Shifted Riemann

**manjohn12** · March 15th 2009, 01:20 PM

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.

**ThePerfectHacker** · March 15th 2009, 03:33 PM

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 $\smallint_a^b g = I + k(b-a)$ .

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