Definite integrals and the fundamental theorem

Accumulation over an interval

A definite integral measures accumulated change over an interval. It is written

$int_a^b f(x),dx$

This represents the signed area between the graph of $f(x)$ and the x-axis from $x=a$ to $x=b$.

In physics, area under a graph often has direct meaning. Area under a velocity-time graph gives displacement. Area under a force-displacement graph gives work. Area under a current-time graph gives charge.

Area from thin slices

The integral can be understood as the limit of adding many thin rectangles. Each rectangle has height $f(x)$ and small width $Delta x$. The approximate area is

$sum f(x_i)Delta x$

As the rectangles become thinner and more numerous, the sum approaches the definite integral:

$int_a^b f(x),dx = lim_{Delta x o 0}sum f(x_i)Delta x$

This is why integration is often described as continuous addition.

Signed area

If $f(x)$ is above the x-axis, the integral contributes positively. If it is below the x-axis, it contributes negatively. This signed nature matters in physics.

For velocity, positive area represents displacement in the positive direction, while negative area represents displacement in the negative direction. Total distance traveled may require integrating speed, not velocity.

The fundamental theorem of calculus

The fundamental theorem of calculus connects derivatives and integrals. If $F'(x)=f(x)$, then

$int_a^b f(x),dx = F(b)-F(a)$

This means we can evaluate a definite integral by finding an antiderivative and subtracting its values at the endpoints.

For example,

$int_0^3 2x,dx$

An antiderivative of $2x$ is $x^2$, so

$int_0^3 2x,dx = 3^2 - 0^2 = 9$

Displacement from velocity

If velocity is $v(t)$, displacement from $t=a$ to $t=b$ is

$Delta x = int_a^b v(t),dt$

If velocity is sometimes negative, the integral accounts for direction.

For example, if a velocity-time graph has positive area of $10 m$ and negative area of $3 m$, the displacement is $7 m$. The distance traveled is $13 m$ if the object actually moved both ways.

Work from force

If force varies with position, work is

$W = int_{x_i}^{x_f} F(x),dx$

For a spring force magnitude that grows with stretch, area under the force-displacement graph gives the energy stored or work done.

Average value

A definite integral can also find the average value of a function over an interval:

f_{avg}=rac{1}{b-a}int_a^b f(x),dx

In physics, this may represent average velocity, average force, average power, or average density over a region.

The big idea

Definite integrals measure accumulated quantity. They are continuous sums and signed areas. The fundamental theorem of calculus makes them practical by connecting accumulation to antiderivatives. In physics, definite integrals turn rates into total changes and variable effects into accumulated results.