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Calculus 2 : Riemann Sums

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Example Questions

Example Question #21 : Riemann Sums

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{0}^{10.5}(17x^{2})dx\\&\text{Using }3\text{ intervals.}\end{align*}$ $\displaystyle \begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{0}^{10.5}(17x^{2})dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

3644.38 $\displaystyle 3644.38$

$\displaystyle 24417.31$

$\displaystyle 32799.38$

1214.79 $\displaystyle 1214.79$

Correct answer:

3644.38 $\displaystyle 3644.38$

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{0}^{10.5}(17x^{2})dx\\&\text{So the interval is}[0,10.5]\text{ the subintervals have length}\frac{10.5-(0)}{3}=3.5\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[0,\frac{7}{2},7]\\&\int_{0}^{10.5}(17x^{2})dx=3[(0)+(\frac{833}{4})+(833)]\\&\int_{0}^{10.5}(17x^{2})dx=3644.38\end{align*}$ $\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{0}^{10.5}(17x^{2})dx\\&\text{So the interval is}[0,10.5]\text{ the subintervals have length}\frac{10.5-(0)}{3}=3.5\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[0,\frac{7}{2},7]\\&\int_{0}^{10.5}(17x^{2})dx=3[(0)+(\frac{833}{4})+(833)]\\&\int_{0}^{10.5}(17x^{2})dx=3644.38\end{align*}$

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Example Question #71 : Introduction To Integrals

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{4}^{6.4}(11ln(x))dx\\&\text{Using left points over }4\text{ intervals.}\end{align*}$ $\displaystyle \begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{4}^{6.4}(11ln(x))dx\\&\text{Using left points over }4\text{ intervals.}\end{align*}$

Possible Answers:

19.86 $\displaystyle 19.86$

254.40 $\displaystyle 254.40$

408.70 $\displaystyle 408.70$

41.70 $\displaystyle 41.70$

Correct answer:

41.70 $\displaystyle 41.70$

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{4}^{6.4}(11ln(x))dx\\&\text{So the interval is}[4,6.4]\text{ the subintervals have length}\frac{6.4-(4)}{4}=0.6\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[4,\frac{23}{5},\frac{26}{5},\frac{29}{5}]\\&\int_{4}^{6.4}(11ln(x))dx=0.6[(11ln(4))+(11ln(\frac{23}{5}))+(11ln(\frac{26}{5}))+(11ln(\frac{29}{5}))]\\&\int_{4}^{6.4}(11ln(x))dx=41.70\end{align*}$ $\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{4}^{6.4}(11ln(x))dx\\&\text{So the interval is}[4,6.4]\text{ the subintervals have length}\frac{6.4-(4)}{4}=0.6\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[4,\frac{23}{5},\frac{26}{5},\frac{29}{5}]\\&\int_{4}^{6.4}(11ln(x))dx=0.6[(11ln(4))+(11ln(\frac{23}{5}))+(11ln(\frac{26}{5}))+(11ln(\frac{29}{5}))]\\&\int_{4}^{6.4}(11ln(x))dx=41.70\end{align*}$

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Example Question #72 : Introduction To Integrals

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{5}^{7.1}(20tan(3x))dx\\&\text{Using }3\text{ intervals.}\end{align*}$ $\displaystyle \begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{5}^{7.1}(20tan(3x))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

70.62 $\displaystyle 70.62$

649.66 $\displaystyle 649.66$

374.26 $\displaystyle 374.26$

190.66 $\displaystyle 190.66$

Correct answer:

70.62 $\displaystyle 70.62$

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{7.1}(20tan(3x))dx\\&\text{So the interval is}[5,7.1]\text{ the subintervals have length}\frac{7.1-(5)}{3}=0.7\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[5,\frac{57}{10},\frac{32}{5}]\\&\int_{5}^{7.1}(20tan(3x))dx=0.7[(20tan(15))+(20tan(\frac{171}{10}))+(20tan(\frac{96}{5}))]\\&\int_{5}^{7.1}(20tan(3x))dx=70.62\end{align*}$ $\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{7.1}(20tan(3x))dx\\&\text{So the interval is}[5,7.1]\text{ the subintervals have length}\frac{7.1-(5)}{3}=0.7\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[5,\frac{57}{10},\frac{32}{5}]\\&\int_{5}^{7.1}(20tan(3x))dx=0.7[(20tan(15))+(20tan(\frac{171}{10}))+(20tan(\frac{96}{5}))]\\&\int_{5}^{7.1}(20tan(3x))dx=70.62\end{align*}$

