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 \(\displaystyle \int_{0}^{3}x^3sin(x^3)dx\)

So the interval is \(\displaystyle [0,3]\), the subintervals have length \(\displaystyle \frac{3-0}{4}=0.75\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [0.375,1.125,1.875,2.625]\)

\(\displaystyle {\int_{0}^{3}x^3sin(x^3)dx\approx (0.75)[0.375^3sin(0.375^3)+1.125^3sin(1.125^3)+1.875^3sin(1.875^3)+2.625^3sin(2.625^3)]}\)

\(\displaystyle \int_{0}^{3}x^3sin(x^3)dx\approx -6.802\)

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 \(\displaystyle \int_{0.1}^{0.22}\frac{tan(x)}{x^2}dx\)

So the interval is \(\displaystyle [0.1,0.22]\), the subintervals have length \(\displaystyle \frac{0.22-0.1}{3}=0.04\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [0.12,0.16,0.2]\)

\(\displaystyle \int_{0.1}^{0.22}\frac{tan(x)}{x^2}dx\approx (0.04)[\frac{tan(0.12)}{0.12^2}+\frac{tan(0.16)}{0.16^2}+\frac{tan(0.2)}{0.2^2}]\)

\(\displaystyle \int_{0.1}^{0.22}\frac{tan(x)}{x^2}dx\approx 0.790\)

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 \(\displaystyle \int_{1.5}^{1.8}\frac{x^4}{tan(x)}dx\)

So the interval is \(\displaystyle [1.5,1.8]\), the subintervals have length \(\displaystyle \frac{1.8-1.5}{3}=0.1\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [1.55,1.65,1.75]\)

\(\displaystyle \int_{1.5}^{1.8}\frac{x^4}{tan(x)}dx\approx (0.1)[\frac{1.55^4}{tan(1.55)}+\frac{1.65^4}{tan(1.65)}+\frac{1.75^4}{tan(1.75)}]\)

\(\displaystyle \int_{1.5}^{1.8}\frac{x^4}{tan(x)}dx\approx -0.217\)

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 \(\displaystyle \int_{7.8}^{7.86}\frac{x^3}{2tan(x)}dx\)

So the interval is \(\displaystyle [7.8,7.86]\), the subintervals have length \(\displaystyle \frac{7.86-7.8}{3}=0.02\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [7.81,7.83,7.85]\)

\(\displaystyle \int_{7.8}^{7.86}\frac{x^3}{2tan(x)}dx\approx (0.02)[\frac{7.81^3}{2tan(7.81)}+\frac{7.83^3}{2tan(7.83)}+\frac{7.85^3}{2tan(7.85)}]\)

\(\displaystyle \int_{7.8}^{7.86}\frac{x^3}{2tan(x)}dx\approx 0.344\)

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 \(\displaystyle \int_{1.5}^{2.1}\frac{x^x}{tan(x)}dx\)

So the interval is \(\displaystyle [1.5,2.1]\), the subintervals have length \(\displaystyle \frac{2.1-1.5}{3}=0.2\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [1.6,1.8,2]\)

\(\displaystyle \int_{1.5}^{2.1}\frac{x^{x}}{tan(x)}dx\approx (0.2)[\frac{1.6^{1.6}}{tan(1.6)}+\frac{1.8^{1.8}}{tan(1.8)}+\frac{2^{2}}{tan(2)}]\)

\(\displaystyle \int_{1.5}^{2.1}\frac{x^{x}}{tan(x)}dx\approx -0.513\)

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 \(\displaystyle \int_{0}^{6}(x!)^{-\sqrt{x}}dx\)

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 \(\displaystyle [1,3,5]\)

\(\displaystyle \int_{0}^{6}(x!)^{-\sqrt{x}}dx\approx (2)[(1!)^{-\sqrt{1}}+(3!)^{-\sqrt{3}}+(5!)^{-\sqrt{5}}]\)

\(\displaystyle \int_{0}^{6}(x!)^{-\sqrt{x}}dx\approx 2.090\)

To find the average of a function over a given interval of values \(\displaystyle [a,b]\), the most precise method is to use an integral as follows:

\(\displaystyle \frac{1}{b-a}\int_{a}^{b} f(x)dx\)

Now for functions that are difficult or impossible to integrate,a Riemann sum can be used to approximate the value. 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 equal to the subinterval length  \(\displaystyle \frac{b-a}{n}\), and variable heights \(\displaystyle f(x_i)\), which depend on the function value at a given point  \(\displaystyle x_i\).

