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

Question

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

Answer 1

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 $[2,3]$.

Answer 2

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