# Shifted Riemann

• Mar 15th 2009, 01:20 PM
manjohn12
Shifted Riemann
Let $f$ be Riemann integrable on $[a,b]$. Let $k$ be any real number, and let $g$ be the function given by $g(x) = f(x) + k$. Prove that $g$ is Riemann integrable on $[a,b]$ and that $\smallint_{a}^{b} g = k(b-a) + \smallint_{a}^{b} f$. What is the geometric interpretation.

Proof. Let $\epsilon >0$. Let $P$ be a partition of $[a,b]$ with $||P|| < \delta$. Let $x_{i}^{*}$ be such that $x_{i-1} \leq x_{i}^{*} \leq x_i$. We know that $|\mathcal{R}(f,P)-I| < \epsilon$. Now $\mathcal{R}(g,P) = \sum_{i=1}^{n} [f(x_{i}^{*})+k](x_{i}-x_{i-1})$. So we just expand this out and see that $\mathcal{R}(g,P) = \mathcal{R}(f,P)+ k(b-a)$?

And the geometric interpretation is that we are increasing the each of the rectangle's height by $k$. And we are doing this for $b-a$ rectangles.
• Mar 15th 2009, 03:33 PM
ThePerfectHacker
Quote:

Originally Posted by manjohn12
Let $f$ be Riemann integrable on $[a,b]$. Let $k$ be any real number, and let $g$ be the function given by $g(x) = f(x) + k$. Prove that $g$ is Riemann integrable on $[a,b]$ and that $\smallint_{a}^{b} g = k(b-a) + \smallint_{a}^{b} f$. What is the geometric interpretation.

Proof. Let $\epsilon >0$. Let $P$ be a partition of $[a,b]$ with $||P|| < \delta$. Let $x_{i}^{*}$ be such that $x_{i-1} \leq x_{i}^{*} \leq x_i$. We know that $|\mathcal{R}(f,P)-I| < \epsilon$. Now $\mathcal{R}(g,P) = \sum_{i=1}^{n} [f(x_{i}^{*})+k](x_{i}-x_{i-1})$. So we just expand this out and see that $\mathcal{R}(g,P) = \mathcal{R}(f,P)+ k(b-a)$?

This is good so far. You get $\mathcal{R}(g,P) = \mathcal{R}(f,P) + k(b-a)$.
Therefore, $|\mathcal{R}(g,P) - I - k(b-a)| = |\mathcal{R}(f,P) - I| < \epsilon$. Thus, we see that $\smallint_a^b g = I + k(b-a)$.