Chapter 11.2, Problem 15E

Solution

To find: The six subintervals of the interval $\left[1,4\right]$ .

The six subintervals between $\left[1,4\right]$ are $s=\left\{\left(1,\frac{3}{2}\right)\left(\frac{3}{2},2\right)\left(2,\frac{5}{2}\right)\left(\frac{5}{2},3\right)\left(3,\frac{7}{2}\right)\left(\frac{7}{2},4\right)\right\}$ .

Given information:

Interval $\left[1,4\right]$ .and number of subintervals which is $6$

Formula used:

  $\Delta x=\frac{b-a}{n}$

Here, $\left[a,b\right]=\left[1,4\right]$ and $n=6$

Calculation:

Substitute the values of $a$ , $b$ and $n$ in the formula. $n$

  $\begin{array}{c}\Delta x=\frac{4-1}{6}\\ =\frac{1}{2}\end{array}$

Starting with $a$ , add $\Delta x$ to obtain the endpoint of the subinterval. The endpoint of the previous subinterval is the starting point of the next and repeat the process.

Therefore, the six subintervals of interval $\left[1,4\right]$ are $s=\left\{\left(1,\frac{3}{2}\right)\left(\frac{3}{2},2\right)\left(2,\frac{5}{2}\right)\left(\frac{5}{2},3\right)\left(3,\frac{7}{2}\right)\left(\frac{7}{2},4\right)\right\}$ .