Subdividing [0, 1] into 10 equally spaced intervals of length [tex]\Delta x=\frac{1-0}{10}=\frac1{10}[/tex] gives the partition
[tex][0,1] = \left[0,\dfrac1{10}\right] \cup \left[\dfrac1{10},\dfrac2{10}\right] \cup \cdots \cup \left[\dfrac9{10},1\right][/tex]
The [tex]i[/tex]-th subinterval has left and right endpoints, respectively, given by
[tex]\ell_i = \dfrac{i-1}{10} \text{ and } r_i = \dfrac i{10}[/tex]
where [tex]i\in\{1,2,3,\ldots,10\}[/tex].
The midpoint of the [tex]i[/tex]-th interval is the average of these,
[tex]m_i = \dfrac{\ell_i+r_i}2 = \dfrac{2i-1}{20} \in \left\{\dfrac1{20},\dfrac3{20},\dfrac5{20},\ldots,\dfrac{19}{20}\right\}[/tex]
We approximate the area under [tex]f(x)[/tex] over [0, 1] by the Riemann sum,
[tex]\displaystyle \int_0^1 f(x) \, dx \approx \sum_{i=1}^{10} f(m_i) \Delta x \\\\ ~~~~~~~~ = \frac1{10} \sum_{i=1}^{10} \bigg(5\sin(\pi m_i) + 2.5 \cos(4\pi m_i)\bigg) \\\\ ~~~~~~~~ = \frac{\sin\left(\frac\pi{20}\right) + \sin\left(\frac{3\pi}{20}\right) + \cdots + \sin\left(\frac{19\pi}{20}\right)}2 \\\\ ~~~~~~~~~~~~ + \dfrac{\cos\left(\frac\pi5\right) + \cos\left(\frac{3\pi}5\right) + \cdots + \cos\left(\frac{19\pi}5\right)}4 \\\\ ~~~~~~~~ \approx 3.19623[/tex]
(D)
