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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
- Average value of f on [a, b] is the integral divided by the interval length.
- Displacement = ∫ v dt. Total distance = ∫ |v| dt.
- Future position = starting position + ∫ v dt.
- Area between curves: integrate (top − bottom) with respect to x, or (right − left) with respect to y.
- Volumes with known cross-sections: integrate the cross-section's area A(x).
- Disc and washer methods: integrate π·(outer radius² − inner radius²).
Formulas and rules
Common mistakes
- Using π(R − r)² instead of π(R² − r²) for washers.
- Giving displacement when the question asks for total distance.
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.
Average value of f(x) = x on [0, 4].
- 2
- 8
- 4
- 1
Show the answer and steps
Answer: A) 2
- Average value = [1/(b − a)]∫ₐᵇ f dx.
- (1/4)∫₀⁴ x dx = (1/4)(8).
- = 2.
Area between y = x and y = x² on [0, 1].
- 1/2
- 1/3
- 1/6
- 5/6
Show the answer and steps
Answer: C) 1/6
- On [0, 1], x is above x².
- ∫₀¹ (x − x²) dx = 1/2 − 1/3.
- = 1/6.
v(t) = 2t and s(0) = 3. Find s(2).
- 4
- 3
- 10
- 7
Show the answer and steps
Answer: D) 7
- s(2) = s(0) + ∫₀² v dt.
- ∫₀² 2t dt = 4.
- s(2) = 3 + 4 = 7.