If we want to estimate the area under the curve from  $\displaystyle x=0$  to  $\displaystyle x=4$  and are told to use  $\displaystyle n=2$,  this means we estimate the area using two rectangles that will each be two units wide and whose height is the value of the function at the midpoint of the interval. We have a rectangle from  $\displaystyle x=0$  to  $\displaystyle x=2$,  whose height is the value of the function at  $\displaystyle x=1$,  and a rectangle from  $\displaystyle x=2$  to  $\displaystyle x=4$,  whose height is the value of the function at  $\displaystyle x=3$.  First we can find the value of the function at these midpoints, and then add the areas of the two rectangles, which gives us the following:

$\displaystyle f(1)=\frac{1}{2}(1)^3+4=\frac{9}{2}$

$\displaystyle f(3)=\frac{1}{2}(3)^3+4=\frac{35}{2}$

$\displaystyle A=2\left(\frac{9}{2}\right)+2\left(\frac{35}{2}\right)=44$

If we are told to use $\displaystyle n=4$  rectangles from  $\displaystyle x=0$  to  $\displaystyle x=4$,  this means we have a rectangle from  $\displaystyle x=0$  to  $\displaystyle x=1$,  a rectangle from  $\displaystyle x=1$  to  $\displaystyle x=2$,  a rectangle from  $\displaystyle x=2$  to  $\displaystyle x=3$,  and a rectangle from  $\displaystyle x=3$  to  $\displaystyle x=4$.  We can see that the width of each rectangle is  $\displaystyle 1$  because we have an interval that is  $\displaystyle 4$  units long for which we are using  $\displaystyle 4$  rectangles to estimate the area under the curve. The height of each rectangle is the value of the function at the midpoint for its interval, so first we find the height of each rectangle and then add together their areas to find our answer:

$\displaystyle f(0.5)=\frac{1}{4}(0.5)^2+3=\frac{49}{16}$

$\displaystyle f(1.5)=\frac{1}{4}(1.5)^2+3=\frac{57}{16}$

$\displaystyle f(2.5)=\frac{1}{4}(2.5)^2+3=\frac{73}{16}$

$\displaystyle f(3.5)=\frac{1}{4}(3.5)^2+3=\frac{97}{16}$

$\displaystyle A=(1)\left(\frac{49}{16}\right)+(1)\left(\frac{57}{16}\right)+(1)\left(\frac{73}{16}\right)+(1)\left(\frac{97}{16}\right)=\frac{276}{16}=\frac{69}{4}$

The problem becomes this:

Addings these rectangles up to approximate the area under the curve is

$\displaystyle A=0.2(1-0.1^2)+0.2(1-0.3^2)+0.2(1-0.5^2)+0.2(1-0.7^2)+0.2(1-0.9^2)$

$\displaystyle A=0.2[(1-0.1^2)+(1-0.3^2)+(1-0.5^2)+(1-0.7^2)+(1-0.9^2)]$

$\displaystyle A=0.2[5-0.1^2-0.3^2-0.5^2-0.7^2-0.9^2]$

$\displaystyle A=0.2[3.35]$

$\displaystyle A=0.67$

We begin by finding the given change in x: 

$\displaystyle \Delta x=\frac{b-a}{n}=\frac{8}{5}=1.6$

We then define our partition intervals:

$\displaystyle [x_{i-1},x_i]=[0,1.6),[1.6,3.2),[3.2,4.8),[4.8,6.4),[6.4,8]$

We then choose the midpoint in each interval:

$\displaystyle \{\bar{x_i}\}=\{0.8,2.4,4.0,5.6,7.2\}$

Then we find the value of the function at the point.  This is determined through observation of the graph

$\displaystyle f(\bar{x_i})\approx 4.8,4.5,4,1.6,3.4$

Then we simply substitute these values into the formula for the Riemann Sum

$\displaystyle \\A\approx \sum_{i=1}^nf(x_i^*)\Delta x=4.8\cdot 1.6+4.5\cdot 1.6+4\cdot 1.6+1.6\cdot 1.6+3.4\cdot 1.6\\ =1.6(4.8+4.5+4+1.6+3.4)=1.6\cdot 18.3=29.28$

We begin by defining the size of our partitions and the partitions themselves.