Now note that when using the method of Riemann sums to find an average value of a function, the expression changes:

\(\displaystyle \frac{1}{b-a}\int_a^b f(x)dx \approx (\frac{1}{b-a})\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\)

\(\displaystyle \frac{1}{b-a}\int_a^b f(x)dx \approx \frac{1}{n}(f(x_1)+f(x_2)+...+f(x_n))\)

We're asked to approximate the average of \(\displaystyle f(x)=14sin^4(x)\) over the interval \(\displaystyle (0,6)\)

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

\(\displaystyle \frac{1}{6}\int_{0}^{6}14sin^4(x)dx\approx (\frac{1}{3})[14sin^4(1)+14sin^4(3)+14sin^4(5)]\)

\(\displaystyle \frac{1}{6}\int_{0}^{6}14sin^4(x)dx\approx 6.287\)

To find the average of a function over a given interval of values \(\displaystyle [a,b]\), the most precise method is to use an integral as follows:

\(\displaystyle \frac{1}{b-a}\int_{a}^{b} f(x)dx\)

Now for functions that are difficult or impossible to integrate,a Riemann sum can be used to approximate the value. 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 equal to the subinterval length  \(\displaystyle \frac{b-a}{n}\), and variable heights \(\displaystyle f(x_i)\), which depend on the function value at a given point  \(\displaystyle x_i\).

Now note that when using the method of Riemann sums to find an average value of a function, the expression changes:

\(\displaystyle \frac{1}{b-a}\int_a^b f(x)dx \approx (\frac{1}{b-a})\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\)

\(\displaystyle \frac{1}{b-a}\int_a^b f(x)dx \approx \frac{1}{n}(f(x_1)+f(x_2)+...+f(x_n))\)

We're asked to approximate the average of \(\displaystyle f(x)=2^\frac{x}{3}\) over the interval \(\displaystyle (0,10)\)

The subintervals have length \(\displaystyle \frac{10-0}{5}=2\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [1,3,5,7,9]\)

\(\displaystyle \frac{1}{10}\int_{0}^{10}2^{\frac{x}{3}}dx\approx (\frac{1}{5})[2^{\frac{1}{3}}+2^{\frac{3}{3}}+2^{\frac{5}{3}}+2^{\frac{7}{3}}+2^{\frac{9}{3}}]\)

\(\displaystyle \frac{1}{10}\int_{0}^{10}2^\frac{x}{3}dx\approx 3.895\)

To find the average of a function over a given interval of values \(\displaystyle [a,b]\), the most precise method is to use an integral as follows:

\(\displaystyle \frac{1}{b-a}\int_{a}^{b} f(x)dx\)

Now for functions that are difficult or impossible to integrate,a Riemann sum can be used to approximate the value. 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 equal to the subinterval length  \(\displaystyle \frac{b-a}{n}\), and variable heights \(\displaystyle f(x_i)\), which depend on the function value at a given point  \(\displaystyle x_i\).

Now note that when using the method of Riemann sums to find an average value of a function, the expression changes:

\(\displaystyle \frac{1}{b-a}\int_a^b f(x)dx \approx (\frac{1}{b-a})\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\)

\(\displaystyle \frac{1}{b-a}\int_a^b f(x)dx \approx \frac{1}{n}(f(x_1)+f(x_2)+...+f(x_n))\)

We're asked to approximate the average of \(\displaystyle f(x)=tan(\frac{1}{tan(x)})\) over the interval \(\displaystyle (0.5,3.5)\)

The subintervals have length \(\displaystyle \frac{3.5-0.5}{3}=1\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [1,2,3]\)

\(\displaystyle \frac{1}{3}\int_{0.5}^{3.5}tan(\frac{1}{tan(x)})dx\approx (\frac{1}{3})[tan(\frac{1}{tan(1)})+tan(\frac{1}{tan(2)})+tan(\frac{1}{tan(3)})]\)

\(\displaystyle \frac{1}{3}\int_{0.5}^{3.5}tan(\frac{1}{tan(x)})dx\approx -0.214\)

To find the average of a function over a given interval of values \(\displaystyle [a,b]\), the most precise method is to use an integral as follows:

\(\displaystyle \frac{1}{b-a}\int_{a}^{b} f(x)dx\)

Now for functions that are difficult or impossible to integrate,a Riemann sum can be used to approximate the value. 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 equal to the subinterval length  \(\displaystyle \frac{b-a}{n}\), and variable heights \(\displaystyle f(x_i)\), which depend on the function value at a given point  \(\displaystyle x_i\).

Now note that when using the method of Riemann sums to find an average value of a function, the expression changes:

\(\displaystyle \frac{1}{b-a}\int_a^b f(x)dx \approx (\frac{1}{b-a})\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\)

\(\displaystyle \frac{1}{b-a}\int_a^b f(x)dx \approx \frac{1}{n}(f(x_1)+f(x_2)+...+f(x_n))\)

We're asked to approximate the average of \(\displaystyle f(x)=sec(cos(x^2))\) over the interval \(\displaystyle (1.5,4.5)\)

The subintervals have length \(\displaystyle \frac{4.5-1.5}{3}=1\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [2,3,4]\)

\(\displaystyle \frac{1}{3}\int_{1.5}^{4.5}sec(cos(x^2))dx\approx (\frac{1}{3})[sec(cos(2^2))+sec(cos(3^2))+sec(cos(4^2))]\)

\(\displaystyle \frac{1}{3}\int_{1.5}^{4.5}sec(cos(x^2))dx\approx 1.543\)