The form of a Riemann sum follows:

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

Since two mid points are asked for, divide the interval into two: $\displaystyle [1,2.5][2.5,4]$, and find the mid points of those: $\displaystyle [1.75,3.25]$.

Therefore, the approximation of the integral is:

$\displaystyle \frac{4-1}{2}((2+2e^{(1.75-1)^2})+(2+2e^{(3.25-1)^2}))$

$\displaystyle \frac{3}{2}(4+2e^{(0.75)^2}+2e^{(2.25)^2})$

$\displaystyle 485.22$

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 two intervals of length $\displaystyle \frac{4-0}{2}=2$, the midpoints are $\displaystyle [1,3]$.

Therefore, the approximation of the integral is:

$\displaystyle \frac{4-0}{2}(cos(e^1)cos(e^3))$

$\displaystyle -1.16627...$

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 $\displaystyle [2,8]$ can be divided into three intervals of length $\displaystyle \frac{8-2}{3}=2$

$\displaystyle [2,4][4,6][6,8]$, with midpoints of $\displaystyle [3,5,7]$.

Therefore, the approximation of the integral is:

$\displaystyle \frac{8-2}{3}(3^2+5^2+7^2)$

$\displaystyle 166$

The form of a Riemann sum follows:

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

It can be applied to combinations of functions, i.e. $\displaystyle f(x)=h(x)-g(x)$

To find the interval for this problem, find the intersections of the two functions, i.e. the values of $\displaystyle x$ that satisfy

$\displaystyle x=\frac{x^2}{9}$

$\displaystyle x\left(\frac{x}{9}-1\right)=0$

$\displaystyle x=0,9$

Knowing this interval, it can be divided into three intervals of length $\displaystyle \frac{9-0}{3}=3$:

$\displaystyle [0,3][3,6][6,9]$, with midpoints $\displaystyle [1.5,4.5,7.5]$.

So the approximation of the area between functions is:

$\displaystyle \frac{9-0}{3}\left(\left(1.5-\frac{1.5^2}{9}\right)+\left(4.5-\frac{4.5^2}{9}\right)+\left(7.5-\frac{7.5^2}{9}\right)\right)$

$\displaystyle 14.25$

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 $\displaystyle [0,\sqrt{\pi}]$ can be divided into four intervals of length 

$\displaystyle \frac{\sqrt{\pi}-0}{4}=\frac{\sqrt{\pi}}{4}$.

$\displaystyle \left[0,\frac{\sqrt{\pi}}{4}\right]\left[\frac{\sqrt{\pi}}{4},\frac{\sqrt{\pi}}{2}\right]\left[\frac{\sqrt{\pi}}{2},\frac{3\sqrt{\pi}}{4}\right]\left[\frac{3\sqrt{\pi}}{4},\pi\right]$

With the midpoints

$\displaystyle \left[\frac{\sqrt{\pi}}{8},\frac{3\sqrt{\pi}}{8},\frac{5\sqrt{\pi}}{8},\frac{7\sqrt{\pi}}{8}\right]$.

Therefore, the approximation of the integral is:

$\displaystyle \frac{\sqrt{\pi}-0}{4}\left(sin\left(\left(\frac{\sqrt{\pi}}{8}\right)^2\right)+sin\left(\left(\frac{3\sqrt{\pi}}{8}\right)^2\right)+sin\left(\left(\frac{5\sqrt{\pi}}{8}\right)^2\right)+sin\left(\left(\frac{7\sqrt{\pi}}{8}\right)^2\right)\right)$

$\displaystyle 0.926$

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 $\displaystyle [0,2\pi]$ can be divided into four intervals of length $\displaystyle \frac{2\pi-0}{4}=\frac{\pi}{2}$

with mid points $\displaystyle \left[\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\right]$.

