Answer to Question #294794 in Real Analysis for Ojugbele Daniel

Question

Let f(x)=x, show that f is Riemann integrable in the interval [a,b]. Hence find the Riemann integral of f.

Answer 1

Consider the partition $\displaystyle P=\left\{x_o=a,x_1=a+\frac{b-a}{N},x_2=a+2\left(\frac{b-a}{N}\right),\cdots ,b\right\}\text{where }h=\frac{b-a}{N},\ N\in\N\ni$

$\displaystyle x_k=x_0+kh=a+k\left(\frac{b-a}{N}\right),\text{where } k=0,1,\cdots,N$ of $[a,b]$.

Then,

$\displaystyle U(P,f)=\sum^N_{k=1}f(x_k)(x_k-x_{k-1})=\sum^N_{k=1}x_k\left(\frac{b-a}{n}\right)=\sum^N_{k=1}\left[a+k\left(\frac{b-a}{n}\right)\right]\left(\frac{b-a}{n}\right)$

$\displaystyle =\sum^N_{k=1}a\left(\frac{b-a}{N}\right)+\sum^N_{k=1}k\left(\frac{b-a}{N}\right)^2=a\left(\frac{b-a}{N}\right)\sum^N_{k=1}1+\left(\frac{b-a}{N}\right)^2\sum^N_{k=1}k$

$\displaystyle =a\left(\frac{b-a}{N}\right)N+\left(\frac{b-a}{N}\right)^2\frac{N(N+1)}{2}=a(b-a)+\frac{(b-a)^2}{2}\left(1+\frac{1}{N}\right)$

$\displaystyle L(P,f)=\sum^N_{k=1}f(x_{k-1})(x_k-x_{k-1})=\sum^N_{k=1}x_{k-1}\left(\frac{b-a}{n}\right)$

$\displaystyle =\sum^N_{k=1}\left[a+(k-1)\left(\frac{b-a}{n}\right)\right]\left(\frac{b-a}{n}\right)$

$\displaystyle =\sum^N_{k=1}a\left(\frac{b-a}{N}\right)+\sum^N_{k=1}(k-1)\left(\frac{b-a}{N}\right)^2$

$\displaystyle =a\left(\frac{b-a}{N}\right)\sum^N_{k=1}1+\left(\frac{b-a}{N}\right)^2\sum^N_{k=1}(k-1)$

$\displaystyle \displaystyle =a\left(\frac{b-a}{N}\right)\sum^N_{k=1}1+\left(\frac{b-a}{N}\right)^2\sum^{N-1}_{k=0}k$

$\displaystyle =a\left(\frac{b-a}{N}\right)N+\left(\frac{b-a}{N}\right)^2\frac{N(N-1)}{2}$

$\displaystyle =a(b-a)+\frac{(b-a)^2}{2}\left(1-\frac{1}{N}\right)$

Since $\displaystyle \lim_{N\rightarrow \infty}U(P_n,f)=\lim_{N\rightarrow \infty}L(P_n,f)=a(b-a)+\frac{(b-a)^2}{2}=\frac{b^2-a^2}{2}$ this implies that $f(x)=x$ is Riemann integrable on $[a,b]$ with;

$\displaystyle \int^b_af(x)\ dx=\int^b_ax\ dx=\frac{b^2-a^2}{2}$