Numerical Approximations to Definite Integrals - AP Calculus BC

Card 0 of 460

Approximate

$\int_${1}^{2}$ e ^${x^{2}$} \mathrm{d}x$

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

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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. They are

$\left [ 1,1.2 \right ], \left [ 1.2, 1.4 \right ], \left [ 1.4, 1.6 \right ], \left [ 1.6, 1.8 \right ], \left [ 1.8, 2 \right ]$ .

The trapezoidal rule approximates the area of the given integral $\int_${1}^{2}$ e ^${x^{2}$} \mathrm{d}x$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + 2 $f(x_{4}$) + $f(x_{5}$) \right ]$ ,

where

$\Delta x = 0.2$

$f (x) = e ^${x^{2}$}$

and

.

$T =\frac{0.2}{2}$ \left[ f(1) + 2 f(1.2}) + 2 f(1.4) + 2 f(1.6) + 2 f(1.8) + f(2) \right]$

$f(1) = e ^${1^{2}$} \approx 2.7183$

$f(1.2) = e ^${1.2^{2}$} \approx 4.2207$

$f(1.4) = e ^${1.4^{2}$} \approx 7.0993$

$f(1.6) = e ^${1.6^{2}$} \approx 12.9358$

$f(1.8) = e ^${1.8^{2}$} \approx 25.5337$

$f(2) = e ^${2^{2}$} \approx 54.5982$

So

$T \approx $\frac{0.2}{2}$ \left [2.7183 + 2 \cdot 4.2207 + 2 \cdot 7.0993 + 2 \cdot 12.9358 + 2 \cdot 25.5337 + 54.5982 \right ]$

$\approx 0.1 \left (2.7183 + 8.4414 + 14.1986 + 25.8716 + 51.0674 + 54.5982 \right )$

$\approx 0.1 \cdot 156.8955 \approx 15.690$

Find the Left Riemann sum of the function

on the interval divided into four sub-intervals.

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The interval divided into four sub-intervals gives rectangles with vertices of the bases at

For the Left Riemann sum, we need to find the rectangle heights which values come from the left-most function value of each sub-interval, or f(0), f(2), f(4), and f(6).

Because each sub-interval has a width of 2, the Left Riemann sum is

Given a function , find the Left Riemann Sum of the function on the interval divided into three sub-intervals.

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In order to find the Riemann Sum of a given function, we need to approximate the area under the line or curve resulting from the function using rectangles spaced along equal sub-intervals of a given interval. Since we have an interval divided into sub-intervals, we'll be using rectangles with vertices at .

To approximate the area under the curve, we need to find the areas of each rectangle in the sub-intervals. We already know the width or base of each rectangle is because the rectangles are spaced units apart. Since we're looking for the Left Riemann Sum, we want to find the heights of each rectangle by taking the values of each leftmost function value on each sub-interval, as follows:

Putting it all together, the Left Riemann Sum is

.

Given a function $y=\frac{1}{4}$x+2$ , find the Left Riemann Sum of the function on the interval divided into four sub-intervals.

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In order to find the Riemann Sum of a given function, we need to approximate the area under the line or curve resulting from the function using rectangles spaced along equal sub-intervals of a given interval. Since we have an interval divided into sub-intervals, we'll be using rectangles with vertices at .

To approximate the area under the curve, we need to find the areas of each rectangle in the sub-intervals. We already know the width or base of each rectangle is because the rectangles are spaced unit apart. Since we're looking for the Left Riemann Sum, we want to find the heights of each rectangle by taking the values of each leftmost function value on each sub-interval, as follows:

$f(0)=\frac{1}{4}$(0)+2=0+2=2$

$f(1)=\frac{1}{4}$(1)+2=\frac{1}{4}$+\frac{8}{4}$=\frac{9}{4}$$

$f(2)=\frac{1}{4}$(2)+2=\frac{1}{2}$+\frac{4}{2}$=\frac{5}{2}$$

$f(3)=\frac{1}{4}$(3)+2=\frac{3}{4}$+\frac{8}{4}$=\frac{11}{4}$$

Putting it all together, the Left Riemann Sum is

$A=bh=1(2+\frac{9}{4}$+\frac{5}{2}$+\frac{11}{4}$)=\frac{8}{4}$+\frac{9}{4}$+\frac{10}{4}$+\frac{11}{4}$=\frac{38}{4}$=\frac{19}{2}$$