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Example Question #73 : Introduction To Integrals

$\begin{align*}&\text{Using }3\text{ intervals, and the method of left point Riemann sums approximate the integral:}\\&\int_{1}^{12.4}(-14x^{2})dx\end{align*}$ $\displaystyle \begin{align*}&\text{Using }3\text{ intervals, and the method of left point Riemann sums approximate the integral:}\\&\int_{1}^{12.4}(-14x^{2})dx\end{align*}$

Possible Answers:

$\displaystyle -5213.60$

-766.71 $\displaystyle -766.71$

$\displaystyle -51614.64$

$\displaystyle -10948.56$

Correct answer:

$\displaystyle -5213.60$

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{1}^{12.4}(-14x^{2})dx\\&\text{So the interval is}[1,12.4]\text{ the subintervals have length}\frac{12.4-(1)}{3}=3.8\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[1,\frac{24}{5},\frac{43}{5}]\\&\int_{1}^{12.4}(-14x^{2})dx=3.8[(-14)+(-\frac{8064}{25})+(-\frac{25886}{25})]\\&\int_{1}^{12.4}(-14x^{2})dx=-5213.60\end{align*}$ $\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{1}^{12.4}(-14x^{2})dx\\&\text{So the interval is}[1,12.4]\text{ the subintervals have length}\frac{12.4-(1)}{3}=3.8\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[1,\frac{24}{5},\frac{43}{5}]\\&\int_{1}^{12.4}(-14x^{2})dx=3.8[(-14)+(-\frac{8064}{25})+(-\frac{25886}{25})]\\&\int_{1}^{12.4}(-14x^{2})dx=-5213.60\end{align*}$

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Example Question #74 : Introduction To Integrals

$\begin{align*}&\text{Using }3\text{ intervals, and the method of left point Riemann sums approximate the integral:}\\&\int_{-1}^{6.8}(-3tan(3x))dx\end{align*}$ $\displaystyle \begin{align*}&\text{Using }3\text{ intervals, and the method of left point Riemann sums approximate the integral:}\\&\int_{-1}^{6.8}(-3tan(3x))dx\end{align*}$

Possible Answers:

10.79 $\displaystyle 10.79$

314.74 $\displaystyle 314.74$

87.43 $\displaystyle 87.43$

13.88 $\displaystyle 13.88$

Correct answer:

87.43 $\displaystyle 87.43$

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-1}^{6.8}(-3tan(3x))dx\\&\text{So the interval is}[-1,6.8]\text{ the subintervals have length}\frac{6.8-(-1)}{3}=2.6\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[-1,\frac{8}{5},\frac{21}{5}]\\&\int_{-1}^{6.8}(-3tan(3x))dx=2.6[(3tan(3))+(-3tan(\frac{24}{5}))+(-3tan(\frac{63}{5}))]\\&\int_{-1}^{6.8}(-3tan(3x))dx=87.43\end{align*}$ $\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-1}^{6.8}(-3tan(3x))dx\\&\text{So the interval is}[-1,6.8]\text{ the subintervals have length}\frac{6.8-(-1)}{3}=2.6\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[-1,\frac{8}{5},\frac{21}{5}]\\&\int_{-1}^{6.8}(-3tan(3x))dx=2.6[(3tan(3))+(-3tan(\frac{24}{5}))+(-3tan(\frac{63}{5}))]\\&\int_{-1}^{6.8}(-3tan(3x))dx=87.43\end{align*}$

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Example Question #75 : Introduction To Integrals

$\begin{align*}&\text{Using }4\text{ intervals, and the method of left point Riemann sums approximate the integral:}\\&\int_{2}^{20.8}(15tan(x))dx\end{align*}$ $\displaystyle \begin{align*}&\text{Using }4\text{ intervals, and the method of left point Riemann sums approximate the integral:}\\&\int_{2}^{20.8}(15tan(x))dx\end{align*}$

