### Video Transcript

Suppose that on the closed interval
negative two to five, the values of π lie in the closed interval lowercase π to
capital π. Between which bounds does the
definite integral between negative two and five of π of π₯ with respect to π₯
lie?

We recall one of the comparison
properties of integrals. It tells us that if π of π₯ is
greater than or equal to lowercase π and less than or equal to capital π for
values of π₯ greater than or equal to π and less than or equal to π. Then π times π minus π is less
than or equal to the definite integral between π and π of π of π₯ with respect to
π₯, which in turn is less than or equal to capital π times π minus π. In other words, given that lower
case π is the absolute minimum of π and uppercase π is the absolute maximum of
π, the area under the graph of π is greater than the area of the rectangle with a
height lower case π. But itβs less than the area of the
rectangle with a height uppercase π.

In this example, weβre going to let
π be equal to negative two and π be equal to five. Then, we see that π times five
minus negative two is less than or equal to the definite integral between negative
two and five of π of π₯ with respect to π₯ which, in turn, is less than or equal to
capital π times five minus negative two. Five minus negative two is
seven. And so, we see that our definite
integral must be greater than or equal to seven π and less than or equal to seven
uppercase π.