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

${\begin{aligned}\int _{-1}^{2}(x-1)\,dx&=\lim _{N\to \infty }\sum _{i=1}^{N}f(-1+3i/N){\frac {3}{N}}\\&=\lim _{N\to \infty }\sum _{i=1}^{N}(-1+3i/N-1){\frac {3}{N}}\\&=\lim _{N\to \infty }{\frac {3}{N}}\sum _{i=1}^{N}\left(-2+{\frac {3i}{N}}\right)\\&=\lim _{N\to \infty }\left(-{\frac {6}{N}}\sum _{i=1}^{N}1+{\frac {9}{N^{2}}}\sum _{i=1}^{N}i\right)\end{aligned}}$

Since

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

we can continue our computation

${\begin{aligned}\int _{-1}^{2}(x-1)\,dx&=\lim _{N\to \infty }\left(-{\frac {6}{N}}N+{\frac {9}{N^{2}}}{\frac {N(N+1)}{2}}\right)\\&=-6+{\frac {9}{2}}\lim _{N\to \infty }{\frac {N+1}{N}}\\&=-6+{\frac {9}{2}}\\&=-{\frac {3}{2}}.\end{aligned}}$

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

${\begin{aligned}\int _{-1}^{2}(x-1)\,dx&=\left.{\frac {x^{2}}{2}}-x\right|_{-1}^{2}\\&={\frac {4}{2}}-2-\left({\frac {1}{2}}+1\right)\\&=-{\frac {3}{2}}.\end{aligned}}$