1. ## Shifted Riemann

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

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

And the geometric interpretation is that we are increasing the each of the rectangle's height by $\displaystyle k$. And we are doing this for $\displaystyle b-a$ rectangles.

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

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