A Reimann sum integral approximation over an interval $\displaystyle [a,b]$ with $\displaystyle n$ subintervals follows the form:

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

It is essentially a sum of $\displaystyle n$ rectangles each with a base of length $\displaystyle \frac{b-a}{n}$ and variable heights $\displaystyle f(x_i)$, which depend on the function value at a given point  $\displaystyle x_i$.

So the interval is $\displaystyle [\pi,4\pi]$, the subintervals have length $\displaystyle \frac{4\pi-\pi}{3}=\pi$, and since we are using the midpoints of each interval, the x-values are $\displaystyle [0.5\pi,1.5\pi,2.5\pi]$

$\displaystyle \int_{\pi}^{4\pi} \frac{sin(x)}{x^2}dx \approx \pi[\frac{sin(0.5\pi)}{(0.5\pi)^2}+\frac{sin(1.5\pi)}{(1.5\pi)^2}+\frac{sin(2.5\pi)}{(2.5\pi)^2}]$

$\displaystyle \int_{\pi}^{4\pi} \frac{sin(x)}{x^2}dx \approx \frac{836}{225\pi}$

A Reimann sum integral approximation over an interval $\displaystyle [a,b]$ with $\displaystyle n$ subintervals follows the form:

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

It is essentially a sum of $\displaystyle n$ rectangles each with a base of length $\displaystyle \frac{b-a}{n}$ and variable heights $\displaystyle f(x_i)$, which depend on the function value at a given point  $\displaystyle x_i$.

Our integral is $\displaystyle \int_5^{11} \frac{ln^2(x)}{x^3}$

So the interval is $\displaystyle [5,11]$, the subintervals have length $\displaystyle \frac{11-5}{3}=2$, and since we are using the midpoints of each interval, the x-values are $\displaystyle [6,8,10]$

$\displaystyle \int_5^{11} \frac{ln^2(x)}{x^3} \approx 2[\frac{ln^2(6)}{6^3}+\frac{ln^2(8)}{8^3}+\frac{ln^2(10)}{10^3}]$

$\displaystyle \int_5^{11} \frac{ln^2(x)}{x^3} \approx 0.0572$

A Reimann sum integral approximation over an interval $\displaystyle [a,b]$ with $\displaystyle n$ subintervals follows the form:

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

It is essentially a sum of $\displaystyle n$ rectangles each with a base of length $\displaystyle \frac{b-a}{n}$ and variable heights $\displaystyle f(x_i)$, which depend on the function value at a given point  $\displaystyle x_i$.

The function given is $\displaystyle \int_{10}^{16}\frac{cos(x^2)}{x}dx$

So the interval is $\displaystyle [10\pi,16\pi]$, the subintervals have length $\displaystyle \frac{16\pi-10\pi}{3}=2\pi$, and since we are using the midpoints of each interval, the x-values are $\displaystyle [11\pi,13\pi,15\pi]$

$\displaystyle \int_{10}^{16}\frac{cos(x^2)}{x}dx \approx 2\pi[\frac{cos((11\pi)^2)}{11\pi}+\frac{cos((13\pi)^2)}{13\pi}+\frac{cos((15\pi)^2)}{15\pi}]$

A Reimann sum integral approximation over an interval $\displaystyle [a,b]$ with $\displaystyle n$ subintervals follows the form:

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

It is essentially a sum of $\displaystyle n$ rectangles each with a base of length $\displaystyle \frac{b-a}{n}$ and variable heights $\displaystyle f(x_i)$, which depend on the function value at a given point  $\displaystyle x_i$.

Our function is $\displaystyle \int_0^6 (x^2-x)dx$

So the interval is $\displaystyle [0,6]$, the subintervals have length $\displaystyle \frac{6-0}{3}=2$, and since we are using the midpoints of each interval, the x-values are $\displaystyle [1,3,5]$

$\displaystyle \int_0^6 (x^2-x)dx\approx 2[(1^2-1)+(3^2-3)+(5^2-5)]$

$\displaystyle \int_0^6 (x^2-x)dx \approx 52$

A Reimann sum integral approximation over an interval $\displaystyle [a,b]$ with $\displaystyle n$ subintervals follows the form:

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

It is essentially a sum of $\displaystyle n$ rectangles each with a base of length $\displaystyle \frac{b-a}{n}$ and variable heights $\displaystyle f(x_i)$, which depend on the function value at a given point  $\displaystyle x_i$.

Our integral is $\displaystyle \int_0^{12} \frac{x+4}{x+8}dx$

So the interval is $\displaystyle [0,12]$, the subintervals have length $\displaystyle \frac{12-0}{3}=4$, and since we are using the midpoints of each interval, the x-values are $\displaystyle [2,6,10]$

$\displaystyle \int_0^{12} \frac{x+4}{x+8}dx \approx 4[\frac{2+4}{2+8}+\frac{6+4}{6+8}+\frac{10+4}{10+8}]$

$\displaystyle \int_0^{12} \frac{x+4}{x+8}dx \approx \frac{2636}{315}$

A Riemann sum integral approximation over an interval $\displaystyle [a,b]$ with $\displaystyle n$ subintervals follows the form:

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

It is essentially a sum of $\displaystyle n$ rectangles each with a base of length $\displaystyle \frac{b-a}{n}$ and variable heights $\displaystyle f(x_i)$, which depend on the function value at a given point  $\displaystyle x_i$.

