Science:Math Exam Resources/Courses/MATH105/April 2015/Question 01 (d)/Solution 1

We compare the given sum $\sum _{k=1}^{4}f(1+k)\cdot 1=f(2)+f(3)+f(4)+f(5)=\sum _{k=0}^{3}f(2+k)\cdot 1$ to the formula for a left Riemann sum (see hint 1):

$\sum _{k=0}^{n-1}f(a+k\Delta x)\cdot \Delta x,\quad \Delta x={\frac {b-a}{n}}.$

From comparing the two formulas, we see that that $\Delta x=1$ and $n=3+1=4.$

Furthermore, since the first term of the left Riemann sum is $f(a)$, then $a=2$.

Finally, rearranging the equation for $\Delta x,$ we obtain $b=a+n\Delta x=2+4\cdot 1=6$.

In summary, ${\color {blue}a=2,b=6,n=4}.$