Riemann Sum: Midpoint Evaluation - AP Calculus BC

Card 0 of 150

Question

Approximate

$\int_{0}^{\frac{\pi}{2}} x^{2} \cos x ; \mathrm {d}x$

using the midpoint rule with . Round your answer to three decimal places.

Answer

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

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

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

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

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

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

The midpoint rule requires us to calculate:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

we can approximate as

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

Compare your answer with the correct one above

Question

Approximate

$\int_{1}^{2} x^{2} \ln x ; \mathrm {d}x$

using the midpoint rule with . Round your answer to three decimal places.

Answer

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

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

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

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

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

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

The midpoint rule requires us to calculate:

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

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

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

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

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

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

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

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

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

Compare your answer with the correct one above

Question

Approximate

$\int_{1}^{5} \frac{ \ln x ; \mathrm {d}x }{x^2}$

using the midpoint rule with . Round your answer to three decimal places.

Answer

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

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

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

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

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

The midpoint rule requires us to calculate:

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

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

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

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

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

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

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

So

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

Compare your answer with the correct one above

Question

$\begin{align}&\text{Calculate the Riemann sums integral approximation of :}\&\int_{-4}^{14.4}(-20sin(13tan(2x)))dx\&\text{Using midpoints over }4\text{ intervals.}\end{align}$

Answer

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

Compare your answer with the correct one above

Question

$\begin{align}&\text{Using the method of midpoint Riemann sums approximate the integral:}\&\int_{5}^{21}(-16cos(sin(6x)))dx\&\text{Using }4\text{ intervals.}\end{align}$

Answer

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

Compare your answer with the correct one above

Question

$\begin{align}&\text{Using }3\text{ intervals, and the method of midpoint Riemann sums approximate the integral:}\&\int_{2}^{14.6}(16cos(20e^{(3x)}))dx\end{align}$

Answer

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

Compare your answer with the correct one above

Question

$\begin{align}&\text{Using }3\text{ intervals, and the method of midpoint Riemann sums approximate the integral:}\&\int_{-3}^{9.9}(\frac{13}{3^{(5cos(3x))}})dx\end{align}$

Answer

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

Compare your answer with the correct one above

Question

$\begin{align}&\text{Using the method of midpoint Riemann sums approximate the integral:}\&\int_{1}^{9}(19cos(2x^{2}))dx\&\text{Using }4\text{ intervals.}\end{align}$

Answer

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

Compare your answer with the correct one above

Question

$\begin{align}&\text{Using }4\text{ intervals, and the method of midpoint Riemann sums approximate the integral:}\&\int_{1}^{12.6}(-9sin(18e^{(3x)}))dx\end{align}$

Answer

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

Compare your answer with the correct one above

Question

$\begin{align}&\text{Using the method of midpoint Riemann sums approximate the integral:}\&\int_{0}^{10}(8cos(19e^{(x)}))dx\&\text{Using }4\text{ intervals.}\end{align}$