$\begin{align}&$\text{Using the method of left point Riemann sums approximate the integral:}$\&\int_${3}^{3.9}$(-9tan(10sin(5x)))dx\&$\text{Using }$3$\text{ intervals.}$\end{align}$

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$\begin{align}&$\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}^{3.9}$(-9tan(10sin(5x)))dx\&$\text{So the interval is }$[3,3.9]$\text{ the subintervals have length }$$\frac{3.9-(3)}{3}$=\frac{3}{10}$\&$\text{and since we are using the leftpoint of each interval, the x-values are:}$\&[3,$\frac{33}{10}$,$\frac{18}{5}$]\&\int_${3}^{3.9}$(-9tan(10sin(5x)))dx=\frac{3}{10}$[(-9tan(10sin(15)))+(-9tan(10sin($\frac{33}{2}$)))+(-9tan(10sin(18)))]\&\int_${3}^{3.9}$(-9tan(10sin(5x)))dx=9.91\end{align}$

$\begin{align}&$\text{Calculate the Riemann sums integral approximation of :}$\&\int_${-2}^{11.6}$$($\frac{18}{3^{(4tan(4x))}$$})dx\&$\text{Using left points over }$4$\text{ intervals.}$\end{align}$

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$\begin{align}&$\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}^{11.6}$$($\frac{18}{3^{(4tan(4x))}$$})dx\&$\text{So the interval is }$[-2,11.6]$\text{ the subintervals have length }$$\frac{11.6-(-2)}{4}$=\frac{17}{5}$\&$\text{and since we are using the leftpoint of each interval, the x-values are:}$\&[-2,$\frac{7}{5}$,$\frac{24}{5}$,$\frac{41}{5}$]\&\int_${-2}^{11.6}$$($\frac{18}{3^{(4tan(4x))}$$})dx=\frac{17}{5}$[(18\cdot $3^{(4tan(8))}$$)+($\frac{18}{3^{(4tan(\frac{28}$$${5}))}})+($\frac{18}{3^{(4tan(\frac{96}$$${5}))}})+($\frac{18}{3^{(4tan(\frac{164}$${5}))}})]\&\int_${-2}^{11.6}$$($\frac{18}{3^{(4tan(4x))}$$})dx=2200.80\end{align}$

$\begin{align}&$\text{Using the method of left point Riemann sums approximate the integral:}$\&\int_${-4}^{4.4}$$(6cos(11e^{(2x)}$))dx\&$\text{Using }$3$\text{ intervals.}$\end{align}$

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$\begin{align}&$\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}^{4.4}$$(6cos(11e^{(2x)}$))dx\&$\text{So the interval is }$[-4,4.4]$\text{ the subintervals have length }$$\frac{4.4-(-4)}{3}$=\frac{14}{5}$\&$\text{and since we are using the leftpoint of each interval, the x-values are:}$\&[-4,-$\frac{6}{5}$,$\frac{8}{5}$]\&\int_${-4}^{4.4}$$(6cos(11e^{(2x)}$$))dx=\frac{14}{5}$[(6cos(11e^{(-8)}$$))+(6cos(11e^{(-$\frac{12}$${5}$)}))+(6cos(11e^{($\frac{16}${5}$)}))]\&\int_${-4}^{4.4}$$(6cos(11e^{(2x)}$))dx=41.86\end{align}$

$\begin{align}&$\text{Using the method of left point Riemann sums approximate the integral:}$\&\int_${0}^{12}$(11sin(20sin(x)))dx\&$\text{Using }$3$\text{ intervals.}$\end{align}$

