Question

Let f be a differentiable function such that f(1)=π2 and f′(x)=3arctan(x2−3x+2). What is the value of f(3) ?

Answer 1

Answer:

2.899

Step-by-step explanation:

Given that:

$f' (x) =3 tan ^{-1} (x^2 -3x +2)$

$\text{applying integration on both sides with respect to x}$

$\int \limits ^{3}_{1}} f'(x) \ dx = \int ^3_1 3tan ^{-1} (x^2 -3x+2) \ dx$

$f(3)-f(1)= 3 \int ^3_1 3tan ^{-1} (x^2 -3x+2) \ dx$

$f(3) = \dfrac{\pi}{2}+3\int^{3}_{1} tan^{-1} (x^2 -3x+2)dx ---(1)$

$\text{consider } \ \ 3\int^{3}_{1} tan^{-1} (x^2 -3x+2)dx$

$h = \dfrac{b-a}{n}$

$where; n = 10 , a =1, b= 3$

$h = \dfrac{3-1}{10}$

$h = \dfrac{2}{10}$

$h = 0.2$

$\text{The value of x and corresponding } y = 3tan^{-1}(x^2-3x+2) \ are:$

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:

$\int \limits ^3_1 \ 3 tan^{-1} (x^2-3x+2) dx \simeq \dfrac{h}{3}(y_o+y_{10}) +4(y_1+y_3+y_5+y_7+y_9)+2(y_2+y_4+y_6+y_8)]$

$=\dfrac{0.2}{3}\Big[(0+3.3214)+(-1.9039+2.8265+1.4133+6.1259+11.5657)+2(-1.4133-0.9519+1.4133+4.5899)\Big]$

$= 1.32804$

$\text{From}; f(3) = \dfrac{\pi}{2}+1.32804 \\ \\ f(3) = 2.89883633 \\ \\ \mathbf{f(3) \simeq 2.899}$