Inductive load power is reactive, it is given by the
formula:
pL(t)=VL(t)IL(t),
in time domain (instant power);
PL(s)=VL,RMS(s)IL,RMS(s),
in Laplace transform domain (RMS denotes root mean square
amplitude).
VL is the voltage across the inductor L and IL is its current
(current enters in the "+" voltage reference pin, by applying user
convention in which absorbed power is positive).
Power is reactive since voltage and current are always in
quadrature:
VL(s) = s L IL(s),
in Laplace domain (derived from the time-domain formula vL(t)= L
diL(t)/dt).
A real-life inductor will also show an active power term, which
arises from parasitic resistance non-ideality; it can be modeled as
a resistance DCR in series with the inductor itself:
pACT(t)=DCR IL(t)
<<>>
An inductive load such as an induction motor draws power from
the supply with a power factor of less than 1.
Power = voltage x current x power factor.
This happens because the current reaches its peak in the ac
cycle after the voltage, so that for a small part of the cycle
power flows back into the supply from energy stored in the motor's
internal magnetic field. The time-lag is measured in degrees and
called the phase difference. 360 degrees denotes one complete
cycle.
The power factor is the cosine of the phase difference, so that
(for example) a resistive load has no phase difference so that the
power factor is 1, while for a pure inductor the phase difference
is 90 degrees and the power factor is zero.
If the rms voltage and current are expressed in complex-number
form, also known as vectors or phasors, the real power is the real
part of VI*, where the asterisk denotes the complex conjugate.
Another way to calculate the real power is to calculate the
average value of the instantaneous power V x I. If the voltage is
Vcos(wt) and the current is Icos(wt+phi) then those expressions can
be multiplied together and trigonometry formulas used to show that
the power factor is cos(phi) as stated.
Real power is measured from the average value of volts times
amps with an instrument that contains a voltage coil and a current
coil. The force produced is equal to the instantaneous power, and
the instrument measures its average value muliplied by the
time.

1. Junk Mail - By using direct mail, a company risks alienating
the customers it seeks to gain. The use of direct mail is often
associated with a company of lower status and quality.
2. Environmental Impact - Direct marketing typically involves
the heavy use of physical components in order to give consumers an
object they can hold in their hands. This physical side of direct
marketing means that a greater amount of natural resources are
consumed in their production as well as a greater amount of
pollution.
3. Metrics - It is hard when using direct marketing to get any
reliable metrics on its impact. Direct marketing involves a much
longer-term strategy where the results are uncertain and the best
that can be said is that it may be working. It is often very
unclear whether direct marketing is alienating more customers than
it is creating.
4. Time - It is hard to get as immediate of an impact when using
direct marketing, as the advertising product, by its nature, takes
time to reach consumers. Time is often equated with money in
business, and the time ratio in direct marketing is a
disadvantage.
5. Laws - The distribution of fliers and street advertising is
often prohibited by law in certain areas. This is a major
disadvantage of this form of direct marketing as certain customer
bases will be unreachable. If a marketer wishes to reach customers,
for instance, in a high-traffic and wealthy area, this may be
impossible using direct marketing of this kind.
6. Limited Reach - There is an inherently limited geographic
reach to direct marketing involving fliers and street advertising.
By using this form of direct marketing, you will be limited to
consumers in a very specific distribution area and will have to
depend on word of mouth for others.

In order to work out this problem, we need to learn how to apply
the integration method correctly.
The given expression is ∫ 2xln(2x) dx.
Instead of working out with 2x's, we let u = 2x. Then, du = 2 dx
or du/2 = dx. This method is both valid and easy to avoid working
out with too much expressions. You should get:
∫ uln(u) (du/2)
= ½ ∫ uln(u) du
Use integration by parts, which states that:
∫ f(dg) = fg - ∫ g(df)
We let:
f = ln(u). Then, df = 1/u du
dg = u du. Then, g = ∫ u du = ½u²
Using these substitutions, we now have:
½(½u²ln(u) - ½∫ u du)
= ¼(u²ln(u) - ∫ u du)
Finally, by integration, we obtain:
¼ * (u²ln(u) - ½u²) + c
= 1/8 * (2u²ln(u) - u²) + c
= 1/8 * (2(2x)²ln(u) - (2x)²) + c
= 1/8 * (2x)² * (2ln(u) - 1) + c
= ½ * x² * (2ln(2x - 1)) + c

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