   Chapter 7.7, Problem 25E

Chapter
Section
Textbook Problem

# Find the approximations L n , R n , T n and M n for n = 5 , 10, and 20. Then compute the corresponding errors E L , E R , E T , and E M . (Round your answers to six decimal places. You may wish to use the sum command on a computer algebra system.) What observations can you make? In particular, what happens to the errors when n is doubled? ∫ 0 1 x e x   d x

To determine

To evaluate:

The approximation Ln,Rn,Tn and Mn for n=5, 10 and 20 and the compute the corresponding errors EL,ER,ET and EM for the integral 01xexdx

Explanation

Formula used:

Ln=i=0n1f(a+iΔ)ΔRn=i=0n1f(a+(i+1)Δ)ΔMn=i=0n1f(a+(i+12)Δ)ΔTn=i=0n1Δ2[f(a+iΔ)+f(a+(i+1)Δ)]

Δ=ba2, b, a are limits of the integral

Calculations: Solve f(x)=01xexdx

Exact value of the given integral

01xexdx=[xexex]01=1

(a) n=5

Δ=105=15

Ln=F+f2+f3+f4=15.2e.2+.4e.4+.6e.6+.8e.8=.74294

Rn=f+f2+f3+f4+f5=15.2e.2+.4e.4+.6e.6+.8e.8+e=1.28659

Mn=f110+f310+f510+f710+f910=15.1e.1+.3e.3+.5e.5+.7e.7+9e.9=.99262

Tn=2f15+2f25+2f35+2f45+1=1102×.2e.2+2×.4e.4+2×.6e.6+2×.8e.8=1.014770

So,

EL=1.74294=.25706ER=1120659=.28659EM=1.99262=.00738ET=11.0147704=.0147704

(b) n=10

Ln=i=09f(110)Δ=110[f(.0)+f(.1)+f(.2)+f(.3)+f(.4)+f(.5)+f(.6)+f(.7)+f(.8)+f(.9)]=110[.1e.1+.2e.2+.3e.3+.4e.4+.5e.5+.6e.6+.7e.7+.8e.8+.9e.9]=110[.110517+.24408+.404958+.596728+.82436+1.093272+1.409625+1.782432+2.21364]=8.67781210=.8677812

Rn=i=09Δf((i+1)Δ)=110[.1e.1+.2e.2+.3e.3+.4e.4+.5e.5+.6e.6+.7e.7+.8e.8+.9e.9+e]=

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