Question

Use the figures to calculate the left and right Riemann sums for f on the given interval and the given value of n. 3 f(x) = + 1 on (1,5), n=4 0 1 2 3 0 1 2 3 4 5 The left Riemann sum fortis (Round to two decimal places as needed.) The right Riemann sum forf is (Round to two decimal places as needed.)

Answer 1

$\\\mathrm{f\left(x\right)=\frac{3}{x}+1\:,\left(1,5\right)\:n=4}\\ \\ \mathrm{\mathrm{Riemann\:Sum\:for}\:\int _1^5\frac{3}{x}+1dx\:\mathrm{with}\:n=4\:\mathrm{rectangles,\:using\:left\:endpoints}}\\ \\ \mathrm{Riemann\:sum\:using\:left\:endpoints\:of\:subinterval}\\ \\ \mathrm{\int _a^bf\left(x\right)dx\:\approx \Delta \:x\left(f\left(x_0\right)+f\left(x_1\right)+...+f\left(x_{n-1}\right)\right),\:}\\ \\ \mathrm{\mathrm{where}\:\Delta \:x\:=\:\frac{b-a}{n}}\\ \\ \mathrm{\mathrm{Apply\:Riemann\:Formula}}\\ \\ \mathrm{\mathrm{Given:}\:a=1,\:b=5,\:n=4}\\ \\ \mathrm{\Delta x=\frac{5-1}{4}=1}\\ \\ \mathrm{\mathrm{Divide\:the\:interval}\:1\le \:x\le \:5\:\mathrm{into}\:n=4\:\mathrm{subintervals\:of\:length}\:\Delta x=1\:}\\ \\ \mathrm{x_0=1,\:x_1=2,\:x_2=3,\:x_3=4,\:x_4=5}\\ \\ \mathrm{=1\cdot \left(f\left(x_0\right)+f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)\right)}\\ \\$

$\\\mathrm{\mathrm{Calculate\:Sub\:Intervals}}\\ \\ \mathrm{\Rightarrow f\left(x_0\right):}\\ \\ \mathrm{f\left(x_0\right)=f\left(1\right)}\\ \\ \mathrm{=\frac{3}{1}+1=4}\\ \\ 4\\ \\ \mathrm{\Rightarrow f\left(x_1\right)}\\ \\ \mathrm{f\left(x_1\right)=f\left(2\right)}\\ \\ \mathrm{=\frac{3}{2}+1}\\ \\ \mathrm{=\frac{5}{2}}\\ \\ \mathrm{\Rightarrow f\left(x_2\right)}\\ \\ \mathrm{f\left(x_2\right)=f\left(3\right)}\\ \\ \mathrm{=\frac{3}{3}+1}\\ \\ \mathrm{=2}\\ \\ \mathrm{\Rightarrow f\left(x_3\right)}\\ \\ \mathrm{f\left(x_3\right)=f\left(4\right)}\\ \\ \mathrm{=\frac{3}{4}+1}\\ \\ \mathrm{=\frac{7}{4}}\\ \\$

$\bg_white \\\mathrm{\mathrm{=1\cdot \left(4+\frac{5}{2}+2+\frac{7}{4}\right)}\\}\\ \\ \mathrm{=10.25}$

$\\\mathrm{\mathrm{Riemann\:Sum\:for}\:\int _1^5\frac{3}{x}+1dx\:\mathrm{with}\:n=4\:\mathrm{rectangles,\:using\:right\:endpoints}}\\ \\ \mathrm{\mathrm{Riemann\:sum\:using\:right\:endpoints\:of\:subinterval}}\\ \\ \mathrm{\int _a^bf\left(x\right)dx\:\approx \Delta \:x\left(f\left(x_1\right)+f\left(x_2\right)+...+f\left(x_n\right)\right),\:}\\ \\ \mathrm{\mathrm{where}\:\Delta \:x\:=\:\frac{b-a}{n}}\\ \\ \mathrm{\mathrm{Apply\:Riemann\:Formula}}\\ \\ \mathrm{\mathrm{Given:}\:a=1,\:b=5,\:n=4}\\ \\ \mathrm{\Delta x=\frac{5-1}{4}=1}\\ \\ \mathrm{\mathrm{Divide\:the\:interval}\:1\le \:x\le \:5\:\mathrm{into}\:n=4\:\mathrm{subintervals\:of\:length}\:\Delta x=1\:}\\ \\ \mathrm{x_0=1,\:x_1=2,\:x_2=3,\:x_3=4,\:x_4=5}\\ \\ \mathrm{=1\cdot \left(f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)+f\left(x_4\right)\right)}\\ \\$

$\\\mathrm{\mathrm{Calculate\:Sub\:Intervals}}\\ \\ \mathrm{\Rightarrowf\left(x_1\right) }\\ \\ \mathrm{f\left(x_1\right)=f\left(2\right)}\\ \\ \mathrm{3/2+1=5/2}\\ \\ \mathrm{\Rightarrow f\left(x_2\right)}\\ \\ \mathrm{f\left(x_2\right)=f\left(3\right)}\\ \\ \mathrm{3/3+1=2}\\ \\ \mathrm{\Rightarrow f\left(x_3\right)}\\ \\ \mathrm{f\left(x_3\right)=f\left(4\right)={3}/{4}+1=7/4}\\ \\ \mathrm{\Rightarrowf\left(x_4\right) }\\ \\ \mathrm{f\left(x_4\right)=f\left(5\right)}\\ \\ \mathrm{{3}/{5}+1={8}/{5}}\\ \\ \mathrm{=1\cdot \left({5}/{2}+2+{7}/{4}+{8}/{5}\right)}\\ \\ \mathrm{=7.85}\\ \\$