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 \(\displaystyle f(x)=x^{x^2}\) over the interval \(\displaystyle (1,1.8)\)

The subintervals have length \(\displaystyle \frac{1.8-1}{4}=0.2\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [1.1,1.3,1.5,1.7]\)

\(\displaystyle \frac{1}{0.8}\int_{1}^{1.8}x^{x^2}dx\approx (\frac{1}{4})[1.1^{1.1^2}+1.3^{1.3^2}+1.5^{1.5^2}+1.7^{1.7^2}]\)

\(\displaystyle \frac{1}{0.8}\int_{1}^{1.8}x^{x^2}dx\approx 2.451\)

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 \(\displaystyle f(x)=1.5^{x^2+x}\) over the interval \(\displaystyle (2,3.2)\)

The subintervals have length \(\displaystyle \frac{3.2-2}{3}=0.4\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [2.2,2.6,3.0]\)

\(\displaystyle \frac{1}{1.2}\int_{2}^{3.2}1.5^{x^2+x}dx\approx (\frac{1}{3})[1.5^{2.2^2+2.2}+1.5^{2.6^2+2.6}+1.5^{3^2+3}]\)

\(\displaystyle \frac{1}{1.2}\int_{2}^{3.2}1.5^{x^2+x}dx\approx 63.866\)

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  \(\displaystyle f(x)=(\frac{1}{4})^{cos(x)}\) over the interval \(\displaystyle (3,3.6)\)

The subintervals have length \(\displaystyle \frac{3.6-3}{3}=0.2\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [3.1,3.3,3.5]\)

\(\displaystyle \frac{1}{3.6-3}\int_{3}^{3.6}(\frac{1}{4})^{cos(x)}dx\approx (\frac{1}{3})[(\frac{1}{4})^{cos(3.1)}+(\frac{1}{4})^{cos(3.3)}+(\frac{1}{4})^{cos(3.5)}]\)

\(\displaystyle \frac{1}{0.6}\int_{3}^{3.6}(\frac{1}{4})^{cos(x)}dx\approx 3.863\)

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  \(\displaystyle f(x)=2^{csc(x)}\) over the interval \(\displaystyle (2,5)\)

The subintervals have length \(\displaystyle \frac{5-2}{3}=1\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [2.5,3.5,4.5]\)

\(\displaystyle \frac{1}{5-2}\int_{2}^{5}2^{csc(x)}dx\approx (\frac{1}{3})[2^{csc(2.5)}+2^{csc(3.5)}+2^{csc(4.5)}]\)

\(\displaystyle \frac{1}{3}\int_{2}^{5}2^{csc(x)}dx\approx 1.272\)

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  \(\displaystyle f(x)=e^{tan(x)}\) over the interval \(\displaystyle (0,1.5)\)

The subintervals have length \(\displaystyle \frac{1.5-0}{3}=0.5\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [0.25,0.75,1.25]\)

\(\displaystyle \frac{1}{1.5-0}\int_{0}^{1.5}e^{tan(x)}dx\approx (\frac{1}{3})[e^{tan(0.25)}+e^{tan(0.75)}+e^{tan(1.25)}]\)

\(\displaystyle \frac{1}{1.5}\int_{0}^{1.5}e^{tan(x)}dx\approx 8.036\)

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  \(\displaystyle f(x)=sec^3(\frac{1}{x})\) over the interval \(\displaystyle (2,2.6)\)

The subintervals have length \(\displaystyle \frac{2.6-2}{3}=0.2\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [2.1,2.3,2.5]\)

\(\displaystyle \frac{1}{2.6-2}\int_{2}^{2.6}sec^3(\frac{1}{x})dx\approx (\frac{1}{3})[sec^3(\frac{1}{2.1})+sec^3(\frac{1}{2.3})+sec^3(\frac{1}{2.5})]\)

\(\displaystyle \frac{1}{0.6}\int_{2}^{2.6}sec^3(\frac{1}{x})dx\approx 1.348\)

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 approximatethe average of  \(\displaystyle f(x)=csc^2(x^3)\) over the interval \(\displaystyle (1,7)\)

The subintervals have length \(\displaystyle \frac{7-1}{3}=2\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [2,4,6]\)

\(\displaystyle \frac{1}{7-1}\int_{1}^{7}csc^2(x^3)dx\approx (\frac{1}{3})[csc^2(2^3)+csc^2(4^3)+csc^2(6^3)]\)

\(\displaystyle \frac{1}{6}\int_{1}^{7}csc^2(x^3)dx\approx 1.422\)

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  \(\displaystyle f(x)=\frac{sin(x^2)}{tan(x^3)}\) over the interval \(\displaystyle (0.3,0.306)\)

The subintervals have length \(\displaystyle \frac{0.306-0.3}{3}=0.002\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [3.001,3.003,3.005]\)

\(\displaystyle \frac{1}{0.306-0.3}\int_{0.3}^{0.306}\frac{sin(x^2)}{tan(x^3)}dx\approx (\frac{1}{3})[\frac{sin(0.301^2)}{tan(0.301^3)}+\frac{sin(0.303^2)}{tan(0.303^3)}+\frac{sin(0.305^2)}{tan(0.305^3)}]\)

\(\displaystyle \frac{1}{0.006}\int_{0.3}^{0.306}\frac{sin(x^2)}{tan(x^3)}dx\approx 3.295\)

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 \(\displaystyle f(x)=5^{\sqrt{x}}\) over the interval \(\displaystyle (0.2,0.8)\)

The subintervals have length \(\displaystyle \frac{0.8-0.2}{3}=0.2\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [0.3,0.5,0.7]\)

\(\displaystyle \frac{1}{0.8-0.2}\int_{0.2}^{0.8}5^{\sqrt{x}}dx\approx (\frac{1}{3})[5^{\sqrt{0.3}}+5^{\sqrt{0.5}}+5^{\sqrt{0.7}}]\)

\(\displaystyle \frac{1}{0.6}\int_{0.2}^{0.8}5^{\sqrt{x}}dx\approx 3.126\)

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 \(\displaystyle f(x)=tan^4(x)\) over the interval \(\displaystyle (1,4)\)

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

\(\displaystyle \frac{1}{4-1}\int_{1}^{4}tan^4(x)dx\approx (\frac{1}{3})[tan^4(1.5)+tan^4(2.5)+tan^4(3.5)]\)

\(\displaystyle \frac{1}{3}\int_{1}^{4}tan^4(x)dx\approx 13180.56\)