The upper sum of the integral

`int_a^b f(t)dt`

is

`U=sum_{i=0}^{n-1} f(x^+) Delta x` where `Delta x = {b-a}/n` , `f(x^+)=max(f)` on the interval `[x_i,x_{i+1}]` and `x_i=a+i Delta x`

The lower sum is

`L=sum_{i=0}^{n-1} f(x^-) Delta x` where `f(x^-)=min(f)` on the same interval

Since `f(x)=2+sin x` , and the bounds on the integral are `a=0` and `b=pi` ,...

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The upper sum of the integral

`int_a^b f(t)dt`

is

`U=sum_{i=0}^{n-1} f(x^+) Delta x` where `Delta x = {b-a}/n` , `f(x^+)=max(f)` on the interval `[x_i,x_{i+1}]` and `x_i=a+i Delta x`

The lower sum is

`L=sum_{i=0}^{n-1} f(x^-) Delta x` where `f(x^-)=min(f)` on the same interval

Since `f(x)=2+sin x` , and the bounds on the integral are `a=0` and `b=pi` , then

`Delta x=pi/n`

`x_i={i pi}/n`

For n=4, this means we have:

`U_4=(f(pi/4)+f(pi/2)+f(pi/2)+f({3pi}/4))pi/4`

`=pi/4(8+2sqrt2/2+1)`

`=pi/4(9+sqrt2)`

`approx 8.18`

`L_4=(f(0)+f(pi/4)+f({3pi}/4)+f(pi))pi/4`

`=pi/4(4+2sqrt2/2)`

`=pi/4(4+sqrt2)`

`approx 3.39`

Noticing the symmetry of the function, we see that we really only need to double the sum up to `x=pi/2` .

When n=8, we should have more refined values and get after the doubling:

`U_8=pi/4(f(pi/8)+f(pi/4)+f({3pi}/8)+f(pi/2))`

`approx 8.65`

Similarly,

`L_8=pi/4(f(0)+f(pi/8)+f(pi/4)+f({3pi}/8))`

`approx 6.29`

**The upper and lower bounds for n=4 are 8.18, 3.39. The upper and lower bounds for n=8 are 8.65, 6.29.**