The interval $\displaystyle \left [ 1,5 \right ]$ is 4 units in width; the interval is divided evenly into four subintervals $\displaystyle \Delta x = 4\div 4 = 1$ units in width, with their midpoints shown:

$\displaystyle \left [ 1,2 \right ] : x _{1} = 1.5$

$\displaystyle \left [2,3 \right ]: x _{2} = 2.5$

$\displaystyle \left [3,4 \right ] : x _{3} = 3.5$

$\displaystyle \left [ 4,5 \right ] : x _{4} = 4.5$

The midpoint rule requires us to calculate:

$\displaystyle M = \Delta x \left [ f (x_{1} )+ f (x_{2}) + f (x_{3}) + f (x_{4}) \right ]$

where $\displaystyle \Delta x = 1$ and $\displaystyle f(x) = \frac{ \ln x }{x^2}$

Evaluate $\displaystyle f(x) = \frac{ \ln x }{x^2}$ for each of $\displaystyle x = x_ { 1 }, x_ { 2 }, x_ { 3 }, x_ { 4 }$:

$\displaystyle f(1.5) = \frac{ \ln 1.5 }{1.5^2} \approx \frac{0.4055}{2.25} \approx 0.1802$

$\displaystyle f(2.5) = \frac{ \ln 2.5 }{2.5^2} \approx \frac{0.9163}{6.25} \approx 0.1466$

$\displaystyle f(3.5) = \frac{ \ln 3.5 }{3.5^2} \approx \frac{1.2528}{12.25} \approx 0.1023$

$\displaystyle f(4.5) = \frac{ \ln 4.5 }{4.5^2} \approx \frac{1.5041}{20.25} \approx 0.0743$

So

$\displaystyle M =1 \left (0.1802 + 0.1466 +0.1023 + 0.0743 \right )$

$\displaystyle = 0.503$

The interval $\displaystyle \left [ 1,2 \right ]$ is 1 unit in width; the interval is divided evenly into five subintervals $\displaystyle \Delta x = 1 \div 5 = 0.2$ units in width, with their midpoints shown: 

$\displaystyle \left [ 1, 1.2 \right ]; x_{1} = 1.1$

$\displaystyle \left [ 1.2, 1.4 \right ]; x_{2} = 1.3$

$\displaystyle \left [1.4, 1.6 \right ]; x_{3} =1.5$

$\displaystyle \left [ 1.6, 1.8 \right ]; x_{4} = 1.7$

$\displaystyle \left [1.8, 2 \right ];x_{5} = 1.9$

The midpoint rule requires us to calculate:

$\displaystyle M = \Delta x \left ( f (x_{1} )+ f (x_{2}) + f (x_{3}) + f (x_{4}) + f (x_{5}) \right )$

where $\displaystyle \Delta x = 0.2$ and $\displaystyle f(x) =x^{2} \ln x$

Evaluate $\displaystyle f(x) =x^{2} \ln x$ for each of $\displaystyle x = x_ { 1 }, x_ { 2 }, x_ { 3 }, x_ { 4 }, x_ { 5 }$:

$\displaystyle f(1.1) = 1.1 ^{2} \cdot \ln 1.1 \approx 1.1 ^{2} \cdot 0.0953 \approx 0.1153$

$\displaystyle f(1.3) = 1.3^{2} \cdot \ln 1.3 \approx 1.3 ^{2} \cdot 0.2624 \approx 0.4434$

$\displaystyle f(1.5) = 1.5 ^{2} \cdot \ln 1.5 \approx 1.5 ^{2} \cdot 0.4055 \approx 0.9123$

$\displaystyle f(1.7) = 1.7 ^{2} \cdot \ln 1.7\approx 1.7 ^{2} \cdot 0.5306 \approx 1.5335$

$\displaystyle f(1.9) = 1.9 ^{2} \cdot \ln 1.9 \approx 1.9 ^{2} \cdot 0.6419 \approx 2.3171$

$\displaystyle M \approx 0.2 \cdot \left ( 0.1153 +0.4434 + 0.9123 + 1.5335 + 2.3171 \right ) \approx 1.064$

