Step 1

Trapezoidal rule

\(\displaystyle={\int_{{{0}}}^{{{2}}}}{{\sin}^{{{2}}}{\left({x}\right)}}\ldots\ldots{\left({1}\right)}\)

As we know that accoding to trapezoidal rule

\(\displaystyle{\int_{{{a}}}^{{{b}}}}{f}{\left({x}\right)}{\left.{d}{x}\right.}\approx{\frac{{\triangle{x}}}{{{2}}}}{\left({f{{\left({x}_{{{0}}}\right)}}}+{2}{f{{\left({x}_{{{1}}}\right)}}}+{2}{f}{\left({x}_{{{2}}}\right)}+\ldots+{2}{f{{\left({x}_{{{n}-{1}}}\right)}}}+{f{{\left({x}_{{{n}}}\right)}}}\ldots{\left({2}\right)}\right.}\)

This is the standard equation for the trapezoidal method in which

\(\displaystyle\triangle{x}={\frac{{{b}-{a}}}{{{n}}}}\)

Now comparing equation (2) with equation (1) we get that

a=0,b=2,n=4

\(\begin{cases}\triangle x=\frac{2-0}{4}=\frac{1}{2}\\ & \end{cases}\)

Now we will divide (0,2) interval into n=4 subinterval of length \(\displaystyle\triangle{x}={\frac{{{1}}}{{{2}}}}\)

end points of the function \[\begin{cases}0\\ 0+\frac{1}{2}=\frac{1}{2} & \\ \frac{1}{2}+\frac{1}{2}=1 & \\ 1+\frac{1}{2}=\frac{3}{2} & \\ \frac{3}{2}+\frac{1}{2}=2 & \end{cases}\]

Therefore the end points of the function are \(\displaystyle{a}={0},{\frac{{{1}}}{{{2}}}},{1},{\frac{{{3}}}{{{2}}}},{2}\)

U sing these end points we analysis the function

\(\displaystyle{f}{\left({x}_{{{0}}}\right)}={f{{\left({a}\right)}}}={f}{\left({0}\right)}={0}\)

\(\displaystyle{2}{f}{\left({x}_{{{1}}}\right)}={2}{f{{\left({\frac{{{1}}}{{{2}}}}\right)}}}={2}{{\sin}^{{{2}}}{\left({\frac{{{1}}}{{{2}}}}\right)}}={0.459698}\)

\(\displaystyle{2}{f}{\left({x}_{{{2}}}\right)}={2}{f{{\left({1}\right)}}}={2}{{\sin}^{{{2}}}{\left({1}\right)}}={1.416147}\)

\(\displaystyle{2}{f}{\left({x}_{{{3}}}\right)}={2}{f{{\left({\frac{{{3}}}{{{2}}}}\right)}}}={2}{{\sin}^{{{2}}}{\left({\frac{{{3}}}{{{2}}}}\right)}}={1.9899925}\)

\(\displaystyle{f}{\left({x}_{{{4}}}\right)}={f{{\left({b}\right)}}}={f}{\left({2}\right)}={{\sin}^{{{2}}}{\left({2}\right)}}={0.0826819}\)

Substitute these value in equation (2) we get that

\(\displaystyle={\frac{{{1}}}{{{4}}}}{\left[{0}+{0.459698}+{1.416147}+{1.9899925}+{0.0826819}\right]}\)

\(\displaystyle={1.173165646}\approx{1.1737}\)

Step 2

By midpoint rule

\(\displaystyle={\int_{{{0}}}^{{{2}}}}{{\sin}^{{{2}}}{\left({x}\right)}}\ldots.{\left({1}\right)}\)

According to mid point rule we know that

\(\displaystyle{\int_{{{a}}}^{{{b}}}}{f}{\left({x}\right)}{\left.{d}{x}\right.}\approx\triangle{x}{\left({f}{\left({\frac{{{x}_{{{0}}}+{x}_{{{1}}}}}{{{2}}}}\right)}+{f}{\left({\frac{{{x}_{{{1}}}+{x}_{{{2}}}}}{{{2}}}}\right)}+{f}{\left({\frac{{{x}_{{{2}}}+{x}_{{{3}}}}}{{{2}}}}\right)}+\ldots.+{f}{\left({\frac{{{x}_{{{n}-{1}}}+{x}_{{{n}}}}}{{{2}}}}\right)}\right)}\)

\(\displaystyle\triangle{x}={\frac{{{1}}}{{{2}}}}\)

End points are \(\displaystyle={0},{\frac{{{1}}}{{{2}}}},{1},{\frac{{{3}}}{{{2}}}},{2}\)

Now calculate for these values

\(\displaystyle{f}{\left({\frac{{{x}_{{{0}}}+{x}_{{{1}}}}}{{{2}}}}\right)}={f}{\left({\frac{{{0}+{\frac{{{1}}}{{{2}}}}}}{{{2}}}}\right)}={f}{\left({\frac{{{1}}}{{{4}}}}\right)}={{\sin}^{{{2}}}{\left({\frac{{{1}}}{{{4}}}}\right)}}={0.0612088}\)

similarly calculate for other

\(\displaystyle{f}{\left({\frac{{{x}_{{{1}}}+{x}_{{{2}}}}}{{{2}}}}\right)}={0.464642}\)

\(\displaystyle{f}{\left({\frac{{{x}_{{{2}}}+{x}_{{{3}}}}}{{{2}}}}\right)}={0.900579}\)

\(\displaystyle{f}{\left({\frac{{{x}_{{{3}}}+{x}_{{{4}}}}}{{{2}}}}\right)}={0.968229}\)

Substitute all these value in equation (2) we get that

\(\displaystyle={\frac{{{1}}}{{{2}}}}{\left[{0.0612088}+{0.464642}+{0.900579}+{0.968229}\right]}\)

=1.18764201