Difficulty: Intermediate | Prerequisites: Sections 13.1-13.4 (antiderivatives, substitution, definite integrals). You need to be able to evaluate definite integrals confidently before tackling these application problems.
This is where all the integration machinery pays off. Definite integrals model real quantities: displacement from a velocity function, revenue from a marginal revenue function, total output from a rate function. They also compute geometric areas, both under a single curve and between two curves. These are the problems that show up as word problems and multi-step questions on the exam, combining integration skills with setup and interpretation.
To find displacement, revenue, or total quantity, integrate the rate function over the relevant interval. To find the area between two curves, integrate the top function minus the bottom function between their intersection points (or given bounds). The hardest part is usually the setup, not the integration itself.
Displacement
The net change in position over a time interval, found by integrating velocity: ∫ₐᵇ v(t) dt. In simple terms, it tells you how far (and in which direction) the object ended up from where it started, not the total distance it travelled.
Net change
The total accumulated change over an interval. If you integrate a rate of change, you get the net change in the original quantity. This is sometimes called the net change theorem.
Area under a curve
The region between f(x) and the x-axis from x = a to x = b, computed as ∫ₐᵇ f(x) dx when f(x) ≥ 0. Think of it as the total "space" the curve covers above the x-axis.
Area between curves
The region enclosed between two functions f(x) and g(x), computed as ∫ₐᵇ |f(x) - g(x)| dx. In practice, you determine which function is on top in each sub-interval and integrate (top - bottom).
Intersection points
Where two curves meet, found by setting f(x) = g(x) and solving. These determine the bounds of integration for area-between-curves problems.
Marginal revenue
The derivative of the revenue function, R'(x). Integrating marginal revenue over an interval [a, b] gives the revenue generated by selling units a through b.
Displacement = ∫ₐᵇ v(t) dt
This gives net displacement (can be negative if the object reverses direction)
Example: v(t) = -t² + 8 from t = 2 to t = 6. Integrate: [-t³/3 + 8t] from 2 to 6 = (-72 + 48) - (-8/3 + 16) = -24 - 40/3 = -112/3
Revenue from selling units a through b = ∫ₐᵇ R'(x) dx
Example: marginal revenue is 500e^(-0.001x), revenue from boxes 101 to 5000 = ∫₁₀₁⁵⁰⁰⁰ 500e^(-0.001x) dx
Antiderivative: -500000e^(-0.001x). Evaluate at bounds and subtract.
If f(t) is a rate (units per time), then ∫ₐᵇ f(t) dt is the total quantity accumulated
Example: YouTube upload rate f(t) = 1.1t² - 2.6t + 2.3 million hours/year, total from t = 0 to t = 3 (June 2007 to June 2010)
Area = ∫ₐᵇ f(x) dx when f(x) ≥ 0 on [a, b]
Example: area under y = √x from x = 0 to x = 16 = ∫₀¹⁶ x^(1/2) dx = [(2/3)x^(3/2)] from 0 to 16 = (2/3)(64) = 128/3
Step 1: Find intersection points by setting f(x) = g(x)
Step 2: Determine which function is on top in each sub-interval (plug in a test point)
Step 3: Integrate (top - bottom) over each sub-interval
Area = ∫ₐᵇ [f(x) - g(x)] dx where f(x) ≥ g(x)
If the top and bottom functions swap, you must split the integral at the crossing point
Example: area between y = x² and y = 1 from x = -1 to x = 2. The curves cross at x = -1 and x = 1. From -1 to 1, y = 1 is on top. From 1 to 2, y = x² is on top.
The correct integral is ∫₋₁¹ (1 - x²) dx + ∫₁² (x² - 1) dx
Set the two functions equal and solve for x
Example: f(x) = x² - x + 5 and g(x) = x + 8. Set x² - x + 5 = x + 8, so x² - 2x - 3 = 0, giving (x - 3)(x + 1) = 0, so x = -1 and x = 3
These become the bounds of integration
\text{Displacement} = \int_a^b v(t) \, dt\text{Revenue from units } a \text{ to } b = \int_a^b R'(x) \, dx\text{Area between curves} = \int_a^b [f(x) - g(x)] \, dx \quad \text{where } f(x) \geq g(x)\text{Area under } f(x) = \int_a^b f(x) \, dx \quad \text{where } f(x) \geq 0Displacement integrals are used in every motion-tracking system, from fitness watches to spacecraft navigation. Revenue-from-marginal-revenue calculations are standard in business analytics for projecting earnings over a range of sales. Area between curves appears in economics (consumer and producer surplus), environmental science (comparing pollutant levels between two models), and engineering (stress analysis between design tolerances).
Confusing displacement with total distance. Displacement can be negative (the object moved backward). Total distance requires integrating |v(t)|, which is a different calculation. The exam review asks for displacement, so use the signed integral.
Integrating (bottom - top) and getting a negative area. Area is always positive. If your answer is negative, you probably subtracted in the wrong order. Check which function is on top.
Forgetting to split when curves cross. If y = x² and y = 1 swap positions in the interval, you cannot use a single integral. You must split at the crossing point and take (top - bottom) in each piece.
Using the wrong bounds for revenue problems. "Revenue from selling boxes 101 to 5000" means the limits are 101 and 5000, not 1 and 5000.
⚠️ Area-between-curves problems almost always require you to find intersection points first. Practise solving quadratics quickly.
