From 0-2 \sin(x^{2}) dx using trapezoidal rule with n=4 as well

oppvarmet16

oppvarmet16

Answered question

2021-11-19

From 0-2 sin(x2)dx using trapezoidal rule with n=4 as well as midpoint rule with n=4

Answer & Explanation

Glenn Cooper

Glenn Cooper

Beginner2021-11-20Added 12 answers

Step 1
Trapezoidal rule
=02sin2(x)(1)
As we know that accoding to trapezoidal rule
abf(x)dxx2(f(x0)+2f(x1)+2f(x2)++2f(xn1)+f(xn)(2)
This is the standard equation for the trapezoidal method in which
x=ban
Now comparing equation (2) with equation (1) we get that
a=0,b=2,n=4
{x=204=12
Now we will divide (0,2) interval into n=4 subinterval of length x=12
end points of the function {00+12=1212+12=11+12=3232+12=2
Therefore the end points of the function are a=0,12,1,32,2
U sing these end points we analysis the function
f(x0)=f(a)=f(0)=0
2f(x1)=2f(12)=2sin2(12)=0.459698
2f(x2)=2f(1)=2sin2(1)=1.416147
2f(x3)=2f(32)=2sin2(32)=1.9899925
f(x4)=f(b)=f(2)=sin2(2)=0.0826819
Substitute these value in equation (2) we get that
=14[0+0.459698+1.416147+1.9899925+0.0826819]
=1.1731656461.1737
Step 2
By midpoint rule
=02sin2(x).(1)
According to mid point rule we know that

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