find I= `int_0^2((5-x)/(9-x^2))dx`

Explanation

Let `f(x)=(5-x)/((9-x^(2)))`
`:.f'(x)=-((x-9)(x-1))/((9-x^(2))^(2))`
For `f'(x)=0`, we have `x=1` as `xepsilon[0,2]`
Now `f(0)=5//9, f(1)=1//2, f(2)=3//5`
Therefore, the greatest and the least values of the integrand in the interval `[0,2]` are, respectively, equal to `f(2)=3//5` and `f(1)=1//2`
Hence `(2-0)1/2 lt int_(0)^(2)((5-x)/(9-x^(2)))dx lt (2-0)3/5`,
or `1ltint_(0)^(2)((5-x)/(9-x^(2)))dx lt6/5`.