The interval $\displaystyle \left [ 0, \frac{\pi}{2}\right ]$ is $\displaystyle \frac{\pi}{2}$ units in width; the interval is divided evenly into five subintervals $\displaystyle \Delta x = \frac{\pi}{2} \div 5 = \frac{\pi}{10}$ units in width, with their midpoints shown: 

$\displaystyle \left [ 0, \frac{\pi}{10}\right ]; x_{1} = \frac{\pi}{20}$

$\displaystyle \left [ \frac{\pi}{10},\frac{\pi}{5} \right ]; x_{2} = \frac{3 \pi}{20}$

$\displaystyle \left [ \frac{\pi}{5},\frac{3\pi}{10} \right ]; x_{3} =\frac{5 \pi}{20} = \frac{ \pi}{4}$

$\displaystyle \left [ \frac{3\pi}{10}, \frac{2 \pi}{5} \right ]; x_{4} =\frac{7 \pi}{20}$

$\displaystyle \left [ \frac{2 \pi}{5} , \frac{\pi}{2} \right ];x_{5} = \frac{9 \pi}{20}$

The midpoint rule requires us to calculate:

$\displaystyle M = \Delta x \left ( f (x_{1} )+ f (x_{2}) + f (x_{3}) + f (x_{4}) + f (x_{5}) \right )$

where $\displaystyle \Delta x = \frac{\pi}{10}$ and $\displaystyle f(x) = x^{2} \cos x$

Evaluate $\displaystyle f(x) = x^{2} \cos x$ for each of $\displaystyle x = x_ { 1 }, x_ { 2 }, x_ { 3 }, x_ { 4 }, x_ { 5 }$:

$\displaystyle f\left \( \frac{\pi}{20} \right$ = \left $\frac{\pi}{20} \right$^{2} \cos \left $\frac{\pi}{20} \right$\)

$\displaystyle \approx 0.1571^{2} \cdot \cos 0.1571 \approx 0.1571^{2} \cdot 0.9877 \approx 0.0244$

$\displaystyle f\left \( \frac{3\pi}{20} \right$ = \left $\frac{3\pi}{20} \right$^{2} \cos \left $\frac{3\pi}{20} \right$\)

$\displaystyle \approx 0.4712 ^{2} \cdot \cos 0.4712 \approx 0.1571^{2} \cdot 0.8910 \approx 0.1979$

$\displaystyle f\left \( \frac{\pi}{4} \right$ = \left $\frac{\pi}{4} \right$^{2} \cos \left $\frac{\pi}{4} \right$\)

$\displaystyle \approx 0.7854^{2} \cdot \cos 0.7854 \approx 0.7854^{2} \cdot 0.7071 \approx 0.4362$

$\displaystyle f\left \( \frac{7\pi}{20} \right$ = \left $\frac{7\pi}{20} \right$^{2} \cos \left $\frac{7\pi}{20} \right$\)

$\displaystyle \approx 1.0996^{2} \cdot \cos 1.0996 \approx 1.0996^{2} \cdot 0.4540 \approx 0.5489$

$\displaystyle f\left \( \frac{9\pi}{20} \right$ = \left $\frac{9\pi}{20} \right$^{2} \cos \left $\frac{9\pi}{20} \right$\)

$\displaystyle \approx 1.4137 ^{2} \cdot \cos1.4137 \approx 1.4137^{2} \cdot 0.1564 \approx 0.3126$

Since $\displaystyle \Delta x = \frac{\pi}{10} \approx 0.3142$,

we can approximate $\displaystyle M$ as

$\displaystyle M \approx 0.3142 \cdot \left ( 0.0244 + 0.1979 + 0.4362 + 0.5489 + 0.3126) \right ) \approx 0.478$.

