AP Calculus BC - Riemann Sum: Midpoint Evaluation

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

Explanation

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

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.

None of the other choices are correct.

Explanation

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

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

Approximate

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

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

Cannot be determined

Explanation

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 $x = x_ { 1 }, x_ { 2 }, x_ { 3 }, x_ { 4 }, x_ { 5 }$ :

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

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

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

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

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

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

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

Explanation

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

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

Explanation

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

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

Explanation

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

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

Explanation

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

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

Explanation

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

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

Explanation

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

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

Explanation

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

Riemann Sum: Midpoint Evaluation

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