Question: A farmer wants to divide the lot

> Find all antiderivatives of each following function: f (x) = 9x8

> Suppose that the velocity of a car at time t is υ(t) = 40 + 8/(t + 1)2 kilometers per hour. (a) Compute the area under the velocity curve from t = 1 to t = 9. (b) What does the area in part (a) represent?

> Some food is placed in a freezer. After t hours, the temperature of the food is dropping at the rate of r(t) degrees Fahrenheit per hour, where r(t) = 12 + 4/(t + 3)2. (a) Compute the area under the graph of y = r(t) over the interval 0 ≤ t ≤ 2. (b) What

> Let M(x) be a company’s marginal cost at production level x. Give an economic interpretation of the number 0∫100 M(x)dx.

> Let M(x) be a company’s marginal profit at production level x. Give an economic interpretation of the number 44∫48 M(x)dx.

> Suppose that the marginal profit function for a company is P(x) = 100 + 50x - 3x2 at production level x. (a) Find the extra profit earned from the sale of 3 additional units if 5 units are currently being produced. (b) Describe the answer to part (a) as

> Suppose that the marginal cost function of a handbag manufacturer is C(x) = 3/32 x2 - x + 200 dollars per unit at production level x (where x is measured in units of 100 handbags). (a) Find the total cost of producing 6 additional units if 2 units are c

> After t hours of operation, an assembly line is producing lawn mowers at the rate of r(t) = 21 – 4/5 t mowers per hour. (a) How many mowers are produced during the time from t = 2 to t = 5 hours? (b) Represent the answer to part (a) as an area.

> A helicopter is rising straight up in the air. Its velocity at time t is υ(t) = 2t + 1 feet per second. (a) How high does the helicopter rise during the first 5 seconds? (b) Represent the answer to part (a) as an area.

> Find the area of the region bounded by y = 1/x, y = 4x, and y = x/2, for x ≥ 0. (The region resembles the shaded region in Exercise 29.) Figure 27: Y /y = 8x y y=x x

> Find the area in Fig. 27 of the region bounded by y = 1/x2, y = x, and y = 8x, for x ≥ 0. Figure 27: Y /y = 8x y y=x x

> Determine the following: ∫3e-2x dx

> Find the area of the region between y = x2 and y = 1/x2 (a) from x = 1 to x = 4, (b) from x = 1/2 to x = 4.

> Find the area of the region between y = x2 - 3x and the x-axis (a) from x = 0 to x = 3, (b) from x = 0 to x = 4, (c) from x = -2 to x = 3.

> Find the area of the region bounded by the curves. y = 4/x and y = 5 - x

> Find the area of the region bounded by the curves. y = 8x2 and y = √x

> Find the area of the region bounded by the curves. y = x3 and y = 2x2

> Find the area of the region bounded by the curves. y = x(x2 - 1) and the x-axis

> Find the area of the region bounded by the curves. y = x2 - 1 and y = 3

> Find the area of the region bounded by the curves. y = -x2 + 6x - 5 and y = 2x - 5

> Find the area of the region bounded by the curves. y = 4x(1 - x) and y = 3/4

> Find the area of the region bounded by the curves. y = x2 and y = x

> Determine the following: ∫ (7/2x3 – 3√x) dx

> Find the area of the region between the curves. y = e2x and y = 1 - x from x = 0 to x = 1

> Find the area of the region between the curves. y = ex and y = 1/x2 from x = 1 to x = 2

> Find the area of the region between the curves. y = x(2 - x) and y = 2 from x = 0 to x = 2

> Find the area of the region between the curves. y = x2 - 6x + 12 and y = 1 from x = 0 to x = 4

> Find the area of the region between the curves. y = x2 + 1 and y = -x2 - 1 from x = -1 to x = 1

> Find the area of the region between the curves. y = 2x2 and y = 8 (a horizontal line) from x = -2 to x = 2

> Find the area of the region between the curve and the x-axis. f (x) = e-x + 2 from -1 to 2

> Find the area of the region between the curve and the x-axis. f (x) = ex - 3 from 0 to ln 3

> Find the area of the region between the curve and the x-axis. f (x) = x2 + 6x + 5 from 0 to 1.

> Find the area of the region between the curve and the x-axis. f (x) = x2 - 2x - 3, from 0 to 2.

> Determine the following: ∫ (x - 2x2 + 1/3x) dx

> Find the area of the region between the curve and the x-axis. f (x) = x(x2 - 1), from -1 to 1.

