Science:Math Exam Resources/Courses/MATH101/April 2011/Question 07/Solution 1

Since $f(a+i(b-a)/N)=f(-1+3i/N)=-1+3i/N-1$ we follow the hints, to obtain

$\begin{array}{rl}\underset{-1}{\overset{2}{\int }}(x-1)dx& =\underset{N\to \mathrm{\infty }}{\lim }\sum _{i=1}^{N}f(-1+3i/N)\frac{3}{N}\\ & =\underset{N\to \mathrm{\infty }}{\lim }\sum _{i=1}^N(-1+3i/N-1)\frac{3}{N}\\ & =\underset{N\to \mathrm{\infty }}{\lim }\frac{3}{N}\sum _{i=1}^N(-2+\frac{3i}{N})\\ & =\underset{N\to \mathrm{\infty }}{\lim }(-\frac{6}{N}\sum _{i=1}^{N}1+\frac{9}{N^2}\sum _{i=1}^{N}i)\end{array}$

Since

$\sum _{i=1}^{N}1=N\text{ and }\sum _{i=1}^{N}i=\frac{N(N+1)}{2},$

we can continue our computation

$\begin{array}{rl}\underset{-1}{\overset{2}{\int }}(x-1)dx& =\underset{N\to \mathrm{\infty }}{\lim }(-\frac{6}{N}N+\frac{9}{N^2}\frac{N(N+1)}{2})\\ & =-6+\frac{9}{2}\underset{N\to \mathrm{\infty }}{\lim }\frac{N+1}{N}\\ & =-6+\frac{9}{2}\\ & =-\frac{3}{2}.\end{array}$

You can always compare your answer to the computation of the integral directly and see that the answers match:

$\begin{array}{rl}\underset{-1}{\overset{2}{\int }}(x-1)dx& ={\frac{x^2}{2}-x|}_{-1}^2\\ & =\frac{4}{2}-2-(\frac{1}{2}+1)\\ & =-\frac{3}{2}.\end{array}$