The Riemann sum is then

$\displaystyle \frac{2\pi-0}{4}(e^{cos(\frac{\pi}{4})}+e^{cos(\frac{3\pi}{4})}+e^{cos(\frac{5\pi}{4})}+e^{cos(\frac{7\pi}{4})})$

$\displaystyle \pi(e^{\frac{1}{\sqrt2}}+e^{-\frac{1}{\sqrt2}})$

First, graph the equation to see if it is above the x-axis - positive, or below - negative. Then we can visualize the intervals and their midpoints:

Dividing the region from $\displaystyle x=-2$ to $\displaystyle x=5$ into 7 intervals gives each a length $\displaystyle \Delta x$ of exactly 1. 

We will be calculating the sum $\displaystyle A \approx \sum_{7}^{n=1}f(x_{n})*\Delta x$ where $\displaystyle x{_{n}}$ is the nth interval's midpoint. Since $\displaystyle \Delta x$ is jus 1, in this case we're just adding together this sum:

$\displaystyle f(-1.5)+f(-0.5)+f(0.5)+f(1.5)+f(2.5)+f(3.5)+f(4.5)$

Evaluating at these points and adding the results together yields $\displaystyle 24.9375$

We are looking to approximate the area between 0 and 3, and we are dividing it into 5 intervals, so each interval will have a length of $\displaystyle \Delta x=\frac{3-0}{5}=0.6$.

The midpoints of the 5 intervals are at $\displaystyle x=0.3, 0.9, 1.5, 2.1, 2.7$, so we'll be evaluating the function for these values of x.

The formula for the Riemann midpoint sum is $\displaystyle A \approx \sum_{n=1}^{5}f(x_{n})*\Delta x$.

Note that you get the same answer regardless if you multiply each individual $\displaystyle f(x_{n})$ times $\displaystyle \Delta x$ or calculate the sum first then multiply by 0.6.

In this case, adding together the heights at the midpoints and multiplying by 0.6 yields $\displaystyle 11.7345$. 

Midpoint sums are given by the formula

$\displaystyle \int_{a}^{b}f(x)\approx \frac{b-a}{n}*[f(x_{1})+f(x_2)]$,

where n is the number of steps and the x values are the midpoints of the rectangles. 

Since we have 2 steps, there would be 2 rectangles, from $\displaystyle x: (1,1.5)$ and the other one $\displaystyle x:(1.5, 2)$. The heights of these rectangles are calculated at the midpoint of the two ends, which occur at $\displaystyle x=1.25$ and $\displaystyle x=1.75$

In our case, the rectangle midpoints are at 1.25 and 1.75, since we only have 2 steps. 

Therefore, the value is

$\displaystyle \frac{2-1}{2}*[ln(sin(1.25)+ln(sin(1.75))]=\frac{[ln(sin(1.25)+ln(sin(1.75))]}{2}$

The general formula for a Reimann with n intervals from a to b is:

$\displaystyle \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

The term $\displaystyle \frac{b-a}{n}$ is also the length of an interval, so for this problem, an interval has a length of:

$\displaystyle \frac{2\pi-0}{4}=\frac{\pi}{2}$

And the four midpoints are thus:

$\displaystyle \frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$

And the midpoint Reimann sum is:

$\displaystyle \frac{\pi}{2}\left(cos\left(\left(\frac{\pi}{4}\right)^2+\left(\frac{\pi}{4}\right)+1\right)+cos\left(\left(\frac{3\pi}{4}\right)^2+\left(\frac{3\pi}{4}\right)+1\right)+cos\left(\left(\frac{5\pi}{4}\right)^2+\left(\frac{5\pi}{4}\right)+1\right)+cos\left(\left(\frac{7\pi}{4}\right)^2+\left(\frac{7\pi}{4}\right)+1\right)\right)$

$\displaystyle \\ \frac{\pi}{2}\left(cos\left(\frac{\pi^2}{16}+\left(\frac{\pi}{4}\right)+1\right)+cos\left(\frac{9\pi^2}{16}\right)+\left(\frac{3\pi}{4}\right)+1\right)+cos\left(\left(\frac{25\pi^2}{16}\right)+\left(\frac{5\pi}{4}\right)+1\right)+cos\left(\left(\frac{49\pi^2}{16}\right)+\left(\frac{7\pi}{4}\right)+1\right)\right)$