is my n correct in this problem.
R Ryan Rigdon Junior Member Joined Jun 10, 2010 Messages 246 Jul 19, 2010 #1 is my n correct in this problem. Attachments lastscan.jpg 85.9 KB · Views: 54
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Jul 19, 2010 #2 n = 5 is correct.\displaystyle n \ = \ 5 \ is \ correct.n = 5 is correct. ∫241xdx =˙ .69314718056, (TI−89)\displaystyle \int_{2}^{4}\frac{1}{x}dx \ \dot= \ .69314718056, \ (TI-89)∫24x1dx =˙ .69314718056, (TI−89) Trapezoid Rule when n = 5\displaystyle Trapezoid \ Rule \ when \ n \ = \ 5Trapezoid Rule when n = 5 ∫241xdx = 15[1/2+5/6+5/7+5/8+5/9+1/4] =˙ .695634920635\displaystyle \int_{2}^{4}\frac{1}{x}dx \ = \ \frac{1}{5}[1/2+5/6+5/7+5/8+5/9+1/4] \ \dot= \ .695634920635∫24x1dx = 51[1/2+5/6+5/7+5/8+5/9+1/4] =˙ .695634920635 Hence, .68 ≤ ∫241xdx ≤ .70\displaystyle Hence, \ .68 \ \le \ \int_{2}^{4}\frac{1}{x}dx \ \le \ .70Hence, .68 ≤ ∫24x1dx ≤ .70
n = 5 is correct.\displaystyle n \ = \ 5 \ is \ correct.n = 5 is correct. ∫241xdx =˙ .69314718056, (TI−89)\displaystyle \int_{2}^{4}\frac{1}{x}dx \ \dot= \ .69314718056, \ (TI-89)∫24x1dx =˙ .69314718056, (TI−89) Trapezoid Rule when n = 5\displaystyle Trapezoid \ Rule \ when \ n \ = \ 5Trapezoid Rule when n = 5 ∫241xdx = 15[1/2+5/6+5/7+5/8+5/9+1/4] =˙ .695634920635\displaystyle \int_{2}^{4}\frac{1}{x}dx \ = \ \frac{1}{5}[1/2+5/6+5/7+5/8+5/9+1/4] \ \dot= \ .695634920635∫24x1dx = 51[1/2+5/6+5/7+5/8+5/9+1/4] =˙ .695634920635 Hence, .68 ≤ ∫241xdx ≤ .70\displaystyle Hence, \ .68 \ \le \ \int_{2}^{4}\frac{1}{x}dx \ \le \ .70Hence, .68 ≤ ∫24x1dx ≤ .70
R Ryan Rigdon Junior Member Joined Jun 10, 2010 Messages 246 Jul 19, 2010 #3 i got the same answer as you. had to round off to three decimal places. thought you would like to see my work. thanx for the help again. Attachments lastscan.jpg 133.3 KB · Views: 44
i got the same answer as you. had to round off to three decimal places. thought you would like to see my work. thanx for the help again.
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Jul 19, 2010 #4 You are welcome, good show.\displaystyle You \ are \ welcome, \ good \ show.You are welcome, good show. Note, when n = 6, Simpson′s Rule:\displaystyle Note, \ when \ n \ = \ 6, \ Simpson's \ Rule:Note, when n = 6, Simpson′s Rule: ∫241xdx = 19[1/2+12/7+3/4+4/3+3/5+12/11+1/4] =˙ .693167979317\displaystyle \int_{2}^{4}\frac{1}{x}dx \ = \ \frac{1}{9}[1/2+12/7+3/4+4/3+3/5+12/11+1/4] \ \dot= \ .693167979317∫24x1dx = 91[1/2+12/7+3/4+4/3+3/5+12/11+1/4] =˙ .693167979317 An error of only .69314718056−.693167979317 = −.00003255044.\displaystyle An \ error \ of \ only \ .69314718056-.693167979317 \ = \ -.00003255044.An error of only .69314718056−.693167979317 = −.00003255044.
You are welcome, good show.\displaystyle You \ are \ welcome, \ good \ show.You are welcome, good show. Note, when n = 6, Simpson′s Rule:\displaystyle Note, \ when \ n \ = \ 6, \ Simpson's \ Rule:Note, when n = 6, Simpson′s Rule: ∫241xdx = 19[1/2+12/7+3/4+4/3+3/5+12/11+1/4] =˙ .693167979317\displaystyle \int_{2}^{4}\frac{1}{x}dx \ = \ \frac{1}{9}[1/2+12/7+3/4+4/3+3/5+12/11+1/4] \ \dot= \ .693167979317∫24x1dx = 91[1/2+12/7+3/4+4/3+3/5+12/11+1/4] =˙ .693167979317 An error of only .69314718056−.693167979317 = −.00003255044.\displaystyle An \ error \ of \ only \ .69314718056-.693167979317 \ = \ -.00003255044.An error of only .69314718056−.693167979317 = −.00003255044.