AP Calculus BC - Numerical Approximations to Definite Integrals

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

Explanation

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{0}^{12}(11sin(20sin(x)))dx\&\text{So the interval is }[0,12]\text{ the subintervals have length }\frac{12-(0)}{3}=4\&\text{and since we are using the *left*point of each interval, the x-values are:}\&[0,4,8]\&\int_{0}^{12}(11sin(20sin(x)))dx=4[(0)+(11sin(20sin(4)))+(11sin(20sin(8)))]\&\int_{0}^{12}(11sin(20sin(x)))dx=11.66\end{align*}$

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

Explanation

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{-2}^{11.6}(\frac{18}{3^{(4tan(4x))}})dx\&\text{So the interval is }[-2,11.6]\text{ the subintervals have length }\frac{11.6-(-2)}{4}=\frac{17}{5}\&\text{and since we are using the *left*point of each interval, the x-values are:}\&[-2,\frac{7}{5},\frac{24}{5},\frac{41}{5}]\&\int_{-2}^{11.6}(\frac{18}{3^{(4tan(4x))}})dx=\frac{17}{5}[(18\cdot 3^{(4tan(8))})+(\frac{18}{3^{(4tan(\frac{28}{5}))}})+(\frac{18}{3^{(4tan(\frac{96}{5}))}})+(\frac{18}{3^{(4tan(\frac{164}{5}))}})]\&\int_{-2}^{11.6}(\frac{18}{3^{(4tan(4x))}})dx=2200.80\end{align*}$

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

Explanation

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

Explanation

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{-2}^{11.6}(\frac{18}{3^{(4tan(4x))}})dx\&\text{So the interval is }[-2,11.6]\text{ the subintervals have length }\frac{11.6-(-2)}{4}=\frac{17}{5}\&\text{and since we are using the *left*point of each interval, the x-values are:}\&[-2,\frac{7}{5},\frac{24}{5},\frac{41}{5}]\&\int_{-2}^{11.6}(\frac{18}{3^{(4tan(4x))}})dx=\frac{17}{5}[(18\cdot 3^{(4tan(8))})+(\frac{18}{3^{(4tan(\frac{28}{5}))}})+(\frac{18}{3^{(4tan(\frac{96}{5}))}})+(\frac{18}{3^{(4tan(\frac{164}{5}))}})]\&\int_{-2}^{11.6}(\frac{18}{3^{(4tan(4x))}})dx=2200.80\end{align*}$

$\begin{align*}&\text{Calculate the right-side Riemann sums integral approximation of :}\&\int_{-1}^{15}(3x + 5sin(2e^{(2x)}))dx\&\text{Using points over }4\text{ intervals.}\end{align*}$

Explanation

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{-1}^{15}(3x + 5sin(2e^{(2x)}))dx\&\text{So the interval is }[-1,15]\text{ the subintervals have length }\frac{15-(-1)}{4}=4\&\text{and since we are using the *right*point of each interval, the x-values are:}\&[3,7,11,15]\&\int_{-1}^{15}(3x + 5sin(2e^{(2x)}))dx\approx4[(3(3) + 5sin(2e^{(2(3))}))+(3(7) + 5sin(2e^{(2(7))}))+(3(11) + 5sin(2e^{(2(11))}))+(3(15) + 5sin(2e^{(2(15))}))]\&\int_{-1}^{15}(3x + 5sin(2e^{(2x)}))dx\approx431.64\end{align*}$

$\begin{align*}&\text{Calculate the right-side Riemann sums integral approximation of :}\&\int_{-1}^{15}(3x + 5sin(2e^{(2x)}))dx\&\text{Using points over }4\text{ intervals.}\end{align*}$

Explanation

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{-1}^{15}(3x + 5sin(2e^{(2x)}))dx\&\text{So the interval is }[-1,15]\text{ the subintervals have length }\frac{15-(-1)}{4}=4\&\text{and since we are using the *right*point of each interval, the x-values are:}\&[3,7,11,15]\&\int_{-1}^{15}(3x + 5sin(2e^{(2x)}))dx\approx4[(3(3) + 5sin(2e^{(2(3))}))+(3(7) + 5sin(2e^{(2(7))}))+(3(11) + 5sin(2e^{(2(11))}))+(3(15) + 5sin(2e^{(2(15))}))]\&\int_{-1}^{15}(3x + 5sin(2e^{(2x)}))dx\approx431.64\end{align*}$

Approximate

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

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

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. 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

$x_{0} = 1, x_{1} = 1.2,...x_{5} = 2$ .

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

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

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

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

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

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

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

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

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

$\approx 0.1 \cdot 156.8955 \approx 15.690$

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

Explanation

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{0}^{3}(-16sin(2x^{2}))dx\&\text{So the interval is }[0,3]\text{ the subintervals have length }\frac{3-(0)}{3}=1\&\text{and since we are using the *left*point of each interval, the x-values are:}\&[0,1,2]\&\int_{0}^{3}(-16sin(2x^{2}))dx=1[(0)+(-16sin(2))+(-16sin(8))]\&\int_{0}^{3}(-16sin(2x^{2}))dx=-30.38\end{align*}$

Find the Left Riemann sum of the function

on the interval divided into four sub-intervals.

Explanation

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

Find the Left Riemann sum of the function

on the interval divided into four sub-intervals.

Explanation

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

Numerical Approximations to Definite Integrals

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