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 evaluating $\displaystyle \int_0^{0.15} \tan(\cos(\sin(x^3)))$

So the interval is $\displaystyle [0,0.15]$, the subintervals have length $\displaystyle \frac{0.15-0}{3}=0.05$, and since we are using the midpoints of each interval, the x-values are $\displaystyle [0.025,0.075,0.125]$

$\displaystyle {\int_0^{0.15} \tan(\cos(\sin(x^3))) \approx (0.05)[\tan(\cos(\sin(0.025^3)))+\tan(\cos(\sin(0.075^3)))+\tan(\cos(\sin(0.125^3)))}]$

$\displaystyle {\int_0^{0.15} \tan(\cos(\sin(x^3))) \approx 0.234$

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 asked to approximate $\displaystyle \int_{10}^{40}e^{\tan(x)}dx$

So the interval is $\displaystyle [10,40]$, the subintervals have length $\displaystyle \frac{40-10}{3}=10$, and since we are using the midpoints of each interval, the x-values are $\displaystyle [15,25,35]$

$\displaystyle \int_{10}^{40}e^{\tan(x)}dx\approx (10)[e^{\tan(15)}+e^{\tan(25)}+e^{\tan(35)}]$

$\displaystyle \int_{10}^{40}e^{\tan(x)}dx\approx 49.16$

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 asked to approximate $\displaystyle \int_{-4}^{4} \tan(x^3+x+1)dx$

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

$\displaystyle \int_{-4}^{4} \tan(x^3+x+1)dx \approx (4)[\tan((-2)^3+(-2)+1)+\tan(2^3+2+1)]$

$\displaystyle \int_{-4}^{4} \tan(x^3+x+1)dx \approx -901.99$

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 asked to approximate $\displaystyle \int_{-1}^{5} \frac{1}{\cos^2(x^2)}dx$

So the interval is $\displaystyle [-1,5]$, the subintervals have length $\displaystyle \frac{5-(-1)}{3}=2$, and since we are using the midpoints of each interval, the x-values are $\displaystyle [0,2,4]$

$\displaystyle \int_{-1}^{5} \frac{1}{\cos^2(x^2)}dx \approx (2)[\frac{1}{\cos^2(0^2)}+\frac{1}{\cos^2(2^2)}+\frac{1}{\cos^2(4^2)}]$

$\displaystyle \int_{-1}^{5} \frac{1}{\cos^2(x^2)}dx \approx 8.86$

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 asked to approximate $\displaystyle \int_{0.19\pi}^{0.37\pi} \tan(\tan(x))dx$

So the interval is $\displaystyle [0.19\pi,0.37\pi]$, the subintervals have length $\displaystyle \frac{0.37\pi-0.19\pi}{3}=0.06\pi$, and since we are using the midpoints of each interval, the x-values are $\displaystyle [0.22\pi,0.28\pi,0.34\pi]$

$\displaystyle \int_{0.19\pi}^{0.37\pi} \tan(\tan(x))dx \approx (0.06\pi)[\tan(\tan(0.22\pi))+\tan(\tan(0.28\pi))+\tan(\tan(0.34\pi))]$

$\displaystyle \int_{0.19\pi}^{0.37\pi} \tan(\tan(x))dx \approx -0.041$

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 asked to approximate $\displaystyle \int_{e}^{e^2}\ln(x^2)dx$

So the interval is $\displaystyle [e,e^2]$, the subintervals have length $\displaystyle \frac{e^2-e}{3}$, and since we are using the midpoints of each interval, the x-values are $\displaystyle [e+\frac{e^2-e}{6},e+\frac{e^2-e}{6}+\frac{e^2-e}{3},e+\frac{e^2-e}{6}+\frac{2(e^2-e)}{3}]$

or, more simply,

$\displaystyle [\frac{e^2+5e}{6},\frac{e^2+e}{2},\frac{5e^2+e}{6}]$

It is possible, of course, to simply use numerical approximations of each of these terms.

