I guess the "5" is supposed to represent the integral sign?
[tex]I=\displaystyle\int_1^4\ln t\,\mathrm dt[/tex]
With [tex]n=10[/tex] subintervals, we split up the domain of integration as
[1, 13/10], [13/10, 8/5], [8/5, 19/10], ... , [37/10, 4]
For each rule, it will help to have a sequence that determines the end points of each subinterval. This is easily, since they form arithmetic sequences. Left endpoints are generated according to
[tex]\ell_i=1+\dfrac{3(i-1)}{10}[/tex]
and right endpoints are given by
[tex]r_i=1+\dfrac{3i}{10}[/tex]
where [tex]1\le i\le10[/tex].
a. For the trapezoidal rule, we approximate the area under the curve over each subinterval with the area of a trapezoid with "height" equal to the length of each subinterval, [tex]\dfrac{4-1}{10}=\dfrac3{10}[/tex], and "bases" equal to the values of [tex]\ln t[/tex] at both endpoints of each subinterval. The area of the trapezoid over the [tex]i[/tex]-th subinterval is
[tex]\dfrac{\ln\ell_i+\ln r_i}2\dfrac3{10}=\dfrac3{20}\ln(ell_ir_i)[/tex]
Then the integral is approximately
[tex]I\approx\displaystyle\sum_{i=1}^{10}\frac3{20}\ln(\ell_ir_i)\approx\boxed{2.540}[/tex]
b. For the midpoint rule, we take the rectangle over each subinterval with base length equal to the length of each subinterval and height equal to the value of [tex]\ln t[/tex] at the average of the subinterval's endpoints, [tex]\dfrac{\ell_i+r_i}2[/tex]. The area of the rectangle over the [tex]i[/tex]-th subinterval is then
[tex]\ln\left(\dfrac{\ell_i+r_i}2\right)\dfrac3{10}[/tex]
so the integral is approximately
[tex]I\approx\displaystyle\sum_{i=1}^{10}\frac3{10}\ln\left(\dfrac{\ell_i+r_i}2\right)\approx\boxed{2.548}[/tex]
c. For Simpson's rule, we find a quadratic interpolation of [tex]\ln t[/tex] over each subinterval given by
[tex]P(t_i)=\ln\ell_i\dfrac{(t-m_i)(t-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+\ln m_i\dfrac{(t-\ell_i)(t-r_i)}{(m_i-\ell_i)(m_i-r_i)}+\ln r_i\dfrac{(t-\ell_i)(t-m_i)}{(r_i-\ell_i)(r_i-m_i)}[/tex]
where [tex]m_i[/tex] is the midpoint of the [tex]i[/tex]-th subinterval,
[tex]m_i=\dfrac{\ell_i+r_i}2[/tex]
Then the integral [tex]I[/tex] is equal to the sum of the integrals of each interpolation over the corresponding [tex]i[/tex]-th subinterval.
[tex]I\approx\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt[/tex]
It's easy to show that
[tex]\displaystyle\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt=\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)[/tex]
so that the value of the overall integral is approximately
[tex]I\approx\displaystyle\sum_{i=1}^{10}\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)\approx\boxed{2.545}[/tex]
