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Let f be a continuous function on the closed
interval [0, 2] . If 2 (less than ir equal to) f(x) (less than or equal to) 4, then the greatest
possible value of (integral from 0 to 2)f(x)dx is

a. 0
b. 2
c. 4
d. 8
e. 16

Let f be a continuous function on the closed interval [0, 2] . If 2 (less than ir equal to) f(x) (less than or equal to) 4, then the greatest possible value of class=

Answer :

xKelvin

Answer:

D) 8

Step-by-step explanation:

We are given a continuous function f on the closed interval [0, 2].

Where:

[tex]2\leq f(x)\leq 4[/tex]

And we want to find the greatest possible value of:

[tex]\displaystyle \int_0^2f(x)\,d x[/tex]

The range restriction tells us that even if f(x) = 2 for all x in the interval [0, 2], the smallest area possible will be 4, since that is the area of the rectangle.

Then in that case, the maximum possible value of the integral must be 8. f(x) cannot exceed 4, and the length of the interval is two units. Thus, the greatest possible value of the integral is 8. This will only occur if f(x) is a horizontal line at y = 4 from x = 0 to x = 2.

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