes of revolution: the disc method

A

  1. π2\frac{\pi}{2}
  2. 9π2\frac{9 \pi}{2}

B

  1. 64π3\frac{64 \pi}{3}
  2. 343π3\frac{343 \pi}{3}

C

  1. π(−1+e2)2\frac{\pi \left(-1 + e^{2}\right)}{2}
  2. π(−1+e6)2\frac{\pi \left(-1 + e^{6}\right)}{2}

D

  1. π2\frac{\pi}{2}
  2. 9π2\frac{9 \pi}{2}
  3. 8π8 \pi

6. Volumes of revolution: the washer method

A

  1. π∫−44(256−x4) dx\pi\displaystyle\int_{-4}^{ 4 } \big(256 - x^4\big)\,dx

B

  1. 162π5\frac{162 \pi}{5}
  2. 2π15\frac{2 \pi}{15}

C

  1. 153π2\frac{153 \pi}{2}
  2. 11π6\frac{11 \pi}{6}

D

  1. 64π15\frac{64 \pi}{15}

E

  1. 162π5\frac{162 \pi}{5}
  2. 2π15\frac{2 \pi}{15}
  3. 64π15\frac{64 \pi}{15}