Question

Let f: [0,1] to R ne a function defined by

f(x)= 1-x^2

Let P1= { 0,1/2,2/3,1}

P2= { 0,1/4,1/2,3/4,1}

be two partition of the interval [0,1]. Calculate L(P2,f) and U(,P1,f)

Answer 1

1) We aim to find $U(P_1,f)$

$P_1 = \{[0, \frac{1}{2}][ \frac{1}{2}, \frac{2}{3} ][ \frac{2}{3},1 ] \}$

The function f, as defined is decreasing for each subinterval of [0,1], thus the supremum of f, is the value of f at the left end point of each subinterval.

Define

$M_i = sup \{ f(x): x \in [x_{i-1}, x_i] \}$

And

$\delta x_i = x_i - x_{i-1}$ for each $i \in \mathbb{N}$

$M_1 = f(0) = 1$

$M_2 = f(\frac{1}{2}) = \frac{3}{4}$

$M_3 = f(\frac{2}{3}) = \frac{5}{9}$

$\delta x_1 = \frac{1}{2}$

$\delta x_2 = \frac{1}{6}$

$\delta x_3 = \frac{1}{3}$

$U(P_1,f) = M_1 \delta x_1 + M_2 \delta x_2 + M_3 \delta x_3 = \frac{1}{2} + \frac{1}{8} + \frac{5}{27} = \frac{175}{216}$

2) We aim to find $L(P_2,f)$

$P_2 = \{[0, \frac{1}{4}][ \frac{1}{4}, \frac{1}{2} ] [\frac{1}{2}, \frac{3}{4}] [ \frac{3}{4},1 ] \}$

The function f, as defined is decreasing for each subinterval of [0,1], thus the infimum of f, is the v alue of f at the right end point of each subinterval

Define,

$m_i = inf \{ f(x): x \in [x_{i-1}, x_i] \}$

$m_1 = f(\frac{1}{4}) = \frac{15}{16}$

$m_2 = f(\frac{1}{2}) = \frac{3}{4}$

$m_3 = f(\frac{3}{4}) = \frac{7}{16}$

$m_4 = f(1) = 0$

$\delta x_i = \frac{1}{4}$ for each i

$L(P_2,f) = m_1 \delta x_1 + m_2 \delta x_2 + m_3 \delta x_3 + m_4 \delta x_4 = \frac{15}{64} + \frac{3}{16} + \frac{7}{64} +0 = \frac{17}{32}$