Question 1. Power companies typically bill customers based on the number of kilowatt-hours used during a single billing period. A kilowatt is a measure of how much power (energy) a customer is using, while a kilowatt-hour is one kilowatt of power being used for one hour For constant power use, the number of kilowatt-hours used is calculated by kilowatt-hours 1. Power companies typically bill customers based on the number of kilowatt-hours used during a single billing period. A kilowatt is a measure of how much power (energy) a customer is using, while a kilowatt-hour is one kilowatt of power being used for one hour For constant power use, the number of kilowatt-hours used is calculated by kilowatt-hours kilowatts time (in hours). Thus, if customers use 5 kilowatts for 30 minutes, they'll have used 5 kilowatts hours 2.5 kilowatt-hours Suppose the power use of a customer over a 30-day period is given by the continuous function P =/(1), where P is kilowatts, 1 is time in hours, and 1-0 corresponds to the beginning of the 30 day period A. Approximate, with a Riemann sum, the total number of kilowatt-hours used by the customer in the 30-day period, and explain why your Riemann sum is an approximation of the desired property B. Derive an expression representing the total number of kilowatt-hours used by the customer in the 30-day period, and explain your reasoning. (This expression should not be an approximation.) C. Consider the following table of data for the function/() 0 2.3 1 2.5 2 2.1 3 3.9 4 3.6 5 5.5 6 4.5 7 5.6 9 1.0 10 1.8 Recall that ft) represents the number of kilowatts being used by a customer at time t hours from the beginning of the billing period. Estimate the number of kilowatt-hours the customer uses in this 10-hour period, and explain your method

Anscier(1)1.(A)Total number of kilowatt-hours used by customer\begin{array}{l}=30 \text { days } \\=30 \times 24 \\=780 \text { howrs. }\end{array}Expression for Riemann sum then the calculation rule is the tolal number of kilo-watt hours used by the castomer is\sigma=\int_{t=1}^{t=720} f(t) d tB) total number of kilowatt-hours used by customer in\begin{aligned}\nabla & =30 \text { days. } \\& =30 \times 24=720 \text { howrs } .\end{aligned}(2)Accoiding to the left Riemann sum, we get.P(0)+P(1)+P(2)+P(3)+\cdots+P(789)= Total kilo-watt howr used by the castomer for sodays.c_{1}Given\begin{array}{ll}t & f(t) \\ 0 & 2.3 \\ 1 & 2.5 \\ 2 & 2.1 \\ 3 & 3.9 \\ 4 & 3.6 \\ 5 & 5.5 \\ 6 & 4.5 \\ 7 & 5.6 \\ 8 & 1.2 \\ 9 & 1.0 \\ 10 & 1.8\end{array}(3)number of lilowatt - hows the customer uses in 10-howr\text { period o } p(t)=2.3+2.5+2.1+3.9+3.6+5.5+4.5+56+1.2+11=32.2 \text { kilo-watt howr. }PART Awe know that according to the standard of the time 1 day =24 hours so from that aspect we can say that30 days =30 \star 24=720 hoursnow if we follow the expression for the Riemann Sum then the calculation should follow the following rulesay the total number of Kilo-watt hours used by the customer can be represented with the symbol \sigmafollowing the Riemann Sum we can write that\sigma=\int_{t=1}^{t=720} f(t) d twe have used the above formula for the reason that we have considered the time interval has remain constant through outPART Bwe know that according to the standard of the time 1 day =24 hours so from that aspect we can say that 30 days =30 \star 24=720 hoursso, the total kilo-watt hour can be calculated following the left Riemann sum for the above purposenow according to the left Riemann Sum we get the value\mathrm{P}(0)+\mathrm{P}(1)+\mathrm{P}(2)+\mathrm{P}(3)++\mathrm{P}(719)= Total Kilo-watt hour used by the customer for the 30 days periodPART CAccording to the given data\begin{array}{l}t_{0}=0, t_{1}=1, t_{2}=2, t_{3}=3, t_{4}=4, t_{5}=5, t_{6}=6, t_{7}=7, t_{8}=8, t_{9}=9, \\t_{10}=10\end{array}so the difference between each time interval or rather we can say the time period through which the customer has used a par ticular amount of power constantly is nothing but 1 hour so the total number of Kilo-watt hour used by the customer in the 10 hours period can only be as followingP(t)=2.3+2.5+2.1+3.9+3.6+5.5+4.5+5.6+1.2+1=32.2kilo-watt hourplease like it 😊 ...