∫151[In(x)/5+x] dx, n=8
This is from 1 to 15, not 15 to 1. If you really need 15 to 1, just negate all the answers here.
If you have a definite integral ∫ba f(x) dx with n subintervals then:
Δx = (b-a)/n
Δx = (15-1)/8 = 7/4 or 1.75
x0=1 = a
x1=1+1.75 = 2.75
x2=1+2(1.75) = 4.5
x3=1+3(1.75) = 6.25
x4=1+4(1.75) = 8
x5=1+5(1.75) = 9.75
x6=1+6(1.75) = 11.5
x7=1+7(1.75) = 13.25
x8=1+8(1.75) = 15 = b
(a)For Trapezoidal Rule:
∫ba f(x) dx ≈ (Δx/2)[f(x0)+2f(x1)+2f(x2)+...+2f(xn-1)+f(xn)]
≈(7/8)[1 +( 2•ln(11/4)/5+11/2) + (2•ln(9/2)/5 + 9)+ ... + (ln(15)/5+15)]
≈117.27986...
(b) For Reimann sum using midpoint of subintervals:
∫ba f(x) dx ≈ Δx[f((x0+x1)/2)+f((x1+x2)/2)+f((x2+x3)/2)+f((x3+x4)/2)+f((x4+x5)/2)+f((x5+x6)/2)+f((x6+x7)/2)+f((x7+x8)/2)]
≈ (7/4)[f(15/8)+f((29/8)+f(43/8)+f(57/8)+f(71/8)+f(85/8)+f(99/8)+f(113/8)]
≈ (7/4)[ln(15/8)+15/8+ln(29/8)+29/8+ln(43/8)+43/8+ln(57/8)+57/8+ln(71/8)+71/8+ln(85/8)+85/8+ln(99/8)+99/8+ln(113/8)+113/8]
≈117.34516...
(c)∫ba f(x) dx ≈ Δx[f(x0)+4f(x1)+2f(x2)+4f(x3)...+4f(xn-2)+f(xn-1)]
≈117.31784...
