Splitting up the interval [1, 4] into 6 equally-spaced subintervals gives the partition
[1, 3/2], [3/2, 2], [2, 5/2], [5/2, 3], [3, 7/2], and [7/2, 4]
each with length (4-1)/6 = 1/2.
The endpoints of each subinteval form arithmetic sequences,
[tex]\ell_i=1+\dfrac{i-1}2=\dfrac12+\dfrac i2[/tex]
[tex]r_i=1+\dfrac i2[/tex]
([tex]\ell[/tex] for left, [tex]r[/tex] for right) with [tex]1\le i\le6[/tex].
The midpoints of each subinterval are, respectively,
5/4, 7/4, 9/4, 11/4, 13/4, and 15/4
which are prescribed by the sequence,
[tex]m_i=\dfrac{\ell_i+r_i}2=\dfrac{\frac32+i}2=\dfrac34+\dfrac i2[/tex]
([tex]m[/tex] for midpoint) where [tex]1\le i\le6[/tex].
The integral of [tex]f(x)=x^2+1[/tex] over the interval [1, 4] is approximated by the sum of the areas of rectangles with length equal to the base of each subinterval and height equal to [tex]f(m_i)[/tex], the value of [tex]f[/tex] at the midpoint of that subinterval:
[tex]\displaystyle\int_1^4(x^2+1)\,\mathrm dx\approx\sum_{i=1}^6f(m_i)\frac{4-1}6=\frac12\sum_{i=1}^6\left(\left(\frac34+\frac i2\right)^2+1\right)=\frac{383}{16}\approx\boxed{23.94}[/tex]
Compare to the actual value of the integral,
[tex]\displaystyle\int_1^4(x^2+1)\,\mathrm dx=24[/tex]
