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Calculus I (Differential & intro Integral) Unit 8: Applications of integration

Integrals turn rates into totals: distance from velocity, area between curves, volume of solids and the average value of a function.

Covers: Average value, accumulation, area between curves, volumes by cross-sections & discs/washers

Study guide

Integrals add up accumulated change. The average value of f on [a, b] is (1/(b − a))∫ₐᵇ f(x) dx, the height of the rectangle with the same area. If f is a rate, ∫ₐᵇ f(t) dt is the net change in the quantity: a starting amount plus the integral of the rate gives the amount later. For motion, the integral of velocity gives displacement, s(b) = s(a) + ∫ₐᵇ v(t) dt, while the integral of speed |v(t)| gives total distance traveled.

The area between two curves is ∫ₐᵇ [top − bottom] dx. Find where the curves intersect to get the limits, and split the integral if the curves switch places. For regions better described left to right, integrate in y: ∫ [right − left] dy.

Volumes come from integrating cross-sectional areas. For a solid with known cross-sections (squares, semicircles, triangles) perpendicular to the x-axis, write the area A(x) of a slice in terms of x and compute ∫ₐᵇ A(x) dx.

For solids of revolution, the disc method spins a region touching the axis: V = π∫ r(x)² dx. The washer method handles a region with a gap: V = π∫ (R(x)² − r(x)²) dx, with outer radius R and inner radius r. When rotating around a line other than an axis, measure radii as distances from that line.

Common cross-section areas, for a base length s: square s², semicircle with diameter s (π/8)s², equilateral triangle (√3/4)s², isosceles right triangle with leg s ½s², and rectangle with height k·s gives k·s². For rotation about the line y = k above the region, the radii are k minus each curve: R = k − (lower curve) and r = k − (upper curve). For an axis below the region, subtract k from each curve instead.

If r(t) is a rate in units per hour, ∫ₐᵇ r(t) dt is the total change in the quantity, in units, from t = a to t = b. An average value has the same units as the function being averaged. If a quantity has an inflow rate E(t) and an outflow rate L(t), the amount at time t is A(0) + ∫₀ᵗ (E(s) − L(s)) ds. It reaches a candidate max or min where E(t) = L(t) and E − L changes sign.

Key ideas

Formulas and rules

f_avg = [1/(b − a)]·∫ₐᵇ f(x) dx
s(b) = s(a) + ∫ₐᵇ v(t) dt
Area = ∫ₐᵇ [top − bottom] dx
Disc: V = π∫ r(x)² dx
Washer: V = π∫ (R² − r²) dx

Common mistakes

Words to know

Average value of f on [a, b]
[1/(b − a)]·∫ₐᵇ f(x) dx
Total distance traveled from velocity
∫ |v(t)| dt
Area between y = f(x) (top) and y = g(x) (bottom)
∫ₐᵇ [f(x) − g(x)] dx
Washer method volume
V = π∫ (R² − r²) dx
Position at time b given s(a) and v(t)
s(b) = s(a) + ∫ₐᵇ v(t) dt
Volume with known square cross-sections of side s(x)
V = ∫ s(x)² dx
Disc method
V = π∫ r(x)² dx
Net change theorem
∫ₐᵇ F′(t) dt = F(b) − F(a)
Displacement
∫ v(t) dt
Area between curves (in y)
∫ (right − left) dy
Cross-section method
V = ∫ A(x) dx
Axis other than an axis of coordinates
Measure radii as distances from the rotation line.
Accumulation from a rate
Amount later = starting amount + ∫ rate dt.

Sample practice questions

41 more questions like these, from level 1 to 5, are in the practice set.

Question 1 (level 1)

Average value of f(x) = x on [0, 4].

  1. 2
  2. 8
  3. 4
  4. 1
Show the answer and steps

Answer: A) 2

  1. Average value = [1/(b − a)]∫ₐᵇ f dx.
  2. (1/4)∫₀⁴ x dx = (1/4)(8).
  3. = 2.
Question 2 (level 2)

Area between y = x and y = x² on [0, 1].

  1. 1/2
  2. 1/3
  3. 1/6
  4. 5/6
Show the answer and steps

Answer: C) 1/6

  1. On [0, 1], x is above x².
  2. ∫₀¹ (x − x²) dx = 1/2 − 1/3.
  3. = 1/6.
Question 3 (level 3)

v(t) = 2t and s(0) = 3. Find s(2).

  1. 4
  2. 3
  3. 10
  4. 7
Show the answer and steps

Answer: D) 7

  1. s(2) = s(0) + ∫₀² v dt.
  2. ∫₀² 2t dt = 4.
  3. s(2) = 3 + 4 = 7.