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

MATH105 April 2015

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Question 01 (d)
$\displaystyle \sum \limits _{k=1}^{4}f(1+k)\cdot 1$ is a left Riemann sum for a function $f(x)$ on the interval $[a,b]$ with $n$ sub intervals. Find the values of $a$, $b$ and $n$.

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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall that a left Riemann sum for $f$ on $[a,b]$ with $n$ subintervals takes the form $\sum _{k=0}^{n-1}f(a+k\Delta x)\cdot \Delta x,$ where $\Delta x={\frac {b-a}{n}}.$ Compare the given sum to this formula.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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, ${\color {blue}a=2,b=6,n=4}.$