$\displaystyle \int_{e}^{e^2}\ln(x^2)dx \approx (\frac{e^2-e}{3})[\ln((\frac{e^2+5e}{6})^2)+\ln((\frac{e^2+e}{2})^2)+\ln((\frac{5e^2+e}{6})^2)]$

$\displaystyle \int_{e}^{e^2}\ln(x^2)dx \approx 14.82$

To make 4 intervals, we need 5 boundary lines. In this case, they are the vertical lines at $\displaystyle \small x = 0, 2, 4, 6, 8$. So the midpoints of each interval are at $\displaystyle \small x = 1, 3, 5, 7$, respectively.

The area of each region is found by $\displaystyle \small f(x)\cdot \Delta x$, where $\displaystyle \small \Delta x$ is the width of the interval (here, 2). So we simply need to find the functional value at each $\displaystyle \small x$-coordinate of the midpoints above, multiply by 2, and sum. Equivalently, we could sum the functional values, then double the sum.

$\displaystyle \small f(1) = 14$

$\displaystyle \small f(3) = 48$

$\displaystyle \small f(5) = 106$

$\displaystyle \small f(7) = 188$

Their sum is $\displaystyle \small 356$, and twice that is $\displaystyle \small 712$.

The wrong answer $\displaystyle \small 504$ comes from using $\displaystyle \small x = 0, 2, 4, 6$ as the midpoints.

The wrong answer $\displaystyle \small 968$ comes from using $\displaystyle \small x = 2, 4, 6, 8$ as the midpoints.

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 asked to approximate $\displaystyle \int_{0.2}^{0.5}e^{(\frac{1}{2})^{-\frac{1}{x}}}dx$

So the interval is $\displaystyle [0.2,0.5]$, the subintervals have length $\displaystyle \frac{0.5-0.2}{3}=0.1$, and since we are using the midpoints of each interval, the x-values are $\displaystyle [0.25,0.35,0.45]$

$\displaystyle \int_{0.2}^{0.5}e^{(\frac{1}{2})^{-\frac{1}{x}}}dx \approx (0.1)[e^{(\frac{1}{2})^{-\frac{1}{0.25}}}+e^{(\frac{1}{2})^{-\frac{1}{0.35}}}+e^{(\frac{1}{2})^{-\frac{1}{0.45}}}]$

$\displaystyle \int_{0.2}^{0.5}e^{(\frac{1}{2})^{-\frac{1}{x}}}dx \approx 8.89 \cdot 10^5$

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 asked to approximate $\displaystyle \int_{0.8}^{3.2} \frac{0.5}{tan(x^3)}dx$

So the interval is $\displaystyle [0.8,3.2]$, the subintervals have length $\displaystyle \frac{3.2-0.8}{3}=0.8$, and since we are using the midpoints of each interval, the x-values are $\displaystyle [1.2,2.0,2.8]$

$\displaystyle \int_{0.8}^{3.2} \frac{0.5}{tan(x^3)}dx\approx (0.8)[\frac{0.5}{tan(1.2^3)}+\frac{0.5}{tan(2^3)}+\frac{0.5}{tan(2.8^3)}]$

$\displaystyle \int_{0.8}^{3.2} \frac{0.5}{tan(x^3)}dx\approx -10.33$

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 asked to approximate $\displaystyle \int_{0.7}^{1.3}\frac{sin(x)}{sin(x^3)}dx$

So the interval is $\displaystyle [0.7,1.3]$, the subintervals have length $\displaystyle \frac{1.3-0.7}{3}=0.2$, and since we are using the midpoints of each interval, the x-values are $\displaystyle [0.8,1.0,1.2]$

$\displaystyle \int_{0.7}^{1.3}\frac{sin(x)}{sin(x^3)}dx\approx (0.2)[\frac{sin(0.8)}{sin(0.8^3)}+\frac{sin(1.0)}{sin(1.0^3)}+\frac{sin(1.2)}{sin(1.2^3)}]$

$\displaystyle \int_{0.7}^{1.3}\frac{sin(x)}{sin(x^3)}dx\approx 0.681$