Calculus : The Classic Edition (with Make the Grade and Infotrac)
Calculus : The Classic Edition (with Make the Grade and Infotrac)
5th Edition
ISBN: 9780534435387
Author: Earl W. Swokowski
Publisher: Brooks/Cole
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Question
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Chapter 5.7, Problem 1E

(a)

To determine

The integral by trapezoidal rule with four intervals.

(a)

Expert Solution
Check Mark

Answer to Problem 1E

The value of integral by trapezoidal rule is 10.75 .

Explanation of Solution

Given:

The integral is 13(x2+1)dx .

Calculation:

Integrate the function by trapezoidal rule as follows:

  abf(x)dxΔx2(f(x0)+2f(x1)+2f(x2)+...+2f(xn1)+f(xn))

We have, a=1,b=3,n=4

Therefore,

  Δx =314=12

End points of integral are calculated as follows:

  f(x0)=f(a)=f(1)=2=22f(x1)=2f(32)=132=6.52f(x2)=2f(2)=10=102f(x3)=2f(52)=292=14.5f(x4)=f(b)=f(3)=10=10

Substitute all the values in trapezoidal integral equation as follows:

  abf(x)dx14(2+6.5+10+14.5+10)=10.75

Thus, the value of integral by trapezoidal rule is 10.75 .

(b)

To determine

The integral by simpson’s rule with four intervals.

(b)

Expert Solution
Check Mark

Answer to Problem 1E

The value of integral by simpson’s rule is 10.67 .

Explanation of Solution

Given:

The integral is 13(x2+1)dx .

Calculation:

Integrate the function by simpson’s rule as follows:

  abf(x)dxΔx3(f(x0)+4f(x1)+2f(x2)+4f(x3)+2f(x4)+...+2f(xn2)+4f(xn1)+f(xn)) We We have, a=1,b=3,n=4

Therefore,

  Δx =314=12

End points of integral are calculated as follows:

  f(x0)=f(a)=f(1)=2=24f(x1)=4f(32)=13=132f(x2)=2f(2)=10=104f(x3)=4f(52)=29=29f(x4)=f(b)=f(3)=10=10

Substitute all the values in simpson’s integral equation as follows:

  abf(x)dx16(2+13+10+29+10)=10.67

Thus, the value of integral by simpson’s rule is 10.67 .

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