riemann and trapezoidal sums from equations\nscore: 0/5 penalty: none\nquestion\nlet f be the function…
riemann and trapezoidal sums from equations\nscore: 0/5 penalty: none\nquestion\nlet f be the function defined by f(x)=3ln(x). if five subintervals of equal length are used, what is the value of the right riemann sum approximation for ∫₂³ 3ln(x)dx? round to the nearest thousandth if necessary.\nanswer attempt 1 out of 3\nsubmit answer
Answer
Explanation:
Step1: Calculate the width of sub - intervals
The interval is from $a = 2$ to $b = 3$, and $n=5$. The width $\Delta x=\frac{b - a}{n}=\frac{3 - 2}{5}=0.2$.
Step2: Determine the right - hand endpoints
The right - hand endpoints $x_i=a + i\Delta x$ for $i = 1,2,\cdots,5$. So $x_1=2 + 0.2=2.2$, $x_2=2+2\times0.2 = 2.4$, $x_3=2 + 3\times0.2=2.6$, $x_4=2+4\times0.2 = 2.8$, $x_5=2+5\times0.2=3$.
Step3: Calculate the right - Riemann sum
The right - Riemann sum $R_5=\sum_{i = 1}^{5}f(x_i)\Delta x$. Since $f(x)=3\ln(x)$ and $\Delta x = 0.2$, we have: [ \begin{align*} R_5&=0.2\times(3\ln(2.2)+3\ln(2.4)+3\ln(2.6)+3\ln(2.8)+3\ln(3))\ &=0.6\times(\ln(2.2)+\ln(2.4)+\ln(2.6)+\ln(2.8)+\ln(3))\ &=0.6\times\ln(2.2\times2.4\times2.6\times2.8\times3)\ &=0.6\times\ln(121.0368)\ &\approx0.6\times4.7907\ &\approx2.874 \end{align*} ]
Answer:
$2.874$