Estimating Area Under Curves: Riemann Sums, Trapezoids & Real-World Rate Problems

1) The graph of velocity (in m/s) vs. time (in s) for a car is given. What does the area under the graph from t=2 to t=5 represent?

The acceleration at t=5.

The net displacement of the car from t=2 to t=5.

The average velocity from t=2 to t=5.

The change in velocity from t=2 to t=5.

The area under a velocity-time graph represents the displacement or distance traveled. Here, since velocity is in m/s and time in s, the area gives meters, which is distance traveled from t=2 to t=5.

Explanation

The area under a velocity-time graph represents the displacement or distance traveled. Here, since velocity is in m/s and time in s, the area gives meters, which is distance traveled from t=2 to t=5.

2) The graph of f(x) = 2x from x=0 to x=3 is a straight line. What is the area under f(x) from x=0 to x=3?

6

9

12

18

The region under f(x)=2x from 0 to 3 is a triangle. The base is 3 and the height is f(3)=6. The area of a triangle is (½)baseheight = (½)*3*6 = 9.

Explanation

The region under f(x)=2x from 0 to 3 is a triangle. The base is 3 and the height is f(3)=6. The area of a triangle is (½)baseheight = (½)*3*6 = 9.

3) If the rate of change of a population is positive over an interval, what can be said about the accumulated change in population?

It is positive.

It is negative.

It is zero.

It cannot be determined without numbers.

A positive rate of change means the population is increasing. The accumulated change, which is the area under the rate curve, will be positive, indicating an increase in population over the interval.

Explanation

A positive rate of change means the population is increasing. The accumulated change, which is the area under the rate curve, will be positive, indicating an increase in population over the interval.

4) If the rate of change of revenue is in dollars per day and time is in days, what unit does the area under the rate graph represent?

Dollars per day

Days

Dollars

Dollars per day squared

The unit of area is the unit of the rate (dollars per day) multiplied by the unit of time (days), resulting in dollars. This represents total revenue over the time period.

Explanation

The unit of area is the unit of the rate (dollars per day) multiplied by the unit of time (days), resulting in dollars. This represents total revenue over the time period.

5) Given the table of f(x): x: 0, 1, 2, 3 f(x): 2, 5, 3, 8 Approximate the ∫₀³ f(x) dx using a left Riemann sum.

10

15

18

21

A left Riemann sum uses the left endpoints of each subinterval. The subintervals are [0,1], [1,2], [2,3]. The left endpoints are 0,1,2 with function values 2,5,3. The width of each subinterval is 1. So the sum is 1*(2+5+3)=10.

Explanation

A left Riemann sum uses the left endpoints of each subinterval. The subintervals are [0,1], [1,2], [2,3]. The left endpoints are 0,1,2 with function values 2,5,3. The width of each subinterval is 1. So the sum is 1*(2+5+3)=10.

6) Approximate the integral from 1 to 3 of (x²+1) dx using a right Riemann sum with 2 subintervals of equal width.

17

15

26

30

The interval [1,3] divided into 2 subintervals gives width = (3-1)/2 = 1. The right endpoints are x=2 and x=3. The function values are f(2)=2²+1=5 and f(3)=3²+1=10. The right Riemann sum is width times the sum of these values: 1*(5+10)=15.

Explanation

The interval [1,3] divided into 2 subintervals gives width = (3-1)/2 = 1. The right endpoints are x=2 and x=3. The function values are f(2)=2²+1=5 and f(3)=3²+1=10. The right Riemann sum is width times the sum of these values: 1*(5+10)=15.

7) For a concave up function, which approximation method typically gives an overestimate of the definite integral?

Left Riemann sum

Right Riemann sum

Midpoint Riemann sum

Trapezoidal rule

For a concave up function, the trapezoidal rule overestimates the area because the trapezoids include area above the curve. The midpoint rule typically underestimates for concave up functions.

Explanation

For a concave up function, the trapezoidal rule overestimates the area because the trapezoids include area above the curve. The midpoint rule typically underestimates for concave up functions.

8) Which definite integral is represented by the limn 🠒∞ ∑i=1n of (2 + (3i/n))³ * (3/n)?

∫₀³ (2+x)³ dx

∫25 2+x³ dx

∫₀³ x³ dx

∫25 (2+x)³ dx

The term (2 + (3i/n)) suggests the sample points are xᵢ = 3i/n, but we have 2 added to that. So the function is (2+x)³. As i goes from 1 to n, xᵢ goes from 3/n to 3, so in the limit, the interval is [0,3]. Therefore, the limit represents the ∫₀³ (2+x)³ dx.

Explanation

The term (2 + (3i/n)) suggests the sample points are xᵢ = 3i/n, but we have 2 added to that. So the function is (2+x)³. As i goes from 1 to n, xᵢ goes from 3/n to 3, so in the limit, the interval is [0,3]. Therefore, the limit represents the ∫₀³ (2+x)³ dx.

9) Approximate ∫₀³ f(x) dx using the trapezoidal rule and the following values of f(x): x: 0,1,2,3 f(x): 1,4,9,16

20.5

21.5

22.5

23.5

The trapezoidal rule with equal spacing: width = 1. The formula is (width/2) times [f(first) + 2*(sum of interior values) + f(last)]. So, area = (½)*[1 + 2*(4+9) + 16] = (½)*[1+2*13+16] = (½)*[1+26+16] = (½)*43 = 21.5.

Explanation

The trapezoidal rule with equal spacing: width = 1. The formula is (width/2) times [f(first) + 2*(sum of interior values) + f(last)]. So, area = (½)*[1 + 2*(4+9) + 16] = (½)*[1+2*13+16] = (½)*[1+26+16] = (½)*43 = 21.5.

