Answered: Find the area of the shaded area using…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Related questions

Question

Find the area of the shaded area using the trapezoidal rule with a step size of 2 and 1 and the Romberg technique with a step size of two

2
$4
거
-9
5
6
1+
H
4
st
--$-
y=cosh(x)=0.5(ex+ex)

Expert Solution

Step 1: Trapezoidal rule

(a) Step size $1$

$I = \int_{- 2}^{2} \int_{0}^{0.5 (e^{x} + e^{- x})} d y d x = \int_{- 2}^{2} {[y]}_{0}^{0.5 (e^{x} + e^{- x})} d x = \int_{- 2}^{2} \frac{e^{x} + e^{- x}}{2} d x$

Therefore $h = 1$

$I \approx \frac{h}{2} [f (- 2) + 2 (f (- 1) + f (0) + f (1)) + f (2)] = \frac{1}{2} [(\frac{1}{2} (e^{- 2} + e^{2})) + 2 (\frac{1}{2} (e^{- 1} + e^{1}) + \frac{1}{2} (e^{0} + e^{0}) + \frac{1}{2} (e^{1} + e^{- 1})) + (\frac{1}{2} (e^{2} + e^{- 2}))] = \frac{1}{2} [(e^{- 2} + e^{2}) + 2 (e^{- 1} + e^{1} + 1)] = 7.845 u n i t^{2}$

(b) For step size $2$

Therefore $h = 2$

$I \approx \frac{h}{2} [f (- 2) + 2 (f (0)) + f (2)] = \frac{2}{2} [(\frac{1}{2} (e^{- 2} + e^{2})) + 2 (\frac{1}{2} (e^{0} + e^{0})) + (\frac{1}{2} (e^{2} + e^{- 2}))] = \frac{2}{2} [(e^{- 2} + e^{2}) + 2] = e^{- 2} + e^{2} + 2 = 9.52 u n i t s^{2}$