**ln** x =

Integral from

**1** to

**x** 1 / t

**dt** where x must be a

positive number.
Suppose a

definite integral has a variable for its

upper limit of integration, for instance

**∫ ** = integral

**∫**(1,x) (**sin** t) dt

Evaluation the integral gives

-**cos** t with the limits of 1 to x = -**cos** x + **cos** 1 =
-**cos** x + 0.5403....

The answer is an expression involving the upper limit of integration, x. Thus the integral is a function of x. The integral equals the area of the region under the graph of y = **sin** t from t = 1 to t = x. Clearly the area is a function of the value picked for x.