10) Which expression is a Riemann sum for ∫₀2 x² dx with 4 subintervals of equal width using right endpoints?

(0.5)[(0.5)² + (1)² + (1.5)² + (2)²]

(0.5)[(0)² + (0.5)² + (1)² + (1.5)²]

(2)[(0.5)² + (1)² + (1.5)² + (2)²]

(0.5)[(0.5)² + (1)² + (1.5)²]

The interval [0,2] with 4 subintervals gives width = (2-0)/4 = 0.5. The right endpoints are 0.5, 1, 1.5, 2. The Riemann sum using right endpoints is width times the sum of the function values at these points: 0.5*[(0.5)² + (1)² + (1.5)² + (2)²].

Explanation

The interval [0,2] with 4 subintervals gives width = (2-0)/4 = 0.5. The right endpoints are 0.5, 1, 1.5, 2. The Riemann sum using right endpoints is width times the sum of the function values at these points: 0.5*[(0.5)² + (1)² + (1.5)² + (2)²].

11) The lim_{n 🠒∞} the sum from i=1 to n of (5i/n)² * (5/n) can be written as which definite integral?

Integral from 0 to 5 of x² dx

Integral from 0 to 5 of (5x)² dx

∫₀¹ x² dx

∫₀¹ (5x)² dx

The term (5i/n) suggests the sample points are xᵢ = 5i/n, so the interval is [0,5]. The function is (xᵢ)², so it is x². The width is 5/n. Therefore, the limit represents the integral from 0 to 5 of x² dx.

Explanation

The term (5i/n) suggests the sample points are xᵢ = 5i/n, so the interval is [0,5]. The function is (xᵢ)², so it is x². The width is 5/n. Therefore, the limit represents the integral from 0 to 5 of x² dx.

12) Find the area bounded by y = 4 - x² and the x-axis.

16/3

32/3

8/3

64/3

The curve intersects the x-axis when 4-x²=0, so x=-2 and x=2. The area is the integral from -2 to 2 of (4-x²) dx. The antiderivative is 4x - (⅓)x³. Evaluating from -2 to 2: At x=2: 8 - 8/3 = 16/3. At x=-2: -8 + 8/3 = -16/3. So the integral = 16/3 - (-16/3) = 32/3.

Explanation

The curve intersects the x-axis when 4-x²=0, so x=-2 and x=2. The area is the integral from -2 to 2 of (4-x²) dx. The antiderivative is 4x - (⅓)x³. Evaluating from -2 to 2: At x=2: 8 - 8/3 = 16/3. At x=-2: -8 + 8/3 = -16/3. So the integral = 16/3 - (-16/3) = 32/3.

13) Find the area between the curve y = cos(x) and the x-axis from x=0 to x=π.

0

1

2

Pi

cos(x) is positive on [0, π/2] and negative on [π/2, pi]. The area is the sum of the absolute values of the integrals over these intervals. Integral from 0 to π/2 of cos(x) dx = sin(x) from 0 to π/2 = 1-0=1. Integral from π/2 toπ of cos(x) dx = sin(x) from π/2 toπ = 0-1=-1, absolute value is 1. Total area = 1+1=2.

Explanation

cos(x) is positive on [0, π/2] and negative on [π/2, pi]. The area is the sum of the absolute values of the integrals over these intervals. Integral from 0 to π/2 of cos(x) dx = sin(x) from 0 to π/2 = 1-0=1. Integral from π/2 toπ of cos(x) dx = sin(x) from π/2 toπ = 0-1=-1, absolute value is 1. Total area = 1+1=2.

14) Find the area between y = x and y = x³ from x=0 to x=1.

1/4

1/2

3/4

1

The curves intersect at x=0 and x=1. On [0,1], the line y=x is above y=x³. The area is given by the ∫₀¹ (x - x³) dx. The antiderivative is (½)x² - (1/4)x⁴. Evaluating from 0 to 1: (1/2 - 1/4) - 0 = 1/4.

Explanation

The curves intersect at x=0 and x=1. On [0,1], the line y=x is above y=x³. The area is given by the ∫₀¹ (x - x³) dx. The antiderivative is (½)x² - (1/4)x⁴. Evaluating from 0 to 1: (1/2 - 1/4) - 0 = 1/4.

15) The area between the curve y = x² - 4 and the x-axis from x=0 to x=3 is:

23/3

9

25/3

7

The curve crosses the x-axis at x=2 (since x²-4=0). On [0,2], the function is negative, so the area contribution is the integral from 0 to 2 of -(x²-4) dx. On [2,3], the function is positive, so the area is the integral from 2 to 3 of (x²-4) dx. First integral: -( (⅓)x³ - 4x ) from 0 to 2 = -[(8/3-8)-0] = -[8/3-24/3] = -[-16/3] = 16/3. Second integral: (⅓)x³ - 4x from 2 to 3 = (27/3-12) - (8/3-8) = (9-12) - (8/3-24/3) = (-3) - (-16/3) = -3 + 16/3 = -9/3+16/3 = 7/3. Total area = 16/3 + 7/3 = 23/3.

Explanation

The curve crosses the x-axis at x=2 (since x²-4=0). On [0,2], the function is negative, so the area contribution is the integral from 0 to 2 of -(x²-4) dx. On [2,3], the function is positive, so the area is the integral from 2 to 3 of (x²-4) dx. First integral: -( (⅓)x³ - 4x ) from 0 to 2 = -[(8/3-8)-0] = -[8/3-24/3] = -[-16/3] = 16/3. Second integral: (⅓)x³ - 4x from 2 to 3 = (27/3-12) - (8/3-8) = (9-12) - (8/3-24/3) = (-3) - (-16/3) = -3 + 16/3 = -9/3+16/3 = 7/3. Total area = 16/3 + 7/3 = 23/3.