(1)::f(x)=∫_0 ^1 ∣x−2t∣dt, ∫_0 ^3 f(x)dx=? −−−−−−−−−−−−−−−−−−− (2)::f(x)=x^2 ∙∫_1 ^x (dt