Q 68. - Calculus

Question

Use midpoint Riemann sums with the specified numbers

${\int }_0^1{\int }_0^{1}y\sin \left(x^2\right)dxdy.$  Let each sub rectangle be a square

with side length $\frac{1}{3}$unit.

Step-by-Step Solution

Answer

The integral is ${\int }_0^1{\int }_0^{1}y\sin \left(x^2\right)dxdy=0.1525$

Step 1: Given Information

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

Side length is $\frac{1}{3}$ units.

Step 2: Calculation of mid points

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

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

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

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 $x_1^{*}=\frac{0+\frac{1}{3}}{2}=\frac{1}{6},x_2^{*}=\frac{\frac{1}{3}+\frac{2}{3}}{2}=\frac{1}{2}\text{ and } x_3^{*}=\frac{\frac{2}{3}+1}{2}=\frac{5}{6}$

Step 3: Three mid points of y

Similarly as in above part,

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

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

Step 4: Evaluation of Integral

Mid point Reimann sum is

${\int }_0^1{\int }_0^{1}y\sin \left(x^2\right)dxdy=\sum _{j=1}^3\sum _{i=1}^3{y}_j^{*}\sin \left({\left(x_i^{*}\right)}^2\right)\Delta A$

$=\Delta A\sum _{i=1}^3\left(\sum _{i=1}^3{y}_j^{*}\sin \left({\left(x_i^{*}\right)}^2\right)\right)$

$=\frac{1}{9}\times \sum _{j=1}^3\left(y_j^{*}\sin \left({\left(x_1^{*}\right)}^2\right)+y_j^{*}\sin \left({\left(x_2^{*}\right)}^2\right)+y_j^{*}\sin \left({\left(x_3^{*}\right)}^2\right)\right)$

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

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

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

$=0.1525$

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