Q. 41 - Calculus

Question

In Exercises 39–44, write out the sigma notation for the Riemann sum described in such a way that the only letter which appears in the general term of the sum is k. Don’t calculate the value of the sum; just write it down in sigma notation.

$f\left(x\right)=e^x,\left[a,b\right]=\left[1,4\right],midpoint sum,n=6$

Step-by-Step Solution

Answer

The midpoint sum is$\frac{1}{2}\sum _{k=1}^6{e}^{\frac{2k+3}{4}}$

Step 1. Given Information

$f\left(x\right)=e^x,\left[a,b\right]=\left[1,4\right],midpoint sum,n=6$

Step 1. Finding ∆ x , x k and x k - 1

$∆x=\frac{b-a}{n}=\frac{4-1}{n}=\frac{3}{6}=\frac{1}{2}$

$$\begin{gathered} x_k=a+k∆x=1+\frac{k}{2}=\frac{k+2}{2} \\ {x}_{k-1}=a+\left(k-1\right)∆x=\frac{k-1+2}{2}=\frac{k+1}{2} \end{gathered}$$

and $$\begin{gathered} \frac{x_{k-1}+x_k}{2}=\frac{{\displaystyle \frac{k+1}{2}}+{\displaystyle \frac{k+2}{2}}}{2} \\ =\frac{k+1+k+2}{4}=\frac{2k+3}{4} \end{gathered}$$

Step 3. Finding Reimann Sum

We have $f\left(x\right)=e^x$then $f\left(x_k\right)=e^{x_k}$ and $f\left(x_{k-1}\right)=e^{x_{k-1}}$

Now, $$\begin{gathered} \sum _{k=1}^{n}f\left(\frac{x_{k-1}+x_k}{2}\right)∆x=\sum _{k=1}^{n}f\left(\frac{2k+3}{4}\right)\frac{1}{2} \\ =\frac{1}{2}\sum _{k=1}^n{e}^{\frac{2k+3}{4}} \end{gathered}$$