Simpson's rule is solved using the formula

\(\displaystyle \int_{a}^{b}f(x))dx =S_{2m}=\frac{1}{3}(\frac{b-a}{2n})\left [ f(0)+4f(1)+2f(2)+...+2f(m-2)+4f(m-1)+f(m) \right ]\)

where \(\displaystyle n\) is the number of subintervals and \(\displaystyle f(m)\) is the function evaluated at the midpoint.

For this problem, \(\displaystyle (\frac{b-a}{2n})=(\frac{5-1}{10})=0.4\).  

The value of each approximation term is below.

The sum of all the approximation terms is \(\displaystyle -8.23995\) therefore

\(\displaystyle (\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.4)*(-8.23995)=-1.0987\)

Simpson's rule is solved using the formula

\(\displaystyle \int_{a}^{b}f(x))dx=S_{2m}=\frac{1}{3}(\frac{b-a}{2n})\left [ f(0)+4f(1)+2f(2)+...+2f(m-2)+4f(m-1)+f(m) \right ]\)

where \(\displaystyle 2n\) is the number of subintervals and \(\displaystyle f(m)\) is the function evaluated at the midpoint.

For this problem, \(\displaystyle (\frac{b-a}{2n})=(\frac{15-8}{10})=0.7\).  

The value of each approximation term is below.

The sum of all the approximation terms is \(\displaystyle 72.79378\) therefore

\(\displaystyle (\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.7)*(72.79378)=16.9852\)

Simpson's rule is solved using the formula

\(\displaystyle \int_{a}^{b}f(x))dx =S_{2m}=\frac{1}{3}(\frac{b-a}{2n})\left [ f(0)+4f(1)+2f(2)+...+2f(m-2)+4f(m-1)+f(m) \right ]\)

where \(\displaystyle n\) is the number of subintervals and \(\displaystyle f(m)\) is the function evaluated at the midpoint.

For this problem, \(\displaystyle (\frac{b-a}{2n})=(\frac{1+1}{10})=0.2\).  

The value of each approximation term is below.

The sum of all the approximation terms is \(\displaystyle 10.5840\) therefore

\(\displaystyle (\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.2)*(10.5840)=0.7056\)

Simpson's rule is solved using the formula

\(\displaystyle \int_{a}^{b}f(x))dx\approx S_{2m}=\frac{1}{3}(\frac{b-a}{2n})\left [ f(0)+4f(1)+2f(2)+...+2f(m-2)+4f(m-1)+f(m) \right ]\)

where \(\displaystyle n\) is the number of subintervals and \(\displaystyle f(m)\) is the function evaluated at the midpoint.

For this problem, \(\displaystyle (\frac{b-a}{2n})=(\frac{9-0}{10})=0.9\).  

The value of each approximation term is below.

The sum of all the approximation terms is \(\displaystyle 2.3081\) therefore

\(\displaystyle (\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.9)*(2.3081)=0.6924\)

 Trapezoidal approximations are solved using the formula

\(\displaystyle \int_{a}^{b}f(x))dx\approx T_{n}=(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]\)

where \(\displaystyle n\) is the number of subintervals and \(\displaystyle f(m)\) is the function evaluated at the midpoint.

For this problem, \(\displaystyle (\frac{b-a}{2n})=(\frac{7-2}{2*5})=0.5\).  

The value of each approximation term is below.

The sum of all the approximation terms is \(\displaystyle 0.86027754\), therefore

\(\displaystyle (\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.5)*(0.86027754)=0.4301\)

 Trapezoidal approximations are solved using the formula

\(\displaystyle \int_{a}^{b}f(x))dx\approx T_{n}=(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]\)

where \(\displaystyle n\) is the number of subintervals and \(\displaystyle f(m)\) is the function evaluated at the midpoint.

For this problem, \(\displaystyle (\frac{b-a}{2n})=(\frac{2-1}{2*5})=0.1\).  

The value of each approximation term is below.

The sum of all the approximation terms is \(\displaystyle 28.7867\), therefore

\(\displaystyle (\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.1)*(28.786699)=2.8787\)

 Trapezoidal approximations are solved using the formula

\(\displaystyle \int_{a}^{b}f(x))dx\approx =T_{n}=(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]\)

where \(\displaystyle n\) is the number of subintervals and \(\displaystyle f(m)\) is the function evaluated at the midpoint.

For this problem, \(\displaystyle (\frac{b-a}{2n})=(\frac{4-0}{2*5})=0.4\).  

The value of each approximation term is below.

The sum of all the approximation terms is \(\displaystyle 83.89553\), therefore

\(\displaystyle (\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.4)*(83.89553)=33.5582\)

 Trapezoidal approximations are solved using the formula

\(\displaystyle \int_{a}^{b}f(x))dx\approx T_{n}=(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]\)

where \(\displaystyle n\) is the number of subintervals and \(\displaystyle f(m)\) is the function evaluated at the midpoint.

For this problem, \(\displaystyle (\frac{b-a}{2n})=(\frac{2-1}{2*5})=0.1\).  

The value of each approximation term is below.

The sum of all the approximation terms is \(\displaystyle 11.7257431\), therefore

\(\displaystyle (\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.1)*(11.7257431)=1.1726\)

1) n = 2 indicates 2 equal subdivisions. In this case, they are from 0 to 1, and from 1 to 2.

2) Trapezoidal Rule is: \(\displaystyle \int_{a}^{b} f(x)dx \approx \left [ b-a\right ]\left [ \frac{f(a)+f(b)}{2}\right ]\)

3) For n = 2: \(\displaystyle \int_{0}^2 x^{x^{2}} dx \approx [1-0]\left [ \frac{f(0)+f(1)}{2} \right ]+ [2-1]\left [ \frac{f(1)+f(2)}{2} \right ]\)

4) Simplifying: \(\displaystyle \int_{0}^2 x^{x^{2}} dx \approx [1]\left [ \frac{0+1}{2} \right ]+ [1]\left [ \frac{1+16}{2} \right ] = \frac{1}{2}+\frac{17}{2} = \frac{18}{2} = 9.\)