Answer to Question #248695 in Calculus for user1

Question

Show that 𝑈(−𝑓, 𝑝) = −𝐿(𝑓, 𝑝) and 𝐿(−𝑓, 𝑝) = −𝑈(𝑓, 𝑝)

Answer 1

let p is partition on [a,b]

$p=\{a=x_0, x_1, x_2...x_{i-1}, x_i,...x_n=b \}$

is any partition on [a,b]

By definition lower and upper Riemman sum

$L(f,p)=\displaystyle\sum_{i=1}^nm_i\Delta x_i$

$U(f,p)=\displaystyle\sum_{i=1}^nm_i \Delta x_i$

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

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

$intrimum(-f(x))=-supremum \space f(x)$

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

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

$=-m_i$

$L(-f,p)=\displaystyle\sum_{i=1}^nm_i\Delta x_i$

$=-\displaystyle\sum_{i=1}^nm_i\Delta x_i$

$𝐿(−𝑓, 𝑝) = −𝑈(𝑓, 𝑝).......proved$

now,

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

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

$U(-f,p)=\displaystyle\sum_{i=1}^nm_i\Delta x_i$

$=-\displaystyle\sum_{i=1}^nm_i\Delta x_i$

$𝑈(−𝑓, 𝑝) = −𝐿(𝑓, 𝑝)..........proved$