# Show that $f(x) = 2x+3$ is Riemann integrable on interval [2,3] [closed]

I'm trying to show that $f(x) = 2x+3$ is Riemmann integrable on $[2,3]$ and $\int_2^3f(x)dx=8$.

## closed as off-topic by TravisJ, Glorfindel, Sahiba Arora, Stefan Hamcke, LeucippusAug 1 '17 at 14:36

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• What exactly is your question? Are you trying to show that $f(x)=2x+3$ is Riemann integrable using the integrability criterion or? – thedilated Aug 1 '17 at 12:41
• Are you trying to find the value of the integral? Or are you trying to show it is Riemann Integrable? – Naive Aug 1 '17 at 12:51
• Both, showing that function is reimann integrable and then finding value of integral. – Deepak Dahiya Aug 1 '17 at 13:02
• Add your question in the main text (having it in the title only is confusing). – mlc Aug 1 '17 at 13:03
• I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – Sahiba Arora Aug 1 '17 at 13:59

## 2 Answers

You can use the fact that: $\sum_{i=1}^n i = \frac{n(n+1)}{2}$ to find the accurate value of the upper and lower sum. The fact that we're dealing with infinity can be a little bit tricky, but usually (at least that was the case for me) during the lectures we accept that this formula is true even when $n \to \infty$ without a proof. So:

$$L =\lim_{n \to \infty} \sum_{r=1}^{n} \left(2\cdot\left(2 +\frac{(r-1)}{n}\right) + 3 \right) \cdot \frac 1n = \lim_{n \to \infty}\sum_{r=1}^{n} \left(\frac{2r + 7n-2}{n} \right) \cdot \frac 1n$$ $$= \lim_{n \to \infty} \left(\frac{n(n+1)+7n^2 - 2n}{n} \right) \cdot \frac 1n = \lim_{n \to \infty} \frac{8n^2 - 2n}{n^2} = 8$$

Similarly:

$$U =\lim_{n \to \infty} \sum_{r=1}^{n} \left(2\cdot\left(2 +\frac{r}{n}\right) + 3 \right) \cdot \frac 1n = \lim_{n \to \infty}\sum_{r=1}^{n} \left(\frac{2r + 7n}{n} \right) \cdot \frac 1n$$ $$= \lim_{n \to \infty} \left(\frac{n(n+1)+7n^2}{n} \right) \cdot \frac 1n = \lim_{n \to \infty} \frac{8n^2}{n^2} = 8$$

As $U = L = 8$ we have that the function is Riemann Integrable on $[2,3]$.

• How did you solve summation in both sums, the part where 7n becomes 7n^2 – Deepak Dahiya Aug 1 '17 at 13:32
• @user468740 You can notice that we get rid of the summation. In fact you're summing $7n$ $n$ times. Note that $7n$ isn't depending on $r$, so it's pretty much a constant in terms of the summation. – Stefan4024 Aug 1 '17 at 14:22

Show that $\forall \varepsilon$ $\exists P_n: U(f, P_n) - L(f, P_n) < \varepsilon$,

where $U(f, P_n) = \sum M_i\Delta x_i, L(f, P_n) = \sum m_i\Delta x_i$

In your case $P_n: \{x_{i+1} = x_i + \dfrac{1}{n}\}$ should work