Question

approximate the integral 0 to pi/2 xsin(x)dx by computing the Left and Right sums, using the parition (0, pi/6, pi/4, pi/3, pi/2).

Answer 1

since we have partitions, our Δx will be the ; largest length of our partitions as the following: 

π/6 - 0 = π/6 ===> largest 
π/4 - π/6 = π/12 
π/3 - π/4 = π/12 
π/2 - π/3 = π/6 ===> largest 

Δx = π/6 

Right hand sum: 

π/2 
∫ xsin(x)dx ≈ [ f(π/6) * Δx + f(π/4) * Δx + f(π/3) * Δx + f(π/2) * Δx ] 
0 

Δx * [ f(π/6) + f(π/4) + f(π/3) + f(π/2) ] 

(π/6) * [ (π/6) * sin(π/6) + (π/4) * sin(π/4) + (π/3) * sin(π/3) + (π/2) * sin(π/2) ] 

(π/6) * [ (π/6) * (1/2) + (π/4) * (√(2)/2) + (π/3) * (√(3)/2) + (π/2) * 1 ] 

(π/6) * ( (π/12) + (π√(2)/8) + (π√(3)/6) + (π/2) ) ≈ 1.73 

Left hand sum: 

Δx * [ f(0) + f(π/6) + f(π/4) + f(π/3) ] 

(π/6) * [ 0 + (π/6) * sin(π/6) + (π/4) * sin(π/4) + (π/3) * sin(π/3) ] 

(π/6) * [ (π/6) * (1/2) + (π/4) * (√(2)/2) + (π/3) * (√(3)/2) ] 

(π/6) * ( (π/12) + (π√(2)/8) + (π√(3)/6) ) ≈ 0.903 

Actual : 

π/2 
∫ xsin(x)dx = 1 
0