Lower and Upper Riemann sums of sin(x)

[paulca]

Task in real analysis:
P is a uniform partition on [0, π] and is divided into 6 equal subintervals. Show that the lower and upper riemann sums of sin (x) over P is lesser than 1.5 and larger than 2.4 respectively.

My attempt at the solution:
The greates value and the least value of sin x over an subinterval (x_i- x_i-1) is 1 and -1. The upper and lower riemann sums is then:
upper sum: S(P) = ∑_i=1⁶ (x_i- x_i-1) = x₆ - x₀ = π
Lower sum: s(P) = - ∑_i=1⁶ (x_i- x_i-1) = -π

From this one could say that S(P) > 2.4 and s(P) < 1.5, but i don't feel like this is a full answer to the problem and i don't see another approach to solving the problem, so if anyone could give me some clue or tips it would be much appreciated.

 

[pasmith]

Those are not the correct values for the upper and lower sums.

Let $M_i = \sup\{ f(x) : x_{i-1} \leq x \leq x_i\}$ and $m_i = \inf \{ f(x) : x_{i-1} \leq x \leq x_i \}$. Then $$S(P) = \sum_{i=1}^6 M_i(x_i - x_{i-1}), \\<br /> s(P) = \sum_{i=1}^6 m_i(x_i - x_{i-1}).$$ What are the maximum and minimum values of $\sin x$ when $\frac{(i-1)\pi}6 = x_{i-1} \leq x \leq \frac{i\pi}6 = x_i$? They aren't all 1 and -1 respectively (aside from anything else, $\sin(x) \geq 0$ for $x \in [0, \pi]$).

 

[paulca]

Thank you for replying. I saw that myself a while ago and found what the maximum and minimum value of sin x over each of the 6 subintervals such that the upper sum sin x over partition P should be:
S(P) = ∑_i=1⁶ Δx ⋅max(sinx)_i = (π/6) ⋅(1/2 + √3/2 + 1 + 1 + √3/2 + 1/2) = (π/6) ⋅ (3 + √3) ≈ 2,47 > 2,4
and similar for the lower sum:
s(P) = (π/6) ⋅ (1 + √3) ≈ 1,43 < 1,5