Question

The rate of fuel consumption, in gallons per minute, recorded during an airplane flight is given by a twice-differentiable and strictly increasing function R(t)=.002(t−4.72)³ +20.21 Approximate the value of ∫₀⁵⁰ R(t)dt using a left Riemann sum with the five equal subintervals. Is this approximation less than or greater than the value of ∫₀⁵⁰ R(t)dt ? Explain your reasoning.

Answer 1

To approximate the value of the integral ∫₀⁵⁰ R(t)dt using a left Riemann sum with five equal subintervals, we can calculate the width of each subinterval, evaluate the function at the left endpoint of each subinterval, and calculate the sum of the resulting values multiplied by the width.

Step-by-step explanation:

To approximate the value of the integral ∫₀⁵⁰ R(t)dt using a left Riemann sum with five equal subintervals, we can calculate the width of each subinterval first. The width of each subinterval is given by (b-a)/n, where a = 0, b = 50, and n = 5. Thus, the width is (50-0)/5 = 10.

Next, we can calculate the left Riemann sum by evaluating the function R(t) at the left endpoint of each subinterval and multiplying it by the width. This can be done for each subinterval:

R(0) = 0.002(0−4.72)³ +20.21 = 20.21

R(10) = 0.002(10−4.72)³ +20.21 = 20.21

R(20) = 0.002(20−4.72)³ +20.21 = 21.90

R(30) = 0.002(30−4.72)³ +20.21 = 30.11

R(40) = 0.002(40−4.72)³ +20.21 = 45.69

Finally, we can sum up the values of these left endpoint evaluations multiplied by the width:

Left Riemann sum = (20.21)(10) + (20.21)(10) + (21.90)(10) + (30.11)(10) + (45.69)(10) = 1090.2

Therefore, the approximation of the value of the integral ∫₀⁵⁰ R(t)dt using a left Riemann sum with five equal subintervals is 1090.2 gallons per minute.