Answer:
[tex]\int^7_1 \in f(x) dx =7.6214[/tex]
Step-by-step explanation:
From the question we are told that:
Definite integral [tex]\int^7_1 ln(x) dx, n = 4[/tex]
Generally the equation for Trapezoidal Rule is mathematically given by
[tex]\int^b_a f(x) dx = \frac{\triangle x}{2}(f(x_0)+2f(x_1)+2f(x_2)+... 2f(x_{n-1}+f(x_n))[/tex]
Where
[tex]h =\triangle x[/tex]
[tex]h=\frac{b-a}{n}[/tex]
[tex]h=\frac{7-1}{4}[/tex]
[tex]h=1.5[/tex]
Therefore
For [tex]n =4\ \ i.e\ x_0 -x_4[/tex]
The Endpoints are
[tex]a=1[/tex]
[tex]a'=a+h=>1+1.5=>2.5[/tex]
[tex]a''=a'+h=>2.5+1.5=>4[/tex]
[tex]a'''=a''+h=>4+1.5=>5.5[/tex]
[tex]b=7[/tex]
Evaluating trapezoidal function at endpoints
[tex]f(x_0)=f(a)=f(1)=\in(0)=0[/tex]
[tex]2f(x_1)=2f(2.5)=2\in(2.5)=1.833[/tex]
[tex]2f(x_2)=2f(4)=2\in(4)=2.77[/tex]
[tex]2f(x_3)=2f(5.5)=2\in(5.5)=3.41[/tex]
[tex]f(x_0=f(7=\n(7)=1.945[/tex]
Therefore approximation of the integral
[tex]\int^b_a f(x) dx = \frac{\triangle x}{2}(f(x_0)+2f(x_1)+2f(x_2)+... 2f(x_{n-1}+f(x_n))[/tex]
[tex]\int^b_a f(x) dx = \frac{3/2}{2}(f(0)+1.833+2.77+3.41+1.945)[/tex]
[tex]\int^b_a f(x) dx =7.4704[/tex]
b)
Generally the equation for Simpson,s Rule is mathematically given by
[tex]\int^b_a f(x) dx =\frac{\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))[/tex]
Where
[tex]h \triangle x[/tex]
[tex]h=\frac{b-a}{n}[/tex]
[tex]h=\frac{7-1}{4}[/tex]
[tex]h=1.5[/tex]
Therefore
For [tex]n =4 i.e x_0 -x_4[/tex]
The Endpoints are
[tex]a=1[/tex]
[tex]a'=a+h=>1+1.5=>2.5[/tex]
[tex]a''=a'+h=>2.5+1.5=>4[/tex]
[tex]a'''=a''+h=>4+1.5=>5.5[/tex]
[tex]b=7[/tex]
Evaluating trapezoidal function at endpoints
[tex]f(x_0)=f(a)=f(1)= \in(0)=0[/tex]
[tex]4f(x_1)=4f(2.5)=4 \In (2.5)=3.66516[/tex]
[tex]2f(x_2)=2f(4)=2\in(4)=2.77[/tex]
[tex]4f(x_3)=4f(5.5)=4\in(5.5)=6.8189[/tex]
[tex]f(x_4)=f(b)=f(7)=in(7)=1.9456[/tex]
Therefore
[tex]\int^b_a f(x) dx =\frac{\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))[/tex]
[tex]\int^b_a f(x) dx =\frac{1}{2}(0+3.66516+2.77+6.8189+1.9459)[/tex]
[tex]\int^b_a f(x) dx =7.6013[/tex]
Therefore shown graphing utility
[tex]\int^7_1 \in f(x) dx = x\inx-x+c[/tex]
[tex]\int^7_1 \in f(x) dx =7\in(7)-6[/tex]
[tex]\int^7_1 \in f(x) dx =7.6214[/tex]
