Question

Objective The usual procedure for evaluating a definite integral is to find the antiderivative of the integrand and apply the Fundamental Theorem of Calculus. However, if an antiderivative of the integrand cannot be found, then we must settle for a numerical approximation of the integral. The objective of this project is to illustrate the Trapezoidal Rule and Simpson's Rule. Description To get started, read the section 8.6 in the text. In this project we will illustrate and compare Riemann sum, the Trapezoidal Rule, and Simpson's Rule by looking at how they approximate the definite integral: 1 dx = In 2 (1) 0 +1 Part A A1.) Use n 10 subintervals to approximate the integral (1) by the midpoint approximation 10 subintervals to approximate the integral (1) by the trapezoidal approximation A2.) Use n A3.) Use 2n 20 subintervals to approximate the integral (1) by the Simpson's rule In each case, use the exact value of the integral to approximate the absolute error. Express your answers to at least five decimal places. Which method - Riemann sums, the Trapezoidal Rule, and Simpson's Rule, respectively-provides da In 2? Which provides the worst? the best estimation of +1

Answer 1

= In 2 = 0. 6931 47 1 +- X Part A AL) the mid paiut ruli J using Evalnatce Fia) = amd heri force t ub in tur rals eonsist of are mid paintzo theee &ubintcer vals The 17 13 26 26 20 20 Thus dx ) * +( + 20 20 20 20 20 26 2 3 24 29 3 1 20 20 20 20 35 37 39 33 ) 1 20 1 23 25 2 9 2 1 20 35 2 047119 +0.043478 +0.04+ O.037037+ O.03443+ 0032258 +o.030303t O028571 + O.0270 27 t 20 23 0.025641 = 2 x 030346413 0.692834 EYror = 0.692834-0.6931471 1 tAbsolntu 3.13x101-0.000313

A 2 .) the trapuroidal C= E valnat Tule ueing and The undpants f the fubinturrale eondi'dt uloment of thu sit 1 Thus (419+24(t) + 24(금) + 2 4 () + 2. + 2.16 + 210 12 +2. 10 14 + 2. 10 (6 + 2. 2.d 2. 10 17 2. 10 19 13 7 16 0.090909 0.0833} 34 OO76923+ 71429+ 0.0666 67+ O.O625 0.058824+ 0.05TTS + 20 00 2632 0.07 61877 3 0 693773 Errar So Abdelte 0.693773 0693147 1 6. 26 X10 0.000624

2n- 20 =)n1o Evaluatie Mardhy ing +eu Yule 1 and 11 parnts af the subintcer ral ermk6t the fut p The und 4 4uliments Thus, Fo)+f() 4 + 2 + + | 0 410 1b +4. 2 o 10 1 0 40 + - 20 { 083333 0.07142 9+ 40x -1 2, o 4+ O.090909 0.0769 23 t 20 0O66663+ + 0.058 24 + O.052632 S4 40XO 272&18 +20 x O.345955 +1 0.693147-0 644394 10.91272 +6.9191 O.644394 fo Atolutu Errr

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