Q. 57 - Calculus

Question

For the logarithmic value, use Riemann sums with at least four rectangles to find an over approximation and an under-approximation for the value.

$\ln e$

Step-by-Step Solution

Answer

Using left and right sums with four rectangles is $0.8999\le \ln e\le 1.149$.

Step 1. Given information.

Consider the given question,

$\ln e$

Step 2. Use Riemann sums for left sum definition.

Consider the function,

$f\left(x\right)=\frac{1}{t}$

The left-sum defined for n rectangles on $\left[a,b\right]$ is $\sum _{k=1}^n f\left(x_{k-1}\right)△x$.

Where, $$\begin{gathered} △x=\frac{b-a}{n}, \\ {x}_k=a+k△x \end{gathered}$$

Now $△x$,

$$\begin{gathered} =\frac{2.8-0}{14} \\ =\frac{28}{140} \\ =0.2 \end{gathered}$$

Step 3. Find the left sum.

Consider the given question,

$$\begin{gathered} x_k=0+k\left(0.2\right) \\ =0.2k \end{gathered}$$

In the left sum, $x_{k-1}$ is the leftmost point in the interval $\left[x_{k-1},x_k\right]$.

So, $$\begin{gathered} x_{k-1}=\left(k-1\right)0.2 \\ =0.2k-0.2 \end{gathered}$$

The left sum is given below,

$$\begin{gathered} \sum _{k=0}^{2.8} \ln \left(0.2k-0.2\right)\left(0.2\right) \\ \approx 0.8999 \end{gathered}$$

Step 5. Find the right sum.

The right-sum defined for n rectangles on $\left[a,b\right]$ is $\sum _{k=1}^n f\left(x\right)△x$.

The right sum is given below,

$$\begin{gathered} =\sum _{k=0}^{2.8} \ln \left(0.2k-0.2\right)\left(0.2\right) \\ \approx 1.149 \end{gathered}$$

Therefore, the value of $\ln e$ is $0.8999\le \ln e\le 1.149$.