Question Number 125997 by bramlexs22 last updated on 16/Dec/20
$${Find}\:{the}\:{Riemann}\:{sum}\:{for}\:{the} \\ $$$${given}\:{function}\:{with}\:{the}\:{specified} \\ $$$${number}\:{of}\:{intervals}\:{using}\:{left} \\ $$$${endpoints}\:{f}\left({x}\right)=\:\mathrm{4ln}\:{x}+\mathrm{2}{x}\:;\:\mathrm{1}\leqslant{x}\leqslant\mathrm{4} \\ $$$${n}=\mathrm{7}\:.\:{Round}\:{your}\:{answer}\:{to}\:{two} \\ $$$${decimal}\:{places}\:? \\ $$
Answered by liberty last updated on 16/Dec/20
![We want to use seven rectangles to equals width to estimate the area under f(x)=4ln x+2x over the interval 1≤x≤4 . Δx = ((4−1)/7) = (3/7) then devide the interval [ 1 ,4 ] into 7 equals pieces . Gives x=1, x=((10)/7) ,x=((13)/7) ,...,x=((25)/7) the areas : (3/7)f(1)+(3/7)f(((10)/7))+(3/7)f(((13)/7))+...+(3/7)f(((25)/7)) = (3/7) { 4ln 1+2+4ln ((10)/7)+((20)/7)+4ln ((13)/7)+((26)/7)+...+4ln ((25)/7)+((50)/7) } ≈ 22.66](https://www.tinkutara.com/question/Q126021.png)
$${We}\:{want}\:{to}\:{use}\:{seven}\:{rectangles}\:{to}\:{equals} \\ $$$${width}\:{to}\:{estimate}\:{the}\:{area}\:{under}\:{f}\left({x}\right)=\mathrm{4ln}\:{x}+\mathrm{2}{x} \\ $$$${over}\:{the}\:{interval}\:\mathrm{1}\leqslant{x}\leqslant\mathrm{4}\:. \\ $$$$\:\Delta{x}\:=\:\frac{\mathrm{4}−\mathrm{1}}{\mathrm{7}}\:=\:\frac{\mathrm{3}}{\mathrm{7}} \\ $$$${then}\:{devide}\:{the}\:{interval}\:\left[\:\mathrm{1}\:,\mathrm{4}\:\right]\:{into}\:\mathrm{7}\:{equals}\: \\ $$$${pieces}\:.\:{Gives}\:{x}=\mathrm{1},\:{x}=\frac{\mathrm{10}}{\mathrm{7}}\:,{x}=\frac{\mathrm{13}}{\mathrm{7}}\:,…,{x}=\frac{\mathrm{25}}{\mathrm{7}} \\ $$$${the}\:{areas}\::\:\frac{\mathrm{3}}{\mathrm{7}}{f}\left(\mathrm{1}\right)+\frac{\mathrm{3}}{\mathrm{7}}{f}\left(\frac{\mathrm{10}}{\mathrm{7}}\right)+\frac{\mathrm{3}}{\mathrm{7}}{f}\left(\frac{\mathrm{13}}{\mathrm{7}}\right)+…+\frac{\mathrm{3}}{\mathrm{7}}{f}\left(\frac{\mathrm{25}}{\mathrm{7}}\right) \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{7}}\:\left\{\:\mathrm{4ln}\:\mathrm{1}+\mathrm{2}+\mathrm{4ln}\:\frac{\mathrm{10}}{\mathrm{7}}+\frac{\mathrm{20}}{\mathrm{7}}+\mathrm{4ln}\:\frac{\mathrm{13}}{\mathrm{7}}+\frac{\mathrm{26}}{\mathrm{7}}+…+\mathrm{4ln}\:\frac{\mathrm{25}}{\mathrm{7}}+\frac{\mathrm{50}}{\mathrm{7}}\:\right\} \\ $$$$\:\approx\:\mathrm{22}.\mathrm{66}\: \\ $$