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 \(\displaystyle \int_{0.9}^{0.96}cos(4^{tan(x)})dx\)

So the interval is \(\displaystyle [0.9,0.96]\), the subintervals have length \(\displaystyle \frac{0.96-0.9}{3}=0.02\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [0.91,0.93,0.95]\)

\(\displaystyle \int_{0.9}^{0.96}cos(4^{tan(x)})dx\approx (0.02)[cos(4^{tan(0.91)})+cos(4^{tan(0.93)})+cos(4^{tan(0.95)})]\)

\(\displaystyle \int_{0.9}^{0.96}cos(4^{tan(x)})dx\approx 0.054\)

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 \(\displaystyle \int_{3.5}^{4.3}(cos(x))^{cos(x)}dx\)

So the interval is \(\displaystyle [3.5,4.3]\), the subintervals have length \(\displaystyle \frac{4.3-3.5}{4}=0.2\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [3.6,3.8,4.0,4.2]\)

\(\displaystyle \int_{3.5}^{4.3}(cos(x))^{cos(x)}dx\approx (0.2)[(cos(3.6))^{cos(3.6)}+(cos(3.8))^{cos(3.8)}+(cos(4.0))^{cos(4.0)}+(cos(4.2))^{cos(4.2)}]\)

\(\displaystyle \int_{3.5}^{4.3}(cos(x))^{cos(x)}dx\approx-0.513+0.735i\)

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 \(\displaystyle \int_{0.2}^{1.4}(sin(x))^{\frac{1}{cos(x)}}dx\)

So the interval is \(\displaystyle [0.2,1.4]\), the subintervals have length \(\displaystyle \frac{1.4-0.2}{3}=0.4\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [0.4,0.8,1.2]\)

\(\displaystyle \int_{0.2}^{1.4}(sin(x))^{\frac{1}{cos(x)}}dx\approx (0.4)[(sin(0.4))^{\frac{1}{cos(0.4)}}+(sin(0.8))^{\frac{1}{cos(0.8)}}+(sin(1.2))^{\frac{1}{cos(1.2)}}]\)

\(\displaystyle \int_{0.2}^{1.4}(sin(x))^{\frac{1}{cos(x)}}dx\approx 0.721\)

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 \(\displaystyle \int_{2.6}^{3.2}2.7^{sin(x^2)}dx\)

So the interval is \(\displaystyle [2.6,3.2]\), the subintervals have length \(\displaystyle \frac{3.6-2.6}{3}=0.2\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [2.7,2.9,3.1]\)

\(\displaystyle \int_{2.6}^{3.2}2.7^{sin(x^2)}dx\approx (0.2)[2.7^{sin(2.7^2)}+2.7^{sin(2.9^2)}+2.7^{sin(3.1^2)}]\)

\(\displaystyle \int_{2.6}^{3.2}2.7^{sin(x^2)}dx\approx 1.095\)

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 \(\displaystyle \int_{0.5}^{2.9}(sin(x^{tan(x^{cos(x)})}))dx\)

So the interval is \(\displaystyle [0.5,2.9]\), the subintervals have length \(\displaystyle \frac{2.9-0.5}{2}=1.2\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [1.1,2.3]\)

\(\displaystyle \int_{0.5}^{2.9}(sin(x^{tan(x^{cos(x)})}))\approx (1.2)[(sin(1.1^{tan(1.1^{cos(1.1)})}))dx+(sin(2.3^{tan(2.3^{cos(2.3)})}))]\)

\(\displaystyle \int_{0.5}^{2.9}(sin(x^{tan(x^{cos(x)})}))dx\approx 2.296\)

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 \(\displaystyle \int_{1}^{100} tan(\sqrt[3]{x})dx\)

So the interval is \(\displaystyle [1,100]\), the subintervals have length \(\displaystyle \frac{100-1}{3}=33\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [17.5,50.5,83.5]\)

\(\displaystyle \int_{1}^{100} tan(\sqrt[3]{x})dx\approx (33)[tan(\sqrt[3]{17.5})+tan(\sqrt[3]{50.5})+tan(\sqrt[3]{83.5})]\)

\(\displaystyle \int_{1}^{100} tan(\sqrt[3]{x})dx\approx93.248\)

A Riemann sum integral approximation over an interval \(\displaystyle [a,b]\) with \(\displaystyle n\) subintervals follo\(\displaystyle \int_{0.0003}^{0.0006}14^{cos(\frac{1}{x})}dx\approx 0.00012\)ws 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 \(\displaystyle \int_{0.0003}^{0.0006}14^{cos(\frac{1}{x})}dx\)

So the interval is \(\displaystyle [0.0003,0.0006]\), the subintervals have length \(\displaystyle \frac{0.0006-0.0003}{3}=0.0001\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [0.00035,0.00045,0.00055]\)

\(\displaystyle \int_{0.0003}^{0.0006}14^{cos(\frac{1}{x})}dx\approx (0.0001)[14^{cos(\frac{1}{0.00035})}+14^{cos(\frac{1}{0.00045})}+14^{cos(\frac{1}{0.00055})}]\)

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 \(\displaystyle \int_{1}^{7}tan(cos(x^2+x))dx\)

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

\(\displaystyle {\int_{1}^{7}tan(cos(x^2+x))dx\approx (2)[tan(cos(2^2+2))+tan(cos(4^2+4))+tan(cos(6^2+6))]}\)

\(\displaystyle \int_{1}^{7}tan(cos(x^2+x))dx\approx 2.877\)

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 \(\displaystyle \int_{3}^{15}x!dx\)

So the interval is \(\displaystyle [3,15]\), the subintervals have length \(\displaystyle \frac{15-3}{3}=4\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [5,9,13]\)

\(\displaystyle \int_{3}^{15}x!dx \approx (4)[5!+9!+13!]\)

\(\displaystyle \int_{3}^{15}x!dx \approx 24909535200\)

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 \(\displaystyle f(x)=\sqrt[4]{cos(x^2)}\) over the interval \(\displaystyle (-0.6,0.6)\)

The subintervals have length \(\displaystyle \frac{0.6-(-0.6)}{4}=0.3\), and since we are using the midpoints of each interval, the x-values are \(\displaystyle [-0.45,-0.15,0.15,0.45]\)

\(\displaystyle {\frac{1}{1.2}\int_{-0.6}^{0.6}\sqrt[4]{cos(x^2)}dx\approx (\frac{1}{4})[\sqrt[4]{cos((-0.45)^2)}+\sqrt[4]{cos((-0.15)^2)}+\sqrt[4]{cos(0.15^2)}+\sqrt[4]{cos(0.45^2)}]}\)

\(\displaystyle \frac{1}{1.2}\int_{-0.6}^{0.6}\sqrt[4]{cos(x^2)}dx\approx 0.997\)