Q. 42 - 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.

$f\left(x\right)=\ln \left(x\right),\left[a,b\right]=\left[2,5\right],left sum,n=100$

Step-by-Step Solution

Answer

The left sum is $=0.03\sum _{k=1}^{100}\ln \left(2+0.03\left(k-1\right)\right)$

Step 1. Given Information

The given function is $f\left(x\right)=\ln \left(x\right),\left[a,b\right]=\left[2,5\right],left sum,n=100$

Step 2. Finding ∆ x and x k

$∆x=\frac{b-a}{n}=\frac{5-2}{100}=\frac{3}{100}=0.03$

$$\begin{gathered} x_k=a+k∆x=2+0.03k \\ {x}_{k-1}=a+\left(k-1\right)∆x=2+0.03\left(k-1\right) \end{gathered}$$

Step 3. Finding Reimann Sum

We have $f\left(x\right)=\ln \left(x\right)$then $f\left(x_{k-1}\right)=\ln \left(x_{k-1}\right)$

Now, $$\begin{gathered} \sum _{k=1}^{100}f\left(x_{k-1}\right)∆x=\sum _{k=1}^{100}\ln \left(2+0.03\left(k-1\right)\right)\left(0.03\right) \\ =0.03\sum _{k=1}^{100}\ln \left(2+0.03\left(k-1\right)\right) \end{gathered}$$