The integral we are given is $\displaystyle \int_1^4 x^{x-1}dx$

So the interval is $\displaystyle [1,4]$, the subintervals have length $\displaystyle \frac{4-1}{3}=1$, and since we are using the midpoints of each interval, the x-values are $\displaystyle [1.5,2.5,3.5]$

$\displaystyle \int_1^4 x^{x-1}dx \approx 1(1.5^{1.5-1}+2.5^{2.5-1}+3.5^{3.5-1})$

$\displaystyle \int_1^4 x^{x-1}dx \approx (1.5^{0.5}+2.5^{1.5}+3.5^{2.5})$

$\displaystyle \int_1^4 x^{x-1}dx \approx 28.1$

The more points/intervals that are selected, the closer the Riemann sum approximation becomes to the actual integral value, so n=16 will give a closer approximation than n=8:

Now, assuming the amount of points/intervals are fixed, but the actual widths can differ, it would be prudent to have narrower intervals at those regions where the graph is steepest. Relatively flatter regions of the function do not lead to the approximation overshooting or undershooting as much as the steeper areas, as can be seen in the above figure.

Since the function $\displaystyle x^x$ becomes progressively steeper, a better approximation for a fixed number of intervals would have the intervals grow thinner as $\displaystyle x$ increases.

A Riemann sum integral approximation over an interval $\displaystyle [a,b]$ with $\displaystyle n$ subintervals follows the form:

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

It is essentially a sum of $\displaystyle n$ rectangles each with a base of length $\displaystyle \frac{b-a}{n}$ and variable heights $\displaystyle f(x_i)$, which depend on the function value at a given point  $\displaystyle x_i$.

The integral we're given is $\displaystyle \int_{10\pi}^{16\pi}e^{\frac{cos(x)}{x}}dx$

So the interval is $\displaystyle [10\pi,16\pi]$, the subintervals have length $\displaystyle \frac{16\pi-10\pi}{3}=2\pi$, and since we are using the midpoints of each interval, the x-values are $\displaystyle [11\pi,13\pi,15\pi]$

$\displaystyle \int_{10\pi}^{16\pi}e^{\frac{cos(x)}{x}}dx \approx 2\pi[e^{\frac{cos(11\pi)}{11\pi}}+e^{\frac{cos(13\pi)}{13\pi}}}+e^{\frac{cos(15\pi)}{15\pi}}}]$

$\displaystyle \int_{10\pi}^{16\pi}e^{\frac{cos(x)}{x}}dx \approx 18.39$

To find the area between two functions, the method traditionally used is to find the difference of the integral values between the larger and smaller function over the specified interval. However, neither of these functions can be integrated, so a method of approximation is useful.

A Riemann sum integral approximation over an interval $\displaystyle [a,b]$ with $\displaystyle n$ subintervals follows the form:

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

It is essentially a sum of $\displaystyle n$ rectangles each with a base of length $\displaystyle \frac{b-a}{n}$ and variable heights $\displaystyle f(x_i)$, which depend on the function value at a given point  $\displaystyle x_i$.

We're told that we're integrating from 1 to 2.5, so the interval is $\displaystyle [1,2.5]$, the subintervals have length $\displaystyle \frac{2.5-1}{3}=0.5$, and since we are using the midpoints of each interval, the x-values are $\displaystyle [1.25,1.75,2.25]$.

As $\displaystyle f(x)=x^{2x}$ is greater than $\displaystyle g(x)=x^x$ over the given interval, our approximation looks as follows:

$\displaystyle {\int_{1}^{2.5}(x^{2x}-x^x)dx \approx 0.5[(1.25^{2(1.25)}-1.25^{1.25})+(1.75^{2(1.75)}-1.75^{1.75})+(2.25^{2(2.25)}-2.25^{2.25})]}$

$\displaystyle {\int_{1}^{2.5}(x^{2x}-x^x)dx \approx 18.55$

The more points/intervals that are selected, the closer the Reimann sum approximation becomes to the actual integral value, so n=20 will give a closer approximation than n=10:

Now, assuming the amount of points/intervals are fixed, but the actual widths can differ, it would be prudent to have narrower intervals at those regions where the graph is steepest. Relatively flatter regions of the function do not lead to the approximation overshooting or undershooting as much as the steeper areas, as can be seen in the above figure.

Consider the function $\displaystyle \frac{e^x}{x^2}$ over the interval $\displaystyle x=[10,100]$. The steepness of this graph at a point can be found by taking the function's derivative.

The quotient rule of derivatives states:

$\displaystyle d\frac{u}{v}=\frac{vdu-udv}{v^2}$

So the derivative is

$\displaystyle \frac{x^2e^x-2xe^x}{x^4}$

$\displaystyle \frac{e^x(x-2)}{x^3}$

It might be intuitive to see that the steepness is positive and that it gets progressively greater over the specified interval, although to be precise, we may take the derivative once more to find the rate of change of this steepness and see if it's postive or negative:

$\displaystyle \frac{x^4(2xe^x+x^2e^x-2e^x-2xe^x)-4x^3(x^2e^x-2xe^x)}{x^8}$

$\displaystyle \frac{x^6e^x+6x^4e^x-4x^5e^x}{x^8}$

$\displaystyle \frac{e^x(x^2-4x+6)}{x^4}$

Since this derivative is positive over the specified interval, , and since the slope was shown to be positive as well, the slope is ever-increasing and growing steeper.

The intervals should in turn grow increasingly thinner.