Applications of Integration Flashcards

What is the average value of f(x) = sin x on the interval [0, π]?

Answer: 2/π

The average value is (1/(π−0))∫₀^π sin x dx = (1/π)(2) = 2/π.

A particle moves along a line with velocity v(t) = t² − 4t + 3 for 0 ≤ t ≤ 3. What is the total distance the particle travels on this interval?

Answer: 8/3

v(t) = (t−1)(t−3) is positive on (0,1) and negative on (1,3). The distance is ∫₀¹ v dt − ∫₁³ v dt = 4/3 + 4/3 = 8/3.

What is the area of the region enclosed by the graphs of y = x and y = x²?

Answer: 1/6

The graphs intersect at x = 0 and x = 1, and y = x lies above y = x² between them. The area is ∫₀¹ (x − x²) dx = 1/2 − 1/3 = 1/6.

The region bounded by y = √x, the x-axis, and the line x = 4 is revolved about the x-axis. What is the volume of the resulting solid?

Answer: 8π

By the disk method, V = π∫₀⁴ (√x)² dx = π∫₀⁴ x dx = π(16/2) = 8π.

The region enclosed by y = x and y = x² is revolved about the x-axis. What is the volume of the solid generated?

Answer: 2π/15

Use washers with outer radius x and inner radius x²: V = π∫₀¹ (x² − x⁴) dx = π(1/3 − 1/5) = 2π/15.

The base of a solid is the region in the first quadrant bounded by y = 1 − x² and the coordinate axes. Cross-sections perpendicular to the x-axis are squares. What is the volume of the solid?

Answer: 8/15

Each square has side 1 − x², so V = ∫₀¹ (1 − x²)² dx = ∫₀¹ (1 − 2x² + x⁴) dx = 1 − 2/3 + 1/5 = 8/15.

A particle moves along the x-axis with velocity v(t) = 6t for t ≥ 0. If its position at t = 0 is x = 2, what is its position at t = 2?

Answer: 14

x(2) = x(0) + ∫₀² 6t dt = 2 + 3t²|₀² = 2 + 12 = 14.

Water is in a tank at the rate of change V′(t) = 6 − 2t gallons per minute for 0 ≤ t ≤ 6, where V(t) is the volume of water in gallons. If the tank holds 20 gallons at t = 0, what is the maximum volume of water in the tank on this interval?

Answer: 29

V′ changes from positive to negative at t = 3, so the maximum is at t = 3. V(3) = 20 + ∫₀³ (6 − 2t) dt = 20 + 9 = 29.

What is the area of the region bounded by the curves x = y² and x = y + 2?

Answer: 9/2

The curves meet at y = −1 and y = 2, and x = y + 2 is to the right of x = y². The area is ∫₋₁² (y + 2 − y²) dy = 9/2.

Let f(x) = x² on the interval [0, 3]. At what value of c in (0, 3) does f(c) equal the average value of f on [0, 3]?

Answer: √3

The average value is (1/3)∫₀³ x² dx = (1/3)(9) = 3. Setting c² = 3 gives c = √3, which lies in (0, 3).