x = Integral
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.