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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 😊 ...