My Question today is:
Define g : [−3, 2] → $\mathbb{R}$ by g(x) = 6x + 1. Use the definition of the Riemann integral to show that g is integrable on [−3, 2] and determine the value of the integral $\int_{-3}^{2} g(x) dx$.
I understand the definition of a Riemann Integral but cant seem to be able to apply it to this question. Any help will be appreciated.
$g$ is increasing at $[-3,2]$.
let $n>0$ and $$(x_k=-3+k\frac{5}{n})_{k=0,1...n}$$ be a regular partition of $[-3,2]$.
then
$$U(g,n)=\frac{5}{n}\sum_{k=1}^ng(x_i)$$
$$L(g,n)=\frac{5}{n}\sum_{k=1}^ng(x_{i-1})$$
and by telescopage,
$$U(g,n)-L(g,n)=\frac{5}{n}(g(2)-g(-3))$$
$$=\frac{150}{n}.$$
thus
for each $\epsilon>0,\;\;$ if we take
$$N=\lfloor \frac{150}{\epsilon} \rfloor +1,$$
then
$$U(g,N)-L(g,N)<\epsilon.$$
and $\;g\;$ is Riemann integrable at $[-3,2]$.