Video: Subtracting Two Numbers up to 9

Suppose that on [βˆ’2, 5], the values of 𝑓 lie in the interval [π‘š, 𝑀]. Between which bounds does ∫_(βˆ’2) ^(5) 𝑓(π‘₯) dπ‘₯ lie?

01:29

Video Transcript

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

We recall one of the comparison properties of integrals. It tells us that if 𝑓 of π‘₯ is greater than or equal to lowercase π‘š and less than or equal to capital 𝑀 for values of π‘₯ greater than or equal to π‘Ž and less than or equal to 𝑏. Then 𝑀 times 𝑏 minus π‘Ž is less than or equal to the definite integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯, which in turn is less than or equal to capital 𝑀 times 𝑏 minus π‘Ž. In other words, given that lower case π‘š 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 π‘š. But it’s less than the area of the rectangle with a height uppercase 𝑀.

In this example, we’re going to let π‘Ž 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 π‘₯ with respect to π‘₯ 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 π‘š and less than or equal to seven uppercase 𝑀.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.