phy1020.DVI

Figure 4.1: Finding the area under a curve using rectangles (Credit: pleacher.com)

To do this, imagine dividing the area under the curve into a number of very thin rectangles (Figure 4.1).
The thinner the rectangles, the more rectangles we have, and the better the approximation to the actual area
under the curve.
If we go to the limit where the rectangles are infinitesimally narrow, then we will have infinitely many
of them, and the sum of the areas of all the rectangles exactly equals the area under the curve. Adding up
an infinite number of infinitesimal numbers is calledintegration, and typically results in a finite result. If
we have a curvef.x/, then a rectangle atxhas infinitesimal widthdxand finite heightf.x/, so that that
rectangle has area equal to its width times its height, orf.x/dx. We add together an infinite number of them
by integration; the symbol for which is an elongatedS(for “sum”),

R

:

Z

f.x/dx (4.36)

This expression is called anintegral, and the functionf.x/is called theintegrandof the integral. The area
under the curve clearly depends on where the left and right ends of the area are. The area under the curve
f.x/betweenxDaandxDbin indicated by

Zb

a

f.x/dx (4.37)

Equation (4.36) is called anindefinite integral, and Equation (4.37) is called adefinite integral. To compute
a definite integral, we evaluate theindefiniteintegral at the upper boundb, and subtract the indefinite integral
evaluated at the lower bounda:
Zb

a

f.x/dxD

Z

f.x/dx.atxDb/ 

Z

f.x/dx.atxDa/ (4.38)

For example, suppose we want to find the area under the parabolaf.x/Dx^2 betweenxD 1 andxD 3.
This would be

areaD

Z 3

1

x^2 dxD



x^3

3

ˇˇ

ˇ

ˇ

ˇ

3

1

D

33

3



13

3

D

26

3

square units (4.39)

The vertical bar is used to indicate that we evaluate the expression at the top value (3), then subtract the
expression evaluated at the bottom value (1).