Possible Answers:

-44.55 $\displaystyle -44.55$

-27.20 $\displaystyle -27.20$

-258.40 $\displaystyle -258.40$

-568.47 $\displaystyle -568.47$

Correct answer:

-258.40 $\displaystyle -258.40$

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{2}^{20.8}(15tan(x))dx\\&\text{So the interval is}[2,20.8]\text{ the subintervals have length}\frac{20.8-(2)}{4}=4.7\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[2,\frac{67}{10},\frac{57}{5},\frac{161}{10}]\\&\int_{2}^{20.8}(15tan(x))dx=4.7[(15tan(2))+(15tan(\frac{67}{10}))+(15tan(\frac{57}{5}))+(15tan(\frac{161}{10}))]\\&\int_{2}^{20.8}(15tan(x))dx=-258.40\end{align*}$ $\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{2}^{20.8}(15tan(x))dx\\&\text{So the interval is}[2,20.8]\text{ the subintervals have length}\frac{20.8-(2)}{4}=4.7\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[2,\frac{67}{10},\frac{57}{5},\frac{161}{10}]\\&\int_{2}^{20.8}(15tan(x))dx=4.7[(15tan(2))+(15tan(\frac{67}{10}))+(15tan(\frac{57}{5}))+(15tan(\frac{161}{10}))]\\&\int_{2}^{20.8}(15tan(x))dx=-258.40\end{align*}$

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Example Question #76 : Introduction To Integrals

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{3}^{11.1}(3tan(2x))dx\\&\text{Using left points over }3\text{ intervals.}\end{align*}$ $\displaystyle \begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{3}^{11.1}(3tan(2x))dx\\&\text{Using left points over }3\text{ intervals.}\end{align*}$

Possible Answers:

-2.84 $\displaystyle -2.84$

-34.05 $\displaystyle -34.05$

-5.68 $\displaystyle -5.68$

-0.57 $\displaystyle -0.57$

Correct answer:

-5.68 $\displaystyle -5.68$

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{3}^{11.1}(3tan(2x))dx\\&\text{So the interval is}[3,11.1]\text{ the subintervals have length}\frac{11.1-(3)}{3}=2.7\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[3,\frac{57}{10},\frac{42}{5}]\\&\int_{3}^{11.1}(3tan(2x))dx=2.7[(3tan(6))+(3tan(\frac{57}{5}))+(3tan(\frac{84}{5}))]\\&\int_{3}^{11.1}(3tan(2x))dx=-5.68\end{align*}$ $\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{3}^{11.1}(3tan(2x))dx\\&\text{So the interval is}[3,11.1]\text{ the subintervals have length}\frac{11.1-(3)}{3}=2.7\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[3,\frac{57}{10},\frac{42}{5}]\\&\int_{3}^{11.1}(3tan(2x))dx=2.7[(3tan(6))+(3tan(\frac{57}{5}))+(3tan(\frac{84}{5}))]\\&\int_{3}^{11.1}(3tan(2x))dx=-5.68\end{align*}$

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Example Question #77 : Introduction To Integrals

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{0}^{3.6}(-3cos(6x))dx\\&\text{Using left points over }4\text{ intervals.}\end{align*}$ $\displaystyle \begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{0}^{3.6}(-3cos(6x))dx\\&\text{Using left points over }4\text{ intervals.}\end{align*}$

Possible Answers:

-1.51 $\displaystyle -1.51$

-8.60 $\displaystyle -8.60$

-0.18 $\displaystyle -0.18$

-0.52 $\displaystyle -0.52$

Correct answer:

