Numerical Approximations to Definite Integrals - AP Calculus BC

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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.

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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

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

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Question

Approximate

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

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

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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$

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

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

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

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

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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.

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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

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

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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.}$\end{align}$

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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}$(-20sin(13tan(2x)))dx=-159.88\end{align}$

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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.}$\end{align}$

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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}$(-16cos(sin(6x)))dx=-188.28\end{align}$

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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)}$))dx\end{align}$

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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)}$))dx=26.04\end{align}$

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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}$

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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}$

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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.}$\end{align}$

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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}$))dx=-36.91\end{align}$

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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)}$))dx\end{align}$

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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)}$))dx=-73.83\end{align}$

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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.}$\end{align}$

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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)}$))dx=-44.98\end{align}$

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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.}$\end{align}$

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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}$))dx=2.88\end{align}$

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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.}$\end{align}$

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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)}$))dx=-70.62\end{align}$

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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.}$\end{align}$

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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}$(9tan(9tan(6x)))dx=477.77\end{align}$

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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.}$\end{align}$

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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)}$))dx=-3.55\end{align}$

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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.}$\end{align}$

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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}$(-tan(14sin(3x)))dx=4.60\end{align}$

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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.

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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

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

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Question

Approximate

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

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

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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$

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

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

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

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

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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.

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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

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

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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.}$\end{align}$

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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}$(-20sin(13tan(2x)))dx=-159.88\end{align}$

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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.}$\end{align}$

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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}$(-16cos(sin(6x)))dx=-188.28\end{align}$

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