over the interval (3, 5| using ten approximating||Estimate the area under the graph of f(x) =x + 2rectangles and right endpoints.RnRepeat the approximation using left endpoints.LnRound answers to 4 places. Remember not to round too early in your calculations.

Given that f(x)=1x+2 To estimate right and left endpoints: The right Riemann sum (right endpoint approximation) uses the right endpoints of a subinterval: ∫abf(x)dx≈∆xf(x1)+f(x2)+f(x3)+.......+f(xn-1)+f(xn) The left Riemann sum (left endpoint approximation) uses the left endpoints of a subinterval: ∫abf(x)dx≈∆xf(x0)+f(x1)+f(x2)+f(x3)+.......+f(xn-1)where ∆x=b-an Here a = 3, b = 5, n = 10 ∆x=5-310=15 Divide [3,5] into n = 10 subintervals of length ∆x=15. a=3,165,165,175,175,185,185,195,195,4,4,215,215,225,225,235,235,245,245,5=bTo find Right end point: Evaluate functions at the right end point: fx1=f165=526=0.192307692fx2=f175=527=0.185185185fx3=f185=528=0.178571428fx4=f195=529=0.172413793fx5=f4=16=0.166666666fx6=f215=531=0.161290322fx7=f225=532=0.15625fx8=f235=533=0.151515151fx9=f245=534=0.147058823fx10=fb=f(5)=17=0.142857142 Therefore, ∫351x+2dx≈150.192307692+0.185185185+0.178571428+0.172413793+0.166666666+0.161290322+0.15625+0.151515151+0.147058823+0.14857142 =0.33082324 ≈0.3308 Rn=0.3308 To find left end point: f(x0)=f(a)=f(3)=15=0.2fx1=f165=526=0.192307692fx2=f175=527=0.185185185fx3=f185=528=0.178571428fx4=f195=529=0.172413793fx5=f4=16=0.166666666fx6=f215=531=0.161290322fx7=f225=532=0.15625fx8=f235=533=0.151515151fx9=f245=534=0.147058823 Therefore, ∫351x+2dx≈150.2+0.192307692+0.185185185+0.178571428+0.172413793+0.166666666+0.161290322+0.15625+0.151515151+0.147058823 =0.342251822 ≈0.3422 Ln=0.3422For the function f(x)=1x+2 Rn=0.3308 Ln=0.3422