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$\begin{align}&$\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}$(11sin(20sin(x)))dx\&$\text{So the interval is }$[0,12]$\text{ the subintervals have length }$$\frac{12-(0)}{3}$=4\&$\text{and since we are using the leftpoint of each interval, the x-values are:}$\&[0,4,8]\&\int_${0}^{12}$(11sin(20sin(x)))dx=4[(0)+(11sin(20sin(4)))+(11sin(20sin(8)))]\&\int_${0}^{12}$(11sin(20sin(x)))dx=11.66\end{align}$

$\begin{align}&$\text{Using the method of left point Riemann sums approximate the integral:}$\&\int_${0}^{3}$$(-16sin(2x^{2}$))dx\&$\text{Using }$3$\text{ intervals.}$\end{align}$

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$\begin{align}&$\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}^{3}$$(-16sin(2x^{2}$))dx\&$\text{So the interval is }$[0,3]$\text{ the subintervals have length }$$\frac{3-(0)}{3}$=1\&$\text{and since we are using the leftpoint of each interval, the x-values are:}$\&[0,1,2]\&\int_${0}^{3}$$(-16sin(2x^{2}$))dx=1[(0)+(-16sin(2))+(-16sin(8))]\&\int_${0}^{3}$$(-16sin(2x^{2}$))dx=-30.38\end{align}$

$\begin{align}&$\text{Using }$3$\text{ intervals, and the method of left point Riemann sums approximate the integral:}$\&\int_${4}^{12.7}$$(-17cos(20x^{2}$))dx\end{align}$

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$\begin{align}&$\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}^{12.7}$$(-17cos(20x^{2}$))dx\&$\text{So the interval is }$[4,12.7]$\text{ the subintervals have length }$$\frac{12.7-(4)}{3}$=\frac{29}{10}$\&$\text{and since we are using the leftpoint of each interval, the x-values are:}$\&[4,$\frac{69}{10}$,$\frac{49}{5}$]\&\int_${4}^{12.7}$$(-17cos(20x^{2}$))dx=\frac{29}{10}$[(-17cos(320))+(-17cos($\frac{4761}{5}$))+(-17cos($\frac{9604}{5}$))]\&\int_${4}^{12.7}$$(-17cos(20x^{2}$))dx=16.39\end{align}$

$\begin{align}&$\text{Using }$3$\text{ intervals, and the method of left point Riemann sums approximate the integral:}$\&\int_${0}^{9}$$(-$\frac{4}{4^e$^{(6x)}$})dx\end{align}$

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$\begin{align}&$\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}^{9}$$(-$\frac{4}{4^e$^{(6x)}$})dx\&$\text{So the interval is }$[0,9]$\text{ the subintervals have length }$$\frac{9-(0)}{3}$=3\&$\text{and since we are using the leftpoint of each interval, the x-values are:}$\&[0,3,6]\&\int_${0}^{9}$$(-$\frac{4}{4^e$^${(6x)}$})dx=3[(-1)+(-$\frac{4}{4^e$^${(18)}$})+(-$\frac{4}{4^e$^{(36)}$})]\&\int_${0}^{9}$$(-$\frac{4}{4^e$^{(6x)}$})dx=-3.00\end{align}$

$\begin{align}&$\text{Using the method of left point Riemann sums approximate the integral:}$\&\int_${2}^{19.2}$(-3cos(6ln(3x)))dx\&$\text{Using }$4$\text{ intervals.}$\end{align}$

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$\begin{align}&$\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}^{19.2}$(-3cos(6ln(3x)))dx\&$\text{So the interval is }$[2,19.2]$\text{ the subintervals have length }$$\frac{19.2-(2)}{4}$=\frac{43}{10}$\&$\text{and since we are using the leftpoint of each interval, the x-values are:}$\&[2,$\frac{63}{10}$,$\frac{53}{5}$,$\frac{149}{10}$]\&\int_${2}^{19.2}$(-3cos(6ln(3x)))dx=\frac{43}{10}$[(-3cos(6ln(6)))+(-3cos(6ln($\frac{189}{10}$)))+(-3cos(6ln($\frac{159}{5}$)))+(-3cos(6ln($\frac{447}{10}$)))]\&\int_${2}^{19.2}$(-3cos(6ln(3x)))dx=11.80\end{align}$

