Some integrals cannot be calculated exactly. A good example isThe integral exists and is equal to the area A below.

In examples like this it is useful to be able to calculate lower and uppoer bounds for the value of an integral. We can find a lower bound by dividing the interval of integration on theaxis into sunbintervalsand finding the minimum value of the function,on this interval. Form rectangles with base of lengthand heightand area and add these to get a lower bound for the value of the integral,

To find an upper bound find the maximum value of the function, F-i on each subintervalForm rectangles of baseand height F-i and areaand add these to get a upper bound for the value of the integral,

To illstrate, find upper and lower bounds foronby dividing the intervalinto 5 subintervals

A lower bound is found from the diagram below:

An upper bound for the integral is found from the diagram below.

Compare these bound with the actual value