# Integral vs Derivative

• $x(t)=\int_0^tv(s)ds$ solves $\dot x(t) = v(t)$, $x(0)=0$.
• Without mathematics we cannot penetrate deeply into philosophy. Without philosophy we cannot penetrate deeply into mathematics. Without both we cannot penetrate deeply into anything. (Leibniz)

# The Most Basic IVP

The solution of the Initial Value Problem IVP of finding $x:[0,T]\rightarrow R$ such that

• $\dot x(t) = v(t)\quad\mbox{for }0< t\le T,\quad x(0)=0$,   (IVP)

where $v:[0,T]\rightarrow R$ is a given function and $[0,T]$ a given time-interval, is denoted by

• $x(t) = \int_0^tv(s)\, ds,\quad t\in [0,T]$,

and is referred to as the integral or primitive function of $v(t)$. So far the integral $\int_0^tv(s)\, ds$ is just a sign or name of the solution $x(t)$, and it remains to give it a concrete meaning.

We shall see that the S-like integral sign $\int$ can be viewed as indicating a certain form of Summation, which we shall make precise. The integral sign $\int$ was the strike of genius of Leibniz, long before logotypes became the carriers of the inner meaning of companies and organizations.

Changing the names of the variables, we can write

• $x(T) = \int_0^Tv(t) dt$

with $T>0$ representing any given final time for which the value $x(T)$ is desired. We can use the variable $t$ as final time or as integration variable:

• With $t$ as final time, $s$ is the integration variable ranging from $0$ to $t$.
• With $T$ as final time, $t$ is the integration variable ranging from $0$ to $T$.

The Forward Euler method for (IVP), is given by

• $x((n+1)dt)=x(ndt)+v(ndt)dt\quad\mbox{for } n=0,1,2,...,N,\mbox{ with }\, (N+1)dt=T$,

or equally well

• $x((n+1)ds)=x(nds)+v(nds)ds\quad\mbox{for } n=0,1,2,...,N,\mbox{ with }\, (N+1)ds=T$,

with $dt=ds$ the time step.

If we replace $x(nds)$ by $x((n-1)ds)+v((n-1)ds)ds$, and so on, we see that $x((n+1)ds)$ can be expressed as as a sum

• $x((n+1)ds)=\sum_{m=0}^nv(mds)ds =v(0)ds+v(ds)ds+v(2ds)ds+...+v(nds)ds$.

We are thus led to view

• $\int_0^tv(s) ds\quad\mbox{and }\, \sum_{m=0}^nv(mds)\, ds$,

to be similar, which we shall make precise below. We refer to the sum representation of the integral as a Riemann sum. Let us collect our experience with the integral so far:

#### Observation 1:

The integral $x(t)=\int_0^tv(s) ds$ satisfies by definition

• $\dot x(t)=\frac{d}{dt}\int_0^tv(s) ds = v(t)\quad\mbox{for }0.

The integral $x(t)=\int_0^tv(s) ds$ represents a Riemann sum $\sum_{m=0}^nv(mds)\, ds$ with $(n+1)dt =t$.

#### Observation 2:

The solution of (IVP), with possibly non-zero initial value $x^0$, of finding $x:[0,T]\rightarrow R$ such that

• $\dot x(t) = v(t)\quad\mbox{for }0,

is given by

• $x(t)=x^0+\int_0^tv(s) ds$.

This is because the derivative of a constant function (the function $w(t)=x^0$), is zero ($\dot w=0$).

#### Observation 3:

Since $\dot x(t)=v(t)$ and $\dot v(t)=a(t)$ with $x(t)$ distance, $v(t)$ velocity and $a(t)$ acceleration, we can say that

• distance is the integral of velocity,
• velocity is the integral of acceleration.

#### Observation 4:

Sum up in words:

• integration followed by differentiation = doing nothing = identity operation
• differentiation followed by integration = doing nothing = identity operation.

In other words, as explained in more detail in the next chapter:

• integration is the inverse of differentiation
• differentiation is the inverse of integration.

# Interpreting the Integral as an Area

The area $A(v,t)$ bounded by the graph of the function $v:[0,t]\rightarrow R$ and the $s$-axis of a $(v,s)$-coordinate system ($t$-axis in the figure below), can viewed as a sum of rectangular strips of height $v(mdt)$ and width $dt$ (assuming for definiteness that $v(mdt)\ge 0$), and thus

• $A(v,t)=\sum_{m=0}^nv(mds)ds,\quad t=(n+1)ds$.

We are thus led to interprete the integral as an area:

• $\int_0^tv(s)ds=A(v,t)=\,\mbox{area under the graph of}\, v(s)\, \mbox{ on the interval}\, [0,t]$

as illustrated in

Integral as Riemann sum as area under graph.

# The Trapezoidal Rule

Replacing the shaded rectangle area in the above figure with the area of a trapezoid with right vertical of length $v((m+1)dt$ as illustrated in we obtain the alternative Riemman sum approximation

• $\int_0^tv(s) ds\approx \sum_{m=0}^{n-1}\frac{v(mds)+v((m+1)ds)}{2}ds$
• $=\frac{v(0)}{2}ds+\sum_{m=1}^{n-1}v(mds)ds+\frac{v(t)}{2}ds$ ,

which is the Trapezoidal Rule. We compare with a Midpoint Euler method defining the height of therectangle to be the function value at the midpoint of the interval. This value is close to the mean-value of the endpointvalues used in the Trapezoidal Method, which thus is close to Midpoint Euler.

# Not All Integrals are Areas

Note that distance is the integral of velocity but it is not very natural to say that distance is the area under the velocity graph.

Summing up: The integral is defined as the solution to an IVP. Some integrals can be interpreted as areas, but all integrals are not areas. Some cars (integrals) are Volvos (areas) but all cars (integrals) are not Volvos (areas).

There are also Saabs…or were..

Nevertheless, many Calculus books introduce the integral as the area under a graph, based on the pedagocialidea to define a new concept (the integral) in terms of something supposedly more familiar (area), but this is questionable from mathematical point of view and also confusing, when students discover that all integrals are not areas.

To say that an integral is solution to an IVP, is not questionable, because this is what an integral is.

Piecewise linear approximation of the Trapezoidal Rule vs piecewise constant approximation of Euler Midpoint. Piecewise linear approximation is a basic element of computational mathematics including the finite elemeent method, as you will discover below…Simple and profound…

IVP of Usain Bolt