> Find the area of the region between the curve and the x-axis. f (x) = 1 - x2, from -2 to 2

> Let g(x) be the function pictured in Fig. 26. Determine whether 1 > 0 g(x)dx is positive, negative, or zero. Figure 26: 2 1 -2 -3 y y = g(x) +x + 1 2 3 4 5/6 7 8

> Let f (x) be the function pictured in Fig. 25. Determine whether 1 > 0 f (x)dx is positive, negative, or zero. Figure 25: 3 2 1 -1 Y 12 y = f(x) 3 4 5 7 8 8

> Shade the portion of Fig. 24 whose area is given by the integral 0∫1L [ f (x) - g (x)] + 1∫2 [ g (x) - f (x)] dx. Figure 24: 2 y = f(x) 1 y = g(x) 0 У х а 1 2

> Compute the area of the shaded region in two different ways: (a) by using simple geometric formulas; (b) by applying Theorem I. Y f(x) = -x 2- -2 2 0

> Compute the area of the shaded region in two different ways: (a) by using simple geometric formulas; (b) by applying Theorem I. y 1 0 1 (x)=1 1 2 x

> Compute the area of the shaded region in two different ways: (a) by using simple geometric formulas; (b) by applying Theorem I. 2 Y 0 12 f(x) = 2 23 3 4

> Evaluate a Riemann sum to approximate the area under the graph of f (x) on the given interval, with points selected as specified. f (x) = x√(1 + x2); 1 ≤ x ≤ 3, n = 20, midpoints of subintervals f

> Evaluate a Riemann sum to approximate the area under the graph of f (x) on the given interval, with points selected as specified. f (x) = x√(1 + x2); 1 ≤ x ≤ 3, n = 20, midpoints of subintervals

> Determine the following: ∫ (2/√x + 2√x) dx

> Compute the area under the graph of y = 1 / (1 + x2) from 0 to 5.

> The area under the graph of the function e-x2 plays an important role in probability. Compute this area from -1 to 1.

> We show that, as the number of subintervals increases indefinitely, the Riemann sum approximation of the area under the graph of f (x) = x2 from 0 to 1 approaches the value 1/3 , which is the exact value of the area. Partition the interval [0, 1] into n

> We show that, as the number of subintervals increases indefinitely, the Riemann sum approximation of the area under the graph of f (x) = x2 from 0 to 1 approaches the value 1/3 , which is the exact value of the area. Verify the given formula for n = 1, 2

> Estimate the area (in square feet) of the residential lot in Fig. 17. Figure 17: 0 20 60 100 140 160 106 ft 101 ft 100 ft 113 ft

> Use a Riemann sum with n = 5 and midpoints to estimate the area under the graph of f (x) = √(1 - x2) on the interval 0 ≤ x ≤ 1. The graph is a quarter circle, and the area under the graph is .7854

> The graph of the function f (x) = √(1 - x2) on the interval -1 ≤ x ≤ 1 is a semicircle. The area under the graph is 1/2 π(1)2 = π/2 = 1.57080, to five decimal places. Use a R

> Use a Riemann sum with n = 4 and right endpoints to estimate the area under the graph of f (x) = 2x - 4 on the interval 2 ≤ x ≤ 3. Then, repeat with n = 4 and midpoints. Compare the answers with the exact answer, 1

> Use a Riemann sum with n = 4 and left endpoints to estimate the area under the graph of f (x) = 4 - x on the interval 1 ≤ x ≤ 4. Then repeat with n = 4 and midpoints. Compare the answers with the exact answer, 4.5,

> Determine the following: ∫x√x dx

> Use a Riemann sum to approximate the area under the graph of f (x) in Fig. 14 on the given interval, with selected points as specified. Draw the approximating rectangles. 1 ≤ x ≤ 7, n = 3, midpoints of subintervals

> Use a Riemann sum to approximate the area under the graph of f (x) in Fig. 14 on the given interval, with selected points as specified. Draw the approximating rectangles. 4 ≤ x ≤ 9, n = 5, right endpoints Riemann

> Use a Riemann sum to approximate the area under the graph of f (x) in Fig. 14 on the given interval, with selected points as specified. Draw the approximating rectangles. 3 ≤ x ≤ 7, n = 4, left endpoints Riemann s

> Use a Riemann sum to approximate the area under the graph of f (x) in Fig. 14 on the given interval, with selected points as specified. Draw the approximating rectangles. 0 ≤ x ≤ 8, n = 4, midpoints of subintervals

> Use a Riemann sum to approximate the area under the graph of f (x) on the given interval, with selected points as specified. f (x) = ln x; 2 ≤ x ≤ 4, n = 5, left endpoints Riemann sum: Ricmann sum. f(x1) Δx +

> Use a Riemann sum to approximate the area under the graph of f (x) on the given interval, with selected points as specified. f (x) = e-x; 2 ≤ x ≤ 3, n = 5, right endpoints Riemann sum: Ricmann sum. f(x1) Δx +

> Use a Riemann sum to approximate the area under the graph of f (x) on the given interval, with selected points as specified. f (x) = x3; 0 ≤ x ≤ 1, n = 5, right endpoints Riemann sum: Ricmann sum. f(x1) Δx +

> Use a Riemann sum to approximate the area under the graph of f (x) on the given interval, with selected points as specified. f (x) = x3; 1 ≤ x ≤ 3, n = 5, left endpoints Riemann sum: Ricmann sum. f(x1) Δx + f

> Use a Riemann sum to approximate the area under the graph of f (x) on the given interval, with selected points as specified. f (x) = x2; -2 ≤ x ≤ 2, n = 4, midpoints of subintervals Riemann sum: Ricmann sum.