$\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-4}^{14.4}(-20sin(13tan(2x)))dx\\&\text{So the interval is }[-4,14.4]\text{ the subintervals have length }\frac{14.4-(-4)}{4}=\frac{23}{5}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[-\frac{17}{10},\frac{29}{10},\frac{15}{2},\frac{121}{10}]\\&\int_{-4}^{14.4}(-20sin(13tan(2x)))dx=\frac{23}{5}[(20sin(13tan(\frac{17}{5})))+(-20sin(13tan(\frac{29}{5})))+(-20sin(13tan(15)))+(-20sin(13tan(\frac{121}{5})))]\\&\int_{-4}^{14.4}(-20sin(13tan(2x)))dx=-159.88\end{align*}$

$\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{21}(-16cos(sin(6x)))dx\\&\text{So the interval is }[5,21]\text{ the subintervals have length }\frac{21-(5)}{4}=4\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[7,11,15,19]\\&\int_{5}^{21}(-16cos(sin(6x)))dx=4[(-16cos(sin(42)))+(-16cos(sin(66)))+(-16cos(sin(90)))+(-16cos(sin(114)))]\\&\int_{5}^{21}(-16cos(sin(6x)))dx=-188.28\end{align*}$

$\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{2}^{14.6}(16cos(20e^{(3x)}))dx\\&\text{So the interval is }[2,14.6]\text{ the subintervals have length }\frac{14.6-(2)}{3}=\frac{21}{5}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{41}{10},\frac{83}{10},\frac{25}{2}]\\&\int_{2}^{14.6}(16cos(20e^{(3x)}))dx=\frac{21}{5}[(16cos(20e^{(\frac{123}{10})}))+(16cos(20e^{(\frac{249}{10})}))+(16cos(20e^{(\frac{75}{2})}))]\\&\int_{2}^{14.6}(16cos(20e^{(3x)}))dx=26.04\end{align*}$

$\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-3}^{9.9}(\frac{13}{3^{(5cos(3x))}})dx\\&\text{So the interval is }[-3,9.9]\text{ the subintervals have length }\frac{9.9-(-3)}{3}=\frac{43}{10}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[-\frac{17}{20},\frac{69}{20},\frac{31}{4}]\\&\int_{-3}^{9.9}(\frac{13}{3^{(5cos(3x))}})dx=\frac{43}{10}[(\frac{13}{3^{(5cos(\frac{51}{20}))}})+(\frac{13}{3^{(5cos(\frac{207}{20}))}})+(\frac{13}{3^{(5cos(\frac{93}{4}))}})]\\&\int_{-3}^{9.9}(\frac{13}{3^{(5cos(3x))}})dx=7165.44\end{align*}$

$\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{1}^{9}(19cos(2x^{2}))dx\\&\text{So the interval is }[1,9]\text{ the subintervals have length }\frac{9-(1)}{4}=2\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[2,4,6,8]\\&\int_{1}^{9}(19cos(2x^{2}))dx=2[(19cos(8))+(19cos(32))+(19cos(72))+(19cos(128))]\\&\int_{1}^{9}(19cos(2x^{2}))dx=-36.91\end{align*}$

$\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{1}^{12.6}(-9sin(18e^{(3x)}))dx\\&\text{So the interval is }[1,12.6]\text{ the subintervals have length }\frac{12.6-(1)}{4}=\frac{29}{10}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{49}{20},\frac{107}{20},\frac{33}{4},\frac{223}{20}]\\&\int_{1}^{12.6}(-9sin(18e^{(3x)}))dx=\frac{29}{10}[(-9sin(18e^{(\frac{147}{20})}))+(-9sin(18e^{(\frac{321}{20})}))+(-9sin(18e^{(\frac{99}{4})}))+(-9sin(18e^{(\frac{669}{20})}))]\\&\int_{1}^{12.6}(-9sin(18e^{(3x)}))dx=-73.83\end{align*}$

$\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{0}^{10}(8cos(19e^{(x)}))dx\\&\text{So the interval is }[0,10]\text{ the subintervals have length }\frac{10-(0)}{4}=\frac{5}{2}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{5}{4},\frac{15}{4},\frac{25}{4},\frac{35}{4}]\\&\int_{0}^{10}(8cos(19e^{(x)}))dx=\frac{5}{2}[(8cos(19e^{(\frac{5}{4})}))+(8cos(19e^{(\frac{15}{4})}))+(8cos(19e^{(\frac{25}{4})}))+(8cos(19e^{(\frac{35}{4})}))]\\&\int_{0}^{10}(8cos(19e^{(x)}))dx=-44.98\end{align*}$