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.