Integral Calculus

Integration is, at its heart, an area-finding process, and is the reverse operation of differentiation. Some shapes have areas we can write down directly — a rectangle of height $c$ between $x=a$ and $x=b$ simply has area $(b-a)c$ — but an irregularly shaped region under a curve $f(x)$ needs a different approach.

The trick: chop the region under $f(x)$, from $x=a$ to $x=b$, into many thin vertical strips of width $\Delta x$. Each strip is almost a rectangle, of height roughly $f(x_i)$ at that strip's location, so the total area is approximately the sum of the strip areas:

More compactly, dividing the interval into $N$ strips, $A \approx \sum_{n=1}^{N} f(x_n)\Delta x$. As the number of strips $N\to\infty$ (equivalently $\Delta x\to 0$), this sum settles down to an exact value called the definite integral:

The integral $\int_a^b f(x)\,dx$ is exactly the total area under the curve $f(x)$ between $x=a$ and $x=b$.

Two physics examples that use exactly this idea:

1. Work done by a variable force $F(x)$, moving an object in one dimension from $x=a$ to $x=b$: $W = \int_a^b F(x)\,dx$ — the area under the force-vs-position graph. (No dot product is needed here since the motion is already one-dimensional.)
2. Impulse delivered by a force over an interval $t=0$ to $t=t_1$: $I = \int_0^{t_1} F\,dt$ — the area under the force-vs-time graph.

Average velocity, in vector form. Suppose a particle is at point $P$ (position vector $\vec r_1$) at one instant and, after a time interval $\Delta t$, is at point $Q$ (position vector $\vec r_2$). Its displacement is $\Delta\vec r=\vec r_2-\vec r_1$, and its average velocity is

Average velocity is a vector, pointing along the straight-line displacement $\Delta\vec r$ from $P$ to $Q$ (i.e. along the chord, not along whatever curved path was actually followed).

Instantaneous velocity is the limiting value of the average velocity as $\Delta t\to0$, i.e. the rate of change of the position vector with respect to time:

In component form, since $\vec r = x\hat i+y\hat j+z\hat k$,

The magnitude of $\vec v$ is called speed: $v=\sqrt{v_x^2+v_y^2+v_z^2}$, always a positive scalar, with SI unit metre per second (m s$^{-1}$), same as velocity's unit.

Average speed is defined differently from average velocity — it is the total path length travelled divided by the total time taken:

Because path length (distance) can exceed the magnitude of displacement, average speed is always $\geq$ the magnitude of average velocity over the same interval — the two coincide only for straight-line motion in one direction without reversal. …