# Question Video: Subtracting Two Numbers up to 9 Mathematics • Higher Education

Suppose that on [β2, 5], the values of π lie in the interval [π, π]. Between which bounds does β«_(β2) ^(5) π(π₯) dπ₯ lie?

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### 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 itβs less than the area of the rectangle with a height uppercase π.

In this example, weβre 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.