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

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

The find the area under the curve for the function you use the formula $\displaystyle \int_{a}^{b}f(x)dx$.

Riemann's midpoint formula states that 

$\displaystyle \int_{a}^{b}f(x)dx= \frac{b-a}{n}(f(x_{1})+f(x_{2})+f(x_{3})+...+f(x_{n}))$. 

This method breaks the area under the curve into n rectangles. The first term of the equation that is multiplied out find the base length of each rectangle and then the f(x) terms are the heights at the middle of each rectangle. By multiplying the two terms, you find the area and when you add them together you find the approximation of the area under the curve.

We will plug in our values into this formula. a=0, b=4 and n=4. To find the x values to plug into the function we look at the base length of each rectangle.

$\displaystyle \frac{b-a}{n}=\frac{4-0}{4}=1$

so there is a rectangle wall every x=1, thus the midpoints will be x1=0.5, x2=1.5, x3=2.5, x4=3.5.

$\displaystyle \\ \int_{a}^{b}f(x)dx= \frac{4-0}{4}(f(0.5)+f(1.5)+f(2.5)+f(3.5))\\ = 1(1.125+2.375+12.625+37.875)\\ =54.$

The find the area under the curve for the function you use the formula $\displaystyle \int_{a}^{b}f(x)dx$.

Riemann's midpoint formula states that 

$\displaystyle \int_{a}^{b}f(x)dx= \frac{b-a}{n}(f(x_{1})+f(x_{2})+f(x_{3})+...+f(x_{n}))$. 

This method breaks the area under the curve into n rectangles. The first term of the equation that is multiplied out find the base length of each rectangle and then the f(x) terms are the heights at the middle of each rectangle. By multiplying the two terms, you find the area and when you add them together you find the approximation of the area under the curve.

We will plug in our values into this formula. a=1, b=3 and n=4. To find the x values to plug into the function we look at the base length of each rectangle.

$\displaystyle \frac{b-a}{n}=\frac{3-1}{4}=0.5$

So there is a rectangle wall every x=0.5 starting at x=1, thus the midpoints will be x1=1.25, x2=1.75, x3=2.25, x4=2.75.

$\displaystyle \\ \int_{a}^{b}f(x)dx= \frac{3-1}{4}(f(1.25)+f(1.75)+f(2.25)+f(2.75))\\ = 0.5(4+5+6+7)\\=11$

The find the area under the curve for the function you use the formula $\displaystyle \int_{a}^{b}f(x)dx$.

Riemann's midpoint formula states that 

$\displaystyle \int_{a}^{b}f(x)dx= \frac{b-a}{n}(f(x_{1})+f(x_{2})+f(x_{3})+...+f(x_{n}))$. 

This method breaks the area under the curve into n rectangles. The first term of the equation that is multiplied out find the base length of each rectangle and then the f(x) terms are the heights at the middle of each rectangle. By multiplying the two terms, you find the area and when you add them together you find the approximation of the area under the curve.

We will plug in our values into this formula. a=1, b=9 and n=4. To find the x values to plug into the function we look at the base length of each rectangle.

$\displaystyle \frac{b-a}{n}=\frac{9-1}{4}=2$

so there is a rectangle wall every x=2 starting at x=1, thus the midpoints will be x1=2, x2=4, x3=6, x4=8.

$\displaystyle \\ \int_{a}^{b}f(x)dx= \frac{9-1}{4}(f(2)+f(4)+f(6)+f(8))\\ = 2\left(\frac{5}{3}+\frac{5}{9}+\frac{5}{19}+\frac{5}{33}\right)\\ \approx 5.28$

The form of a Riemann sum follows:

$\displaystyle \int_{a}^{b}f(x)dx\approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

The interval $\displaystyle [0,5]$ can be divided into four intervals of length $\displaystyle \frac{5-0}{4}=1.25$

$\displaystyle [0,1.25][1.25,2.5][2.5,3.75][3.75,5]$

Each of these intervals in turn have respective midpoints :

$\displaystyle [0.625,1.875,3.125,4.375]$

Therefore, the approximation of the integral is:

$\displaystyle \frac{5-0}{4}(e^{\frac{0.625^2}{25}}+e^{\frac{1.875^2}{25}}+e^{\frac{3.125^2}{25}}+e^{\frac{4.375^2}{25}})$

$\displaystyle 7.244$

The form of a Riemann sum follows:

$\displaystyle \int_{a}^{b}f(x)dx\approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

Five intervals can be made of length $\displaystyle \frac{14-4}{5}=2$ with midpoints $\displaystyle [5,7,9,11,13]$

Thus, the Reimann sum is:

$\displaystyle \frac{14-4}{5}(2(5)+2(7)+2(9)+2(11)+2(13))$

$\displaystyle 180$

The form of a Riemann sum follows:

$\displaystyle \int_{a}^{b}f(x)dx\approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

The interval $\displaystyle [1,4]$ can be divided into intervals of length $\displaystyle \frac{4-1}{3}=1$ with midpoints $\displaystyle [1.5,2.5,3.5]$.

Therefore, the integral approximation is:

$\displaystyle \frac{4-1}{3}(1.5^{1.5}+2.5^{2.5}+3.5^{3.5})$

$\displaystyle 91.931$

To solve for the midpoint Riemann sum, the function must be evaluated at x=0.5,1.5,2.5, and 3.5. 

$\displaystyle y(0.5)=0.5^3=0.125$

$\displaystyle y(1.5)=1.5^3=3.375$

$\displaystyle y(2.5)=2.5^3=15.625$

$\displaystyle y(3.5)=3.5^3=42.875$

These must each be multiplied by the segment inteval

$\displaystyle = \frac{total interval}{number of segments }= \frac{4}{4} =1$.

These are then summed to yield the midpoint Riemann sum approximation:

$\displaystyle 0.125\cdot 1+3.375\cdot 1+15.625\cdot 1+42.875\cdot 1=62$

This approximation method has a few steps. First find out how wide your rectangles are. Our interval is [1,3] and we need 4 subintervals.

$\displaystyle \frac{(3-1)}{4}=.5$ tells us that each subinterval is 0.5.

Now we need to figure out high each rectangle is. We're going to be finding those y-values at the midpoints of each triangle; namely, at x=1.25, 1.75, 2.25, and 2.75.

We will then plug those x-values into the function given and then multiply by the width of our rectangles (0.5).

Therefore, our equation looks like:

$\displaystyle 0.5(11.5625+13.0625+15.0625+17.5625)$.

This gives us an area of $\displaystyle 28.625$.

The form of a Riemann sum follows:

$\displaystyle \int_{a}^{b}f(x)dx\approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

For   $\displaystyle f(x)=\frac{1}{ln(x^2)}$ for $\displaystyle x=2$ to $\displaystyle 5$, our three midpoints are $\displaystyle [2.5,3.5,4.5]$.

So the approximation of the area under the curve will be:

$\displaystyle \frac{5-2}{3}\left(\frac{1}{ln(2.5^2)}+\frac{1}{ln(3.5^2)}+\frac{1}{ln(4.5^2)}\right)$

$\displaystyle 1.277$