GATE Mechanical Engineering (ME) 2020 Shift 1 Solved Paper

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Question : 50 of 65

Marks: +1, -0

The evaluation of the definite integral

∫↙{-1}↖{1.4} {| x|} d x

by using Simpson's

1 {/} 3^{\;\text "rd "\;}

(one-third) rule with step size

h=0.6

yields

$0.592$
$0.581$
$1.248$
$0.914$

Solution:

f(x)=x(x)

a=-1

b=1.4

h=0.6

number of Interval

=\;{b-a}/{h}=\;{1.4+1}/{0.6}

n=4

$y_0$	$y_1$	$y_2$	$y_3$	$y_4$
$-1$	$-0.16$	$0.04$	$0.64$	A $.96$

By Simpson's

\;{1}/{3}

rd Rule.

∫↙{-1}↖{1.4} x{|x|}=\;{h}/{3}[y_{0}+y_{n}+2\{y_{2}\}+4(y_{1}+y_{3})]

\;\; =\;{0.6}/{3}[-1+1.96+2(0.04)+4(-0.16+0.64)]

\;\; =0.592

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