Video Transcript
Suppose that on the closed interval
negative two to five, the values of 𝑓 lie in the closed interval lowercase 𝑚 to
capital 𝑀. Between which bounds does the
definite integral between negative two and five of 𝑓 of 𝑥 with respect to 𝑥
lie?
We recall one of the comparison
properties of integrals. It tells us that if 𝑓 of 𝑥 is
greater than or equal to lowercase 𝑚 and less than or equal to capital 𝑀 for
values of 𝑥 greater than or equal to 𝑎 and less than or equal to 𝑏. Then 𝑀 times 𝑏 minus 𝑎 is less
than or equal to the definite integral between 𝑎 and 𝑏 of 𝑓 of 𝑥 with respect to
𝑥, which in turn is less than or equal to capital 𝑀 times 𝑏 minus 𝑎. In other words, given that lower
case 𝑚 is the absolute minimum of 𝑓 and uppercase 𝑀 is the absolute maximum of
𝑓, the area under the graph of 𝑓 is greater than the area of the rectangle with a
height lower case 𝑚. But it’s less than the area of the
rectangle with a height uppercase 𝑀.
In this example, we’re going to let
𝑎 be equal to negative two and 𝑏 be equal to five. Then, we see that 𝑀 times five
minus negative two is less than or equal to the definite integral between negative
two and five of 𝑓 of 𝑥 with respect to 𝑥 which, in turn, is less than or equal to
capital 𝑀 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 𝑚 and less than or equal to seven
uppercase 𝑀.
