AP Calculus BC Practice: Applications of Integration (6 Step-by-Step Explanations)

AP Calculus BC · Applications of Integration

Applications of Integration: AP Calculus BC Practice Problems

This collection pulls together several of the more specialized ways AP Calculus BC applies integration outside the standard area-and-volume setting: recovering a vector-valued velocity from acceleration (and reasoning about motion where position and velocity stay perpendicular), computing an average value in a real-world exponential-growth context, working through a rate word problem where the quantity you're integrating isn't given directly, and using a left Riemann sum to estimate a population from a radially symmetric density function.

Vector-valued motion works exactly like ordinary motion, just componentwise — integrate each component of acceleration separately (each picking up its own constant of integration from the initial condition) to recover velocity, and do the same again to recover position. A geometric fact worth knowing cold: if a curve's position vector and velocity vector are perpendicular at every instant, the object is moving along a circle centered at the origin, since d/dt(x²+y²) = 2(x dx/dt+y dy/dt) = 2(r·v), and this derivative being identically zero means x²+y² is constant. For population-density problems where the density depends only on distance from a center point (not on direction), the correct setup uses thin rings (annuli) of circumference 2πx rather than straight strips — this is a different geometric setup from a density that only depends on distance from a straight highway.

Below are 6 AP Calculus BC-style practice problems covering these applications. Work through each problem, select your answer, and check whether you got it right — every problem includes a complete, step-by-step explanation. Once you've worked through these, keep building your skills with thousands more official-style questions in our free AP Calculus BC QBank.

Vector-valued velocity from acceleration

The acceleration of an object in motion is given by the vector a(t) = ⟨3t², 2e^t⟩. If the object's initial velocity was v(0) = (1,3), which is the velocity vector at any time t?

Full Explanation

Integrate each component of acceleration separately, using the matching component of the initial velocity to solve for each constant.

x-component: v_x(t) = ∫3t²dt = t³+C₁. Since v_x(0)=1: C₁=1, so v_x(t)=t³+1.

y-component: v_y(t) = ∫2e^tdt = 2e^t+C₂. Since v_y(0)=3: 2+C₂=3 → C₂=1, so v_y(t)=2e^t+1.

Combining: v(t) = ⟨t³+1, 2e^t+1⟩. The correct answer is C.

A curve where position and velocity stay perpendicular

A particle moves along a curve in such a way that its position vector and velocity vector are perpendicular at all times. If the particle passes through the point (6,8), then the equation of the curve is

Full Explanation

Let r(t) = (x,y) be the position vector and v(t) = (dx/dt,dy/dt) the velocity vector. If they're perpendicular at every instant, their dot product is always zero: x(dx/dt)+y(dy/dt) = 0.

Notice that d/dt(x²+y²) = 2x(dx/dt)+2y(dy/dt), which is exactly twice the dot product of r and v. Since this is identically zero, x²+y² never changes — it's constant. This means the curve is a circle centered at the origin.

Use the given point (6,8) to find the constant: x²+y² = 6²+8² = 36+64 = 100.

The equation of the curve is x²+y² = 100. The correct answer is B.

Recovering position from velocity, working backward from a known point

The velocity of an object is given by v(t) = (6√t, 9). If this object is at the origin when t=1, where was it at t=0?

Full Explanation

Integrate each component to find position, keeping the constants of integration general for now.

x-component: x(t) = ∫6√t dt = 6·(2/3)t^3/2+C₁ = 4t^3/2+C₁. y-component: y(t) = ∫9 dt = 9t+C₂.

Use the condition that the object is at the origin (0,0) when t=1: x(1)=4+C₁=0 → C₁=−4, and y(1)=9+C₂=0 → C₂=−9.

So x(t)=4t^3/2−4 and y(t)=9t−9. At t=0: x(0)=−4, y(0)=−9. The correct answer is A.

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Average value · exponential population growth

Suppose the current world population is 10 billion and the population t years from now is estimated to be P(t) = 10e^0.012t billion people. On the basis of this supposition, the average population of the world, in billions, over the next 20 years will be approximately

Full Explanation

Average value = (1/20)∫₀²⁰10e^0.012tdt. Find the antiderivative: ∫10e^0.012tdt = (10/0.012)e^0.012t.

Evaluate: (1/20)·(10/0.012)[e^0.012t]₀²⁰ = (10/(20·0.012))[e^0.24−1].

Compute e^0.24 ≈ 1.2712, so e^0.24−1 ≈ 0.2712. Then (10/0.24)×0.2712 ≈ 41.667×0.2712.

≈ 11.302. The correct answer is B.

A rate word problem · fuel consumption

An 18-wheeler is traveling at a speed given by v(t) = 90(t+2)/(t+3) mph at time t hours. The fuel economy for the diesel fuel in the truck is given by f(v) = 5+0.02v miles per gallon. The amount, in gallons, of diesel fuel used during the first 3 hours is approximately

Full Explanation

Fuel economy f(v) is in miles per gallon, so the instantaneous rate of fuel consumption (gallons per hour) at any moment is speed divided by fuel economy: rate(t) = v(t)/f(v(t)) (this correctly converts miles/hour ÷ miles/gallon into gallons/hour).

The total gallons used over the first 3 hours is the accumulated integral of this rate: Gallons = ∫₀³ v(t)/f(v(t)) dt = ∫₀³ [90(t+2)/(t+3)] / [5+0.02·90(t+2)/(t+3)] dt.

This integral doesn't simplify to a clean closed form, so it's evaluated numerically (as is typical for this style of BC free-response problem, using a calculator).

The numerical result is approximately 32.5 gallons. The correct answer is C.

Left Riemann sum for a radially symmetric population density

The population density of a city, which is located in the middle of a large prairie, drops dramatically as distance from the center of town increases. This is shown in the following table:

x = distance (in mi) from the center	0	3	6	9	12	15
f(x) = density (hundreds of people/mi²)	60	50	35	20	10	3

Using a left Riemann Sum, we can calculate the population living within a 15-mile radius of the center to be approximately

Full Explanation

Since the density depends only on distance from the center (not on direction), the correct setup uses thin rings at radius x with thickness dx: each ring has area 2π x dx, so the population is ∫₀¹⁵2π x· f(x) dx, approximated here with a left Riemann sum using five subintervals of width 3.

Left endpoints and their contributions 2π x f(x)(3): at x=0, contribution is 0 (since the ring at the very center has zero circumference). At x=3 (f=50): 2π(3)(50)(3)=900π. At x=6 (f=35): 2π(6)(35)(3)=1260π. At x=9 (f=20): 2π(9)(20)(3)=1080π. At x=12 (f=10): 2π(12)(10)(3)=720π.

Sum: 900π+1260π+1080π+720π = 3960π ≈ 12,440.7 (in hundreds of people, since f was given in hundreds per square mile).

Multiply by 100 to get actual people: 12,440.7 × 100 ≈ 1,244,100. The correct answer is B.

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