-1.51 $\displaystyle -1.51$

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{0}^{3.6}(-3cos(6x))dx\\&\text{So the interval is}[0,3.6]\text{ the subintervals have length}\frac{3.6-(0)}{4}=0.9\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[0,\frac{9}{10},\frac{9}{5},\frac{27}{10}]\\&\int_{0}^{3.6}(-3cos(6x))dx=0.9[(-3)+(-3cos(\frac{27}{5}))+(-3cos(\frac{54}{5}))+(-3cos(\frac{81}{5}))]\\&\int_{0}^{3.6}(-3cos(6x))dx=-1.51\end{align*}$ $\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{0}^{3.6}(-3cos(6x))dx\\&\text{So the interval is}[0,3.6]\text{ the subintervals have length}\frac{3.6-(0)}{4}=0.9\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[0,\frac{9}{10},\frac{9}{5},\frac{27}{10}]\\&\int_{0}^{3.6}(-3cos(6x))dx=0.9[(-3)+(-3cos(\frac{27}{5}))+(-3cos(\frac{54}{5}))+(-3cos(\frac{81}{5}))]\\&\int_{0}^{3.6}(-3cos(6x))dx=-1.51\end{align*}$

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Example Question #78 : Introduction To Integrals

$\begin{align*}&\text{Using }4\text{ intervals, and the method of left point Riemann sums approximate the integral:}\\&\int_{4}^{6.8}(3ln(6x))dx\end{align*}$ $\displaystyle \begin{align*}&\text{Using }4\text{ intervals, and the method of left point Riemann sums approximate the integral:}\\&\int_{4}^{6.8}(3ln(6x))dx\end{align*}$

Possible Answers:

171.30 $\displaystyle 171.30$

28.55 $\displaystyle 28.55$

7.32 $\displaystyle 7.32$

251.25 $\displaystyle 251.25$

Correct answer:

28.55 $\displaystyle 28.55$

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{4}^{6.8}(3ln(6x))dx\\&\text{So the interval is}[4,6.8]\text{ the subintervals have length}\frac{6.8-(4)}{4}=0.7\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[4,\frac{47}{10},\frac{27}{5},\frac{61}{10}]\\&\int_{4}^{6.8}(3ln(6x))dx=0.7[(3ln(24))+(3ln(\frac{141}{5}))+(3ln(\frac{162}{5}))+(3ln(\frac{183}{5}))]\\&\int_{4}^{6.8}(3ln(6x))dx=28.55\end{align*}$ $\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{4}^{6.8}(3ln(6x))dx\\&\text{So the interval is}[4,6.8]\text{ the subintervals have length}\frac{6.8-(4)}{4}=0.7\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[4,\frac{47}{10},\frac{27}{5},\frac{61}{10}]\\&\int_{4}^{6.8}(3ln(6x))dx=0.7[(3ln(24))+(3ln(\frac{141}{5}))+(3ln(\frac{162}{5}))+(3ln(\frac{183}{5}))]\\&\int_{4}^{6.8}(3ln(6x))dx=28.55\end{align*}$

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Example Question #79 : Introduction To Integrals

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{3}^{15.9}(13ln(4x))dx\\&\text{Using }3\text{ intervals.}\end{align*}$ $\displaystyle \begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{3}^{15.9}(13ln(4x))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

3252.16 $\displaystyle 3252.16$

258.11 $\displaystyle 258.11$

4498.83 $\displaystyle 4498.83$

542.03 $\displaystyle 542.03$

Correct answer:

542.03 $\displaystyle 542.03$

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{3}^{15.9}(13ln(4x))dx\\&\text{So the interval is}[3,15.9]\text{ the subintervals have length}\frac{15.9-(3)}{3}=4.3\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[3,\frac{73}{10},\frac{58}{5}]\\&\int_{3}^{15.9}(13ln(4x))dx=4.3[(13ln(12))+(13ln(\frac{146}{5}))+(13ln(\frac{232}{5}))]\\&\int_{3}^{15.9}(13ln(4x))dx=542.03\end{align*}$ $\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{3}^{15.9}(13ln(4x))dx\\&\text{So the interval is}[3,15.9]\text{ the subintervals have length}\frac{15.9-(3)}{3}=4.3\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[3,\frac{73}{10},\frac{58}{5}]\\&\int_{3}^{15.9}(13ln(4x))dx=4.3[(13ln(12))+(13ln(\frac{146}{5}))+(13ln(\frac{232}{5}))]\\&\int_{3}^{15.9}(13ln(4x))dx=542.03\end{align*}$

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Calculus 2 : Riemann Sums

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