91Ó°ÊÓ

The given integral is ${∫}_0^1{∫}_0^{1}y\sin \left(x^2\right)dxdy$

Side length is$\frac{1}{3}$units.

The given intervals in $x,y$are $\left[0,1\right],\left[0,1\right]$.

Dimension of sub rectangle is $\frac{1}{3}×\frac{1}{3}$

$m=\frac{1-0}{\frac{1}{3}}=1×\frac{3}{1}=3$

The three $x$subintervals are $\left[0,\frac{1}{3}\right],\left[\frac{1}{3},\frac{2}{3}\right]\text{and}\left[\frac{2}{3},1\right]$

Mid points are$x_1^{*}=\frac{0+\frac{1}{3}}{2}=\frac{1}{6},x_2^{*}=\frac{\frac{1}{3}+\frac{2}{3}}{2}=\frac{1}{2}\text{and}{x}_3^{*}=\frac{\frac{2}{3}+1}{2}=\frac{5}{6}$

Similarly as in above part,

number of subintervals are $n=\frac{1-0}{\frac{1}{3}}=1×\frac{3}{1}=3$

Subintervals are $\left[0,\frac{1}{3}\right],\left[\frac{1}{3},\frac{2}{3}\right]\text{and}\left[\frac{2}{3},1\right]$

Three mid points are$y_1^{*}=\frac{0+\frac{1}{3}}{2}=\frac{1}{6},y_2^{*}=\frac{\frac{1}{3}+\frac{2}{3}}{2}=\frac{1}{2}\text{and}{y}_3^{*}=\frac{\frac{2}{3}+1}{2}=\frac{5}{6}$

Mid point Reimann sum is

${∫}_0^1{∫}_0^{1}y\sin \left(x^2\right)dxdy=\underset{j=1}{\overset{3}{∑}}\underset{i=1}{\overset{3}{∑}}{y}_j^{*}\sin \left({\left(x_i^{*}\right)}^2\right)\mathrm{Δ}A$

$=\mathrm{Δ}A\underset{i=1}{\overset{3}{∑}}\left(\underset{i=1}{\overset{3}{∑}}{y}_j^{*}\sin \left({\left(x_i^{*}\right)}^2\right)\right)$

$=\frac{1}{9}×\underset{j=1}{\overset{3}{∑}}\left(y_j^{*}\sin \left({\left(x_1^{*}\right)}^2\right)+y_j^{*}\sin \left({\left(x_2^{*}\right)}^2\right)+y_j^{*}\sin \left({\left(x_3^{*}\right)}^2\right)\right)$

$=\frac{1}{9}×\left[\begin{array}{l}\left(y_1^{*}\sin \left({\left(x_1^{*}\right)}^2\right)+y_1^{*}\sin \left({\left(x_2^{*}\right)}^2\right)+y_1^{*}\sin \left({\left(x_3^{*}\right)}^2\right)\right)+\\ \left(y_2^{*}\sin \left({\left(x_1^{*}\right)}^2\right)+y_2^{*}\sin \left({\left(x_2^{*}\right)}^2\right)+y_2^{*}\sin \left({\left(x_3^{*}\right)}^2\right)\right)+\\ \left(y_3^{*}\sin \left({\left(x_1^{*}\right)}^2\right)+y_3^{*}\sin \left({\left(x_2^{*}\right)}^2\right)+y_3^{*}\sin \left({\left(x_3^{*}\right)}^2\right)\right)\end{array}\right]$

$=\frac{1}{9}×\left[\begin{array}{l}\left(\frac{1}{6}×0.0277+\frac{1}{6}×0.2474+\frac{1}{6}×0.64\right)+\\ \left(\frac{1}{2}×0.0277+\frac{1}{2}×0.2474+\frac{1}{2}×0.64\right)\\ \left(\frac{5}{6}×0.0277+\frac{5}{6}×0.2474+\frac{5}{6}×0.64\right)\end{array}\right]$

$=\frac{1}{9}×\left[\begin{array}{l}(0.0046+0.0412+0.1067)+\\ (0.01385+0.1237+0.32)+\\ (0.0231+0.2062+0.533)\end{array}\right]$

$=0.1525$

Hence,${∫}_0^1{∫}_0^{1}y\sin \left(x^2\right)dxdy=0.1525$