Let f be a continuous function on the closed interval [0, 2]. If 2 ≤ f(x) ≤ 4, then the greatest possible value of : ∫(0 to 2) f(x)dx is:

Solution:

Given,

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

Where:

2 ≤ f(x) ≤ 4

And we have to find the greatest possible value of:

$\int_{0}^{2}$ f(x) dx

Definition:

A function f(x) is said to be continuous at a point if the following conditions are met.

The function at that point exists as being finite.

The left and right-hand limit of the function is present.

The limit Lim X→a f(x) = f(a) where is the point.

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.

Therefore, the greatest possible value of the integral is 8.

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Let f be a continuous function on the closed interval [0, 2]. If 2 ≤ f(x) ≤ 4, then the greatest possible value of : ∫(0 to 2) f(x)dx is:

Summary:

Let f be a continuous function on the closed interval [0, 2]. If 2 ≤ f(x) ≤ 4, then the greatest possible value of : ∫(0 to 2) f(x)dx is: 8.