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Mathematics 18 Online
OpenStudy (anonymous):

let f(x) be a continuous function on the closed interval [1,4] if 5<=f(x)<=9 on this interval, then the value of definite integral from 4 to 1 f(x)dx cannot be: 12? 18? 27? 15? 21?

OpenStudy (anonymous):

cannot be 12

OpenStudy (anonymous):

why?

OpenStudy (anonymous):

lowest is 15, highest is 27 inclusive

OpenStudy (anonymous):

from 1 to 4 equates to a base of 3

OpenStudy (anonymous):

3*5(lowest number) = 15 3*9(highest number) = 27

OpenStudy (anonymous):

how does 1 to 4 equate to a base of 3?

OpenStudy (anonymous):

4-1

OpenStudy (anonymous):

what does it mean to find the base for the integral?

OpenStudy (anonymous):

integral = area under curve = base * average height

OpenStudy (anonymous):

base as in length against the x-axis

OpenStudy (anonymous):

ohh that makes more sense now!

OpenStudy (zarkon):

\[5\le f(x)\le 9\] so \[\int\limits_{1}^{4}5dx\le \int\limits_{1}^{4}f(x)dx\le \int\limits_{1}^{4}9dx\] \[15\le \int\limits_{1}^{4}f(x)dx\le 27\]

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