what values must the intergral equal?

[vouslavous]

suppose that the function \(\displaystyle f\) satisfies that \(\displaystyle 0\leq f(x)\leq 2\)
for all \(\displaystyle x\) with \(\displaystyle 0\leq x \leq 2\)

Between what two values must \(\displaystyle \int_{0}^{2} f(x) dx\) lie

 

[tkhunny]

It's rather a square, isn't it?

 

[vouslavous]

the answer is two numbers, what does your post mean?

 

[lookagain]

suppose that the function \(\displaystyle f\) satisfies that \(\displaystyle 0\leq f(x)\leq 2\)
for all \(\displaystyle x\) with \(\displaystyle 0\leq x \leq 2\)

Between what two values must \(\displaystyle \int_{0}^{2} f(x) dx\) lie

The values of the function vary in the interval for f(x) belonging to \(\displaystyle [0, 2]\) for the interval of x
that is \(\displaystyle [0, 1]\), so the shape of maximum area can approach a square with side lengths of 2.
The shape of the minimum area can approach 0, by allowing for certain different shapes of
the graph of f(x).

So the area corresponding to the definite integral is:

\(\displaystyle 0 < Area < 4.\)

Then, the two values are \(\displaystyle 0 \ and \ 4.\)

 

[tkhunny]

the answer is two numbers, what does your post mean?

It means, don't let the calculus get in the way of the geometry.