Answer

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

Compare your answer with the correct one above

Question

$\begin{align}&\text{Using the method of midpoint Riemann sums approximate the integral:}\&\int_{4}^{9.4}(-15sin(x^{2}))dx\&\text{Using }3\text{ intervals.}\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{4}^{9.4}(-15sin(x^{2}))dx\&\text{So the interval is }[4,9.4]\text{ the subintervals have length }\frac{9.4-(4)}{3}=\frac{9}{5}\&\text{and since we are using the midpoint of each interval, the x-values are:}\&[\frac{49}{10},\frac{67}{10},\frac{17}{2}]\&\int_{4}^{9.4}(-15sin(x^{2}))dx=\frac{9}{5}[(-15sin(\frac{2401}{100}))+(-15sin(\frac{4489}{100}))+(-15sin(\frac{289}{4}))]\&\int_{4}^{9.4}(-15sin(x^{2}))dx=2.88\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{Using the method of midpoint Riemann sums approximate the integral:}\&\int_{0}^{12.3}(-12sin(19e^{(4x)}))dx\&\text{Using }3\text{ intervals.}\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{0}^{12.3}(-12sin(19e^{(4x)}))dx\&\text{So the interval is }[0,12.3]\text{ the subintervals have length }\frac{12.3-(0)}{3}=\frac{41}{10}\&\text{and since we are using the midpoint of each interval, the x-values are:}\&[\frac{41}{20},\frac{123}{20},\frac{41}{4}]\&\int_{0}^{12.3}(-12sin(19e^{(4x)}))dx=\frac{41}{10}[(-12sin(19e^{(\frac{41}{5})}))+(-12sin(19e^{(\frac{123}{5})}))+(-12sin(19e^{(41)}))]\&\int_{0}^{12.3}(-12sin(19e^{(4x)}))dx=-70.62\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{Using the method of midpoint Riemann sums approximate the integral:}\&\int_{-2}^{4.6}(9tan(9tan(6x)))dx\&\text{Using }3\text{ intervals.}\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{-2}^{4.6}(9tan(9tan(6x)))dx\&\text{So the interval is }[-2,4.6]\text{ the subintervals have length }\frac{4.6-(-2)}{3}=\frac{11}{5}\&\text{and since we are using the midpoint of each interval, the x-values are:}\&[-\frac{9}{10},\frac{13}{10},\frac{7}{2}]\&\int_{-2}^{4.6}(9tan(9tan(6x)))dx=\frac{11}{5}[(-9tan(9tan(\frac{27}{5})))+(9tan(9tan(\frac{39}{5})))+(9tan(9tan(21)))]\&\int_{-2}^{4.6}(9tan(9tan(6x)))dx=477.77\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{Calculate the Riemann sums integral approximation of :}\&\int_{-2}^{2.5}(-cos(11e^{(2x)}))dx\&\text{Using midpoints over }3\text{ intervals.}\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{-2}^{2.5}(-cos(11e^{(2x)}))dx\&\text{So the interval is }[-2,2.5]\text{ the subintervals have length }\frac{2.5-(-2)}{3}=\frac{3}{2}\&\text{and since we are using the midpoint of each interval, the x-values are:}\&[-\frac{5}{4},\frac{1}{4},\frac{7}{4}]\&\int_{-2}^{2.5}(-cos(11e^{(2x)}))dx=\frac{3}{2}[(-cos(11e^{(-\frac{5}{2})}))+(-cos(11e^{(\frac{1}{2})}))+(-cos(11e^{(\frac{7}{2})}))]\&\int_{-2}^{2.5}(-cos(11e^{(2x)}))dx=-3.55\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{Using the method of midpoint Riemann sums approximate the integral:}\&\int_{0}^{4.8}(-tan(14sin(3x)))dx\&\text{Using }4\text{ intervals.}\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{0}^{4.8}(-tan(14sin(3x)))dx\&\text{So the interval is }[0,4.8]\text{ the subintervals have length }\frac{4.8-(0)}{4}=\frac{6}{5}\&\text{and since we are using the midpoint of each interval, the x-values are:}\&[\frac{3}{5},\frac{9}{5},3,\frac{21}{5}]\&\int_{0}^{4.8}(-tan(14sin(3x)))dx=\frac{6}{5}[(-tan(14sin(\frac{9}{5})))+(-tan(14sin(\frac{27}{5})))+(-tan(14sin(9)))+(-tan(14sin(\frac{63}{5})))]\&\int_{0}^{4.8}(-tan(14sin(3x)))dx=4.60\end{align}$

Compare your answer with the correct one above

Question

Approximate

$\int_{0}^{\frac{\pi}{2}} x^{2} \cos x ; \mathrm {d}x$

using the midpoint rule with . Round your answer to three decimal places.

Answer

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

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

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

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

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

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

The midpoint rule requires us to calculate:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

we can approximate as

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

Compare your answer with the correct one above

Question

Approximate

$\int_{1}^{2} x^{2} \ln x ; \mathrm {d}x$

using the midpoint rule with . Round your answer to three decimal places.

Answer

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

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

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

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

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

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

The midpoint rule requires us to calculate:

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

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

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

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

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

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

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

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

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

Compare your answer with the correct one above

Question

Approximate

$\int_{1}^{5} \frac{ \ln x ; \mathrm {d}x }{x^2}$

using the midpoint rule with . Round your answer to three decimal places.

Answer

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

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

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

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

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

The midpoint rule requires us to calculate:

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

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

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

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

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

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

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

So

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

Compare your answer with the correct one above

Question

$\begin{align}&\text{Calculate the Riemann sums integral approximation of :}\&\int_{-4}^{14.4}(-20sin(13tan(2x)))dx\&\text{Using midpoints over }4\text{ intervals.}\end{align}$

Answer

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

Compare your answer with the correct one above

Question

$\begin{align}&\text{Using the method of midpoint Riemann sums approximate the integral:}\&\int_{5}^{21}(-16cos(sin(6x)))dx\&\text{Using }4\text{ intervals.}\end{align}$

Answer

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

Compare your answer with the correct one above

Tap the card to reveal the answer