Integrals of Polynomials

Basic Formula

We shall now prove that polynomials can be integrated using the following Basic Formula: If v(s)=s^p, then with x(0)=0, then we have

  • x(t)= \frac{t^{p+1}}{p+1} = \int_0^ts^p ds,\quad p = 0,1,...

To prove this, we note that for p=0 we have

  • t= \int_0^t 1 ds,

which is the same as

  • \frac{d}{dt}t = \frac{dt}{dt} = 1.

To see this we note that if x(t)=t, then dx(t)=x(t+dt)-x(t)=t+dt -t=dt and thus \frac{dx}{dt}=1.

For p=1 we are supposed to prove that

  • \frac{t^{2}}{2} = \int_0^ts ds,

which is the same as \frac{d}{dt}t^2 = 2t. To see this we note that if x(t)=t^2, then

  • dx(t)=x(t+dt)-x(t)=(t+dt)(t+dt) -t^2=t^2+2tdt+dtdt-t^2=2tdt-dtdt.

If now dt is small then we can argue that dtdt is so small that it can be neglected, and thus \frac{d}{dt}t^2=2t.

For p=2 we are supposed to prove that

  • \frac{t^{3}}{3} = \int_0^ts^2 ds,

which is the same as

  • \frac{d}{dt}t^3 = 3t^2.

To see this, we note that if $x(t)=t^3$, then

  • dx(t)=x(t+dt)-x(t)=(t+dt)(t+dt)(t+dt)-t^2
  • =t^3+3t^2dt+3tdtdt+dtdtdt-t^3=3t^2dt+3tdtdt +dtdtdt.

If now dt is small then we can argue that 3tdtdt and dtdtdt are so small that they can be neglected, and thus \frac{d}{dt}t^3=3t^2. Similarly, one can show the Basic Formula:

  • \frac{d}{dt}t^p=pt^{p-1} \quad\mbox{for }p=1,2,3...

In MTS we show that his formula also holds for negative exponents

  • \frac{d}{dt}t^p=pt^{p-1}\quad\mbox{for }t>0, \quad p=-1,-2,-3,....

The Logarithm

We shall also discover that the case p=0 is special, and gives rise to the logarithm \log(t)
as the solution of

  • \dot u(t)=t^{-1}=\frac{1}{t}, \quad\mbox{for }t>0,\quad u(1)=0,

that is

  • \log(t)=\int_1^t\frac{1}{s}\, ds, \quad\mbox{for }t>0.

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