#### To determine

**To estimate:**

The energy used onDecember 9, 2004.

#### Answer

475,200 Megawatt-hours

#### Explanation

**1) Concept:**

Use the Net Change theorem and the Midpoint Rule.

**2) Theorem and Rule:**

The Net Change theorem: The integral of a rate of change is the net change.

∫abF'(x)dx=Fb-F(a)

The Midpoint Rule:

∫abfxdx≈∑i=1nfxi- ∆x=∆x fx1-+…+fxn-

where ∆x=b-an and xi-=12xi-1+xi midpoint of xi-1, xi.

**3) Calculation:**

Power is the rate of change of energy with respect to time, that is,

Pt=E't

Here, t∈0,24

By the Net Change theorem,

∫08Pt dt≈∫08E'tdt=E8-E(0) is energy used on December 9, 2004 at the time interval.

Approximate the value of the integral using the Midpoint Rule with n=12, a=0, and b=24. Then the width of interval is

∆t=b-an

Substitute values.

∆t=24-012

∆t=2

From the graph, the three subintervals are 0,2, 2,4, 4,6,…, 20,22, 23,24 and the midpoints are

t1-=12t0+t2=120+2=122=1

t2-=12t2+t4=122+4=126=3

t3-=12t4+t6=124+6=1210=5

.

.

.

t11-=12t20+t22=1220+22=1242=21

t12-=12t22+t24=1222+24=1246=23

So, the value of the integral is

∫024Pt dt≈M6=∑i=112Pti- ∆t

=∆tPt1-+Pt2-+Pt3-+…+Pt11-+Pt12-

=∆tP1+P3+P5+…+P21+P23

From the given graph,

P1=16,900

P3=16,400

P5=17,000

.

.

.

P21=21,700

P23=18,900

=∑i=112Pti- ∆t≈216,900+16,400+17,000+19,800+20,700+21,200+20,500+20,500+21,700+22,300+21,700+18,900

=2237,600=475,200

**Conclusion:**

The energy used on that day was approximately 475,200 megawatt-hours.