$\displaystyle \\ \Delta x=\frac{b-a}{n}=\frac{12+4}{8}=2\\$

$\displaystyle [x_{i-1},x_i]=[-4,-2],[-2,0],[0,2],[2,4],[4,6],[6,8],[8,10],[10,12]$

We then choose the midpoint in each interval:

$\displaystyle \bar{x_i}=-3,-1,1,3,5,7,9,11$

Then we find the function value at each point.

$\displaystyle f(\bar{x_i})=-131,-21,1,31,165,499,1129,2151$

We then substitute these values into the Riemann Sum formula.

$\displaystyle A\approx\sum_{i=1}^nf(\bar{x_i})\Delta x=2(-131-21+1+31+165+499+1129+2151)=7648$

$\displaystyle \Delta x =\frac{ 6-0}{3} = 2$

Thus, our intervals are $\displaystyle 0$ to $\displaystyle 2$, $\displaystyle 2$ to $\displaystyle 4$, and $\displaystyle 4$ to $\displaystyle 6$.

The midpoints of each interval are, respectively, $\displaystyle 1$, $\displaystyle 3$, and $\displaystyle 5$.

Next, we evaluate the function at each midpoint.

$\displaystyle f(x)=\frac{1}{2}x^3-7x^2+14x+60$

$\displaystyle f(1)=67.5$

$\displaystyle f(3)=52.5$

$\displaystyle f(5)=17.5$

Finally, we calculate the estimated area using these values and $\displaystyle \Delta x = 2$

$\displaystyle 67.5\cdot 2+52.5\cdot 2+17.5\cdot 2$

$\displaystyle =275$

$\displaystyle \Delta x =\frac{ 3-0}{3} = 1$

Thus, our intervals are $\displaystyle 0$ to $\displaystyle 1$, $\displaystyle 1$ to $\displaystyle 2$, and $\displaystyle 2$ to $\displaystyle 3$.

The midpoints of each interval are, respectively, $\displaystyle 0.5$, $\displaystyle 1.5$, and $\displaystyle 2.5$.

Next, use the data table to take the values the function at each midpoint.

$\displaystyle f(0.5)=0$

$\displaystyle f(1.5)=9$

$\displaystyle f(2.5)=5$

Finally, we calculate the estimated area using these values and $\displaystyle \Delta x = 1$.

$\displaystyle 0\cdot 1+9\cdot 1+5\cdot 1$

$\displaystyle =14$

Midpoint Riemann sum approximations are solved using the formula

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

where $\displaystyle n$ is the number of subintervals and $\displaystyle f(m_{i})$ is the function evaluated at the midpoint.

For this problem, $\displaystyle (\frac{b-a}{n})=(\frac{8-1}{10})=0.7$.  

The approximate value at each midpoint is below.

The sum of all the approximate midpoints values is $\displaystyle 20.6068259$, therefore

$\displaystyle (\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ] =0.7(20.6068259)\approx14.4248$

Midpoint Riemann sum approximations are solved using the formula

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

where $\displaystyle n$ is the number of subintervals and $\displaystyle f(m_{i})$ is the function evaluated at the midpoint.

For this problem, $\displaystyle (\frac{b-a}{n})=(\frac{7-3}{10})=0.4$.  

The approximate value at each midpoint is below.

The sum of all the approximate midpoints values is $\displaystyle 2.116738$, therefore

$\displaystyle (\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ] =0.4(2.116738)\approx0.8467$

Midpoint Riemann sum approximations are solved using the formula

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

where $\displaystyle n$ is the number of subintervals and $\displaystyle f(m_{i})$ is the function evaluated at the midpoint.

For this problem, $\displaystyle (\frac{b-a}{n})=(\frac{\pi /2-0}{10})=\pi/20\approx0.1571$.  

The approximate value at each midpoint is below.

The sum of all the approximate midpoints values is $\displaystyle 6.37275$, therefore

$\displaystyle (\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ] =0.1571(6.37275)\approx1$