Question

Approximate the definite integral using the Trapezoidal Rule and Simpson's Rule. Compare these results with the approximation of the integral using a graphing utility. (Round your answers to four decimal places.)

7∫1ln(x) dx, n = 4

Answer 1

`\int^7_1 \in f(x) dx =7.6214`

Explanation:

From the question we are told that:

Definite integral
`\int^7_1 ln(x) dx, n = 4`

Generally the equation for Trapezoidal Rule is mathematically given by

`\int^b_a f(x) dx = (\triangle x)/(2)(f(x_0)+2f(x_1)+2f(x_2)+... 2f(x_(n-1)+f(x_n))`

Where

`h =\triangle x`

`h=(b-a)/(n)`

`h=(7-1)/(4)`

`h=1.5`

Therefore

For
`n =4\ \ i.e\ x_0 -x_4`

The Endpoints are

`a=1`

`a'=a+h=>1+1.5=>2.5`

`a''=a'+h=>2.5+1.5=>4`

`a'''=a''+h=>4+1.5=>5.5`

`b=7`

Evaluating trapezoidal function at endpoints

`f(x_0)=f(a)=f(1)=\in(0)=0`

`2f(x_1)=2f(2.5)=2\in(2.5)=1.833`

`2f(x_2)=2f(4)=2\in(4)=2.77`

`2f(x_3)=2f(5.5)=2\in(5.5)=3.41`

`f(x_0=f(7=\\(7)=1.945`

Therefore approximation of the integral

`\int^b_a f(x) dx = (\triangle x)/(2)(f(x_0)+2f(x_1)+2f(x_2)+... 2f(x_(n-1)+f(x_n))`

`\int^b_a f(x) dx = (3/2)/(2)(f(0)+1.833+2.77+3.41+1.945)`

`\int^b_a f(x) dx =7.4704`

b)

Generally the equation for Simpson,s Rule is mathematically given by

`\int^b_a f(x) dx =(\triangle x)/(3)(f(x_0)+4f(x_1)+2f(x_2)}+4f(x_3))+....2f(x_(n-2))+4f(x_(n-1))+f(x_n))`

Where

`h \triangle x`

`h=(b-a)/(n)`

`h=(7-1)/(4)`

`h=1.5`

Therefore

For
`n =4 i.e x_0 -x_4`

The Endpoints are

`a=1`

`a'=a+h=>1+1.5=>2.5`

`a''=a'+h=>2.5+1.5=>4`

`a'''=a''+h=>4+1.5=>5.5`

`b=7`

Evaluating trapezoidal function at endpoints

`f(x_0)=f(a)=f(1)= \in(0)=0`

`4f(x_1)=4f(2.5)=4 \In (2.5)=3.66516`

`2f(x_2)=2f(4)=2\in(4)=2.77`

`4f(x_3)=4f(5.5)=4\in(5.5)=6.8189`

`f(x_4)=f(b)=f(7)=in(7)=1.9456`

Therefore

`\int^b_a f(x) dx =(\triangle x)/(3)(f(x_0)+4f(x_1)+2f(x_2)}+4f(x_3))+....2f(x_(n-2))+4f(x_(n-1))+f(x_n))`

`\int^b_a f(x) dx =(1)/(2)(0+3.66516+2.77+6.8189+1.9459)`

`\int^b_a f(x) dx =7.6013`

Therefore shown graphing utility

`\int^7_1 \in f(x) dx = x\inx-x+c`

`\int^7_1 \in f(x) dx =7\in(7)-6`

`\int^7_1 \in f(x) dx =7.6214`