> Use a Riemann sum to approximate the area under the graph of f (x) on the given interval, with selected points as specified. f (x) = x2; 1 ≤ x ≤ 3, n = 4, midpoints of subintervals Riemann sum: Ricmann sum. f

> Determine the following: ∫1/7x dx

> Determine Δx and the midpoints of the subintervals formed by partitioning the given interval into n subintervals. 3 ≤ x ≤ 5; n = 5

> Determine Δx and the midpoints of the subintervals formed by partitioning the given interval into n subintervals. 1 ≤ x ≤ 4; n = 5

> Determine Δx and the midpoints of the subintervals formed by partitioning the given interval into n subintervals. 0 ≤ x ≤ 3; n = 6

> Determine Δx and the midpoints of the subintervals formed by partitioning the given interval into n subintervals. 0 ≤ x ≤ 2; n = 4

> Find the real number b 7 0 so that the area under the graph of y = x2 from 0 to b is equal to the area under the graph of y = x3 from 0 to b.

> Find the real number b > 0 so that the area under the graph of y = x3 from 0 to b is equal to 4.

> Find the area under each of the given curves. y = e3x; x = - 1/3 to x = 0

> Find the area under each of the given curves. y = (x - 3)4; x = 1 to x = 4

> Find the area under each of the given curves. y = √x; x = 0 to x = 4

> Find the area under each of the given curves. y = 3x2 + x + 2ex/2; x = 0 to x = 1

> Determine the following: ∫ (2/x + x/2) dx

> Find the area under each of the given curves. y = 3x2; x = -1 to x = 1

> Find the area under each of the given curves. y = 4x; x = 2 to x = 3

> Draw the region whose area is given by the definite integral. 0∫4√x dx

> Draw the region whose area is given by the definite integral. 0∫4 (8 - 2x) dx

> Draw the region whose area is given by the definite integral. 2∫4 x2 dx

> Use Theorem I to compute the shaded area in Exercise 11. Shaded area in Exercise 11: Theorem 1: y y = x + 1 1 3 y=x Theorem I: Area under a Graph If f(x) is a continuous nonnegative function on the interval a ≤ x ≤ b, then the area under the grap

> Use Theorem I to compute the shaded area in Exercise 8. Shaded area in Exercise 8: Theorem 1: 0 थ्र y = - 0(0 – 3) 3 Theorem I: Area under a Graph If f(x) is a continuous nonnegative function on the interval a ≤ x ≤ b, then the area under the gra

> Use Theorem I to compute the shaded area in Exercise 7. Shaded Area in Exercise 7: Theorem 1: Y 0 f(x)=1/ 1 2 Theorem I: Area under a Graph If f(x) is a continuous nonnegative function on the interval a ≤ x ≤ b, then the area under the graph of f

> Set-up the definite integral that gives the area of the shaded region. Do not evaluate the integral. y 2. 1 0 x+1 I 1 1 3-x 2

> Set-up the definite integral that gives the area of the shaded region. Do not evaluate the integral. y y = x + ² 1 3 y=x

> Determine the following: ∫x * x2 dx

> Set-up the definite integral that gives the area of the shaded region. Do not evaluate the integral. Y -1 이 y=e 2

> Set-up the definite integral that gives the area of the shaded region. Do not evaluate the integral. Y f(x) = ln x 1 2

> Set-up the definite integral that gives the area of the shaded region. Do not evaluate the integral. fi 0 y = -x(x-3) 3 x

> Set-up the definite integral that gives the area of the shaded region. Do not evaluate the integral. Y 0 f(x)=1/ 1 2 x

> Compute the area of the shaded region in two different ways: (a) by using simple geometric formulas; (b) by applying Theorem I. 2. g 1 2 3 = 6 - 2x 3 - Ꮖ

> Compute the area of the shaded region in two different ways: (a) by using simple geometric formulas; (b) by applying Theorem I. 1 y 0 y = 1- x 1 y = x - 1 2

> Compute the area of the shaded region in two different ways: (a) by using simple geometric formulas; (b) by applying Theorem I. -2 22 Y 0 f(x) = x + 2 2

> Evaluate the given integral. 1∫4 (3√t + 4t) dt

> Evaluate the given integral. -1∫2(x2/3 – 2/9x) dx

> Evaluate the given integral. 0∫1 (2x – 3/4) dx

> Determine the following: ∫x/c dx (c a constant ≠ 0)

> A conical-shaped tank is being drained. The height of the water level in the tank is decreasing at the rate h(t) = - t/2 inches per minute. Find the decrease in the depth of the water in the tank during the time interval 2 ≤ t ≤ 4.