#### To determine

**To estimate:**

The total amount of data transmitted during time period midnight to 8:00 A.M.

#### Answer

15,912 Megabits

#### Explanation

**1) Concept:**

Use the Net change theorem and 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:**

Let Mt denote the number of megabits transmitted at the time t(in hours).

Dt measured in megabits/second is the rate of change of number of megabits transmitted at time t.

Dt=M'(t)

Here,t∈0,8

By the Net Change theorem,

∫08Dt dt≈∫08M'tdt=M8-M(0) is number of megabits transmitted during the time interval.

Approximate the value of the integral using the Midpoint Rule with n=4, a=0, and b=8.Then the width of the interval is

∆t=b-an

Substitute values to get

∆t=8-04

∆t=2

From the graph, the four subintervals are 0,2, 2,4, 4,6, 6,8 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

t4-=12t6+t8=126+8=1214=7

So, the value of integral is

∫08Dt dt≈M6=∑i=14Dti- ∆t

=∆tDt1-+Dt2-+Dt3-+Dt4-

=∆tD1+D3+D5+D7

From the given graph,

D1=0.32

D3=0.50

D5=0.56

D7=0.83

=∑i=14Dti- ∆t≈72000.32+0.50+0.56+0.83=15,912

=15,912 megabits

Notice that we have multiplied with a factor of 3600 to convert hours into seconds.

**Conclusion:**

The total amount of data transmitted during time period from midnight to 8:00 A.M. is 15,912 megabits.