$\begin{align}&$\text{Using }$3$\text{ intervals, and the method of left point Riemann sums approximate the integral:}$\&\int_${-1}^{10.7}$$(-275tan(4x)^{2}$)dx\end{align}$

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$\begin{align}&$\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}^{10.7}$$(-275tan(4x)^{2}$)dx\&$\text{So the interval is }$[-1,10.7]$\text{ the subintervals have length }$$\frac{10.7-(-1)}{3}$=\frac{39}{10}$\&$\text{and since we are using the leftpoint of each interval, the x-values are:}$\&[-1,$\frac{29}{10}$,$\frac{34}{5}$]\&\int_${-1}^{10.7}$$(-275tan(4x)^{2}$$)dx=\frac{39}{10}$[(-275tan(4)^{2}$$)+(-275tan($\frac{58}{5}$)^{2}$$)+(-275tan($\frac{136}{5}$)^{2}$)]\&\int_${-1}^{10.7}$$(-275tan(4x)^{2}$)dx=-7340.85\end{align}$

$\begin{align}&$\text{Using the method of left point Riemann sums approximate the integral:}$\&\int_${1}^{6.4}$(14tan(14cos(6x)))dx\&$\text{Using }$3$\text{ intervals.}$\end{align}$

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$\begin{align}&$\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}^{6.4}$(14tan(14cos(6x)))dx\&$\text{So the interval is }$[1,6.4]$\text{ the subintervals have length }$$\frac{6.4-(1)}{3}$=\frac{9}{5}$\&$\text{and since we are using the leftpoint of each interval, the x-values are:}$\&[1,$\frac{14}{5}$,$\frac{23}{5}$]\&\int_${1}^{6.4}$(14tan(14cos(6x)))dx=\frac{9}{5}$[(14tan(14cos(6)))+(14tan(14cos($\frac{84}{5}$)))+(14tan(14cos($\frac{138}{5}$)))]\&\int_${1}^{6.4}$(14tan(14cos(6x)))dx=-394.53\end{align}$

$\begin{align}&$\text{Using }$4$\text{ intervals, and the method of left point Riemann sums approximate the integral:}$\&\int_${3}^{22.6}$$(10tan(8e^{(4x)}$))dx\end{align}$

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$\begin{align}&$\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}^{22.6}$$(10tan(8e^{(4x)}$))dx\&$\text{So the interval is }$[3,22.6]$\text{ the subintervals have length }$$\frac{22.6-(3)}{4}$=\frac{49}{10}$\&$\text{and since we are using the leftpoint of each interval, the x-values are:}$\&[3,$\frac{79}{10}$,$\frac{64}{5}$,$\frac{177}{10}$]\&\int_${3}^{22.6}$$(10tan(8e^{(4x)}$$))dx=\frac{49}{10}$[(10tan(8e^{(12)}$$))+(10tan(8e^{($\frac{158}$${5}$)}))+(10tan(8e^{($\frac{256}$${5}$)}))+(10tan(8e^{($\frac{354}${5}$)}))]\&\int_${3}^{22.6}$$(10tan(8e^{(4x)}$))dx=-168.87\end{align}$

$\begin{align}&$\text{Using }$3$\text{ intervals, and the method of left point Riemann sums approximate the integral:}$\&\int_${5}^{6.5}$(6tan(19cos(5x)))dx\end{align}$

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$\begin{align}&$\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}^{6.5}$(6tan(19cos(5x)))dx\&$\text{So the interval is }$[5,6.5]$\text{ the subintervals have length }$$\frac{6.5-(5)}{3}$=\frac{1}{2}$\&$\text{and since we are using the leftpoint of each interval, the x-values are:}$\&[5,$\frac{11}{2}$,6]\&\int_${5}^{6.5}$(6tan(19cos(5x)))dx=\frac{1}{2}$[(6tan(19cos(25)))+(6tan(19cos($\frac{55}{2}$)))+(6tan(19cos(30)))]\&\int_${5}^{6.5}$(6tan(19cos(5x)))dx=-5.54\end{align}$

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.

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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.

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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$

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

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

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

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

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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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

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

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