Q. 55 - 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.

$\ln 10$

Step-by-Step Solution

Answer

Using left and right sums with $9$ rectangles is $1.928\le \ln 10\le 2.8289$.

Step 1. Given information.

Consider the given question,

$\ln 10$

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 $△x$,

$$\begin{gathered} =\frac{10-1}{9} \\ =\frac{9}{9} \\ =1 \end{gathered}$$

Step 3. Find the left sum.

Consider the given question,

$$\begin{gathered} x_k=1+k\left(1\right) \\ =k+1 \end{gathered}$$

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

So, $x_{k-1}=k$

The left sum is given below,

$$\begin{gathered} =\sum _{k=1}^{10} \frac{1}{k}\left(1\right) \\ =1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10} \\ \approx 2.828897 \end{gathered}$$

Step 4. Use Riemann sums for right sum definition.

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

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

Now $△x$,

$$\begin{gathered} =\frac{10-1}{9} \\ =\frac{9}{9} \\ =1 \end{gathered}$$

Step 5. Find the right sum.

Consider the given question,

$$\begin{gathered} x_k=1+k\left(1\right) \\ =k+1 \end{gathered}$$

The right sum is given below,

$$\begin{gathered} =\sum _{k=1}^{10} \frac{1}{k+1}\left(1\right) \\ =\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11} \\ \approx 1.9287 \end{gathered}$$

Therefore, the value of $\ln 10$ is $1.928\le \ln 10\le 2.8289$.