Respuesta :
Answer:
2.899
Step-by-step explanation:
Given that:
[tex]f' (x) =3 tan ^{-1} (x^2 -3x +2)[/tex]
[tex]\text{applying integration on both sides with respect to x}[/tex]
[tex]\int \limits ^{3}_{1}} f'(x) \ dx = \int ^3_1 3tan ^{-1} (x^2 -3x+2) \ dx[/tex]
[tex]f(3)-f(1)= 3 \int ^3_1 3tan ^{-1} (x^2 -3x+2) \ dx[/tex]
[tex]f(3) = \dfrac{\pi}{2}+3\int^{3}_{1} tan^{-1} (x^2 -3x+2)dx ---(1)[/tex]
[tex]\text{consider } \ \ 3\int^{3}_{1} tan^{-1} (x^2 -3x+2)dx[/tex]
[tex]h = \dfrac{b-a}{n}[/tex]
[tex]where; n = 10 , a =1, b= 3[/tex]
[tex]h = \dfrac{3-1}{10}[/tex]
[tex]h = \dfrac{2}{10}[/tex]
[tex]h = 0.2[/tex]
[tex]\text{The value of x and corresponding } y = 3tan^{-1}(x^2-3x+2) \ are:[/tex]
x 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3
y f(0) f(1.2) f(1.4) f(1.6) f(1.8) f(2.0) f(2.2) f(2.4) f(2.6) f(2.8) f(3)
By simpson's rule:
[tex]\int \limits ^3_1 \ 3 tan^{-1} (x^2-3x+2) dx \simeq \dfrac{h}{3}(y_o+y_{10}) +4(y_1+y_3+y_5+y_7+y_9)+2(y_2+y_4+y_6+y_8)][/tex]
[tex]=\dfrac{0.2}{3}\Big[(0+3.3214)+(-1.9039+2.8265+1.4133+6.1259+11.5657)+2(-1.4133-0.9519+1.4133+4.5899)\Big][/tex]
[tex]= 1.32804[/tex]
[tex]\text{From}; f(3) = \dfrac{\pi}{2}+1.32804 \\ \\ f(3) = 2.89883633 \\ \\ \mathbf{f(3) \simeq 2.899}[/tex]
Trapezoidal rule is used to evaluate the area of the curve by using various strips of equal width which is in the form of trapezoids. The value of [tex]f(3)[/tex] is 2.8989.
Given information
In the given problem [tex]f[/tex] is the function such that,
[tex]f(1)=\dfrac{\pi}{2} \\[/tex]
The differentiated function is,
[tex]f'(x)=3\tan^-(x^2-3x+2)[/tex]
Integrate the above equation with respect to the x,
[tex]\int\limits^3_1 {f'(x)} \, dx =\int\limits^3_1 {3\tan^-(x^2-3x+2)} \, dx\\f(3)-f(1) =\int\limits^3_1 {3\tan^-(x^2-3x+2)} \, dx\\f(3) =3\int\limits^3_1 {\tan^-(x^2-3x+2)} \, dx +f(1)\\f(3) =3\int\limits^3_1 {\tan^-(x^2-3x+2)} \, dx +\dfrac{\pi}{2}[/tex] .........1
Let the above equation is the equation number 1.
Let the integrated function,
[tex]y=3\int\limits^3_1 {\tan^-(x^2-3x+2)} \, dx[/tex]
Solve the above equation using the trapezoidal rule.
Trapezoidal rule-
Trapezoidal rule is used to evaluate the area of the curve by using various strips of equal width which is in the form of trapezoids.
In the given problem,
[tex]h=\dfrac{b-a}{n}[/tex]
Use this formula for the integrated function,
[tex]h=\dfrac{3-1}{10}\\h=0.2[/tex]
Use the Simpson formula to solve the above equation,
[tex]=3\int\limits^3_1 {\tan^-(x^2-3x+2)} \, dx[/tex]
[tex]=\dfrac{h}{3} (y_0+y_{10})+4(y_1+y_3+y_{5}+y_7+y_{9})+2(y_2+y_{4}+y_6+y_{8})[/tex]
Find the value of y by putting the values in the given formula. Thus,
[tex]=\dfrac{0.2}{3} (0+3.3214)+4(-1.9039+2.8266+1.4133+6.1259+11.5657)+2(-1.4133-0.9519+1.4133+4.5899)\\=1.32804[/tex]
Put this value in equation 1,
[tex]f(3) =1.30824 +\dfrac{\pi}{2}\\f(3)=2.8989[/tex]
Thus the value of [tex]f(3)[/tex] is 2.8989.
Learn more about the trapezoidal rule here;
https://brainly.com/question/15323877