⚠️ The multiple-choice question about which integral correctly computes the area tests whether you know to split at crossing points and take (top - bottom) in each sub-interval.
⚠️ Applied word problems (displacement, revenue, upload rates) test reading comprehension as much as calculus. Identify the rate function, the bounds, and what quantity the integral represents before computing.
⚠️ Be comfortable with both positive and negative results from displacement integrals. A negative displacement is a valid answer.
True or false: Displacement can be negative.
Fill in the blank: To find the area between f(x) and g(x), integrate ___ over the interval.
True or false: If two curves cross within [a, b], you can always use a single integral ∫ₐᵇ [f(x) - g(x)] dx for the area.
Fill in the blank: Revenue from selling units 50 to 200 equals ∫^ R'(x) dx.
True or false: The area under y = √x from 0 to 16 requires u-substitution.
Answers: 1. True. 2. (top function - bottom function). 3. False (you must split at the crossing). 4. ∫₅₀²⁰⁰. 5. False (rewrite as x^(1/2) and use the power rule directly).
Q: A particle moves with velocity v(t) = -t² + 8. Find the displacement between t = 2 and t = 6.
A: Displacement = ∫₂⁶ (-t² + 8) dt = [-t³/3 + 8t] from 2 to 6 = (-216/3 + 48) - (-8/3 + 16) = (-72 + 48) - (-8/3 + 16) = -24 - 40/3 = -112/3 ≈ -37.33. The negative sign means the particle moved backward overall.
Q: Calculate the area bounded by y = √x, the x-axis, x = 0, and x = 16.
A: Area = ∫₀¹⁶ x^(1/2) dx = [(2/3)x^(3/2)] from 0 to 16 = (2/3)(64) - 0 = 128/3 ≈ 42.67.
Q: Find the area enclosed by f(x) = x² - x + 5 and g(x) = x + 8.
A: Find intersections: x² - x + 5 = x + 8, so x² - 2x - 3 = 0, giving x = -1 and x = 3. Test x = 0: g(0) = 8 > f(0) = 5, so g is on top. Area = ∫₋₁³ [(x + 8) - (x² - x + 5)] dx = ∫₋₁³ (-x² + 2x + 3) dx = [-x³/3 + x² + 3x] from -1 to 3 = (−9 + 9 + 9) − (1/3 + 1 − 3) = 9 − (−5/3) = 32/3.
Q: Find the area enclosed by y = -x² + 6x + 2 and y = 2x² + 9x - 4.
A: Set equal: -x² + 6x + 2 = 2x² + 9x - 4, so -3x² - 3x + 6 = 0, giving x² + x - 2 = 0, so (x + 2)(x - 1) = 0, x = -2 and x = 1. Test x = 0: top = 2, bottom = -4, so (-x² + 6x + 2) is on top. Area = ∫₋₂¹ [(-x² + 6x + 2) - (2x² + 9x - 4)] dx = ∫₋₂¹ (-3x² - 3x + 6) dx.
Q: Which integral calculates the area between y = x² and y = 1 from x = -1 to x = 2?
A: ∫₋₁¹ (1 - x²) dx + ∫₁² (x² - 1) dx. The curves cross at x = ±1. From -1 to 1, y = 1 is on top. From 1 to 2, y = x² is on top. You must split and take (top - bottom) in each piece.
Q: The marginal revenue for the x-th box is 500e^(-0.001x). Calculate the revenue from selling boxes 101 to 5000.
A: Revenue = ∫₁₀₁⁵⁰⁰⁰ 500e^(-0.001x) dx. Antiderivative: 500/(-0.001) · e^(-0.001x) = -500000e^(-0.001x). Evaluate: [-500000e^(-0.001·5000)] - [-500000e^(-0.001·101)] = -500000e⁻⁵ + 500000e⁻⁰·¹⁰¹ = 500000(e⁻⁰·¹⁰¹ - e⁻⁵) ≈ 500000(0.9039 - 0.0067) ≈ 448,600.
Q: The rate of video uploads to YouTube is f(t) = 1.1t² - 2.6t + 2.3 million hours per year. Estimate the total hours uploaded from June 2007 to June 2010 (t = 0 to t = 3).
A: Total = ∫₀³ (1.1t² - 2.6t + 2.3) dt = [(1.1/3)t³ - 1.3t² + 2.3t] from 0 to 3 = (1.1/3)(27) - 1.3(9) + 2.3(3) - 0 = 9.9 - 11.7 + 6.9 = 5.1 million hours.
Q: Determine the area between y = x² and y = -1 from x = -1 to x = 1.
A: Since x² ≥ -1 for all x, the top function is y = x² and the bottom is y = -1 throughout the interval. Area = ∫₋₁¹ [x² - (-1)] dx = ∫₋₁¹ (x² + 1) dx = [x³/3 + x] from -1 to 1 = (1/3 + 1) - (-1/3 - 1) = 4/3 + 4/3 = 8/3.
Area between curves extends directly from definite integrals (Sections 13.3-13.4) by adding the step of determining which function is on top. In economics courses, this same technique computes consumer surplus and producer surplus. The displacement and revenue problems connect back to the initial-value problems in Section 13.1, where you recovered a function from its derivative.
Displacement, net change theorem, area under a curve, area between curves, area enclosed by curves, intersection points, top minus bottom, splitting the integral, marginal revenue, total revenue from marginal, velocity to displacement, rate to total, consumer surplus, producer surplus, MAT 210, Brief Calculus, ASU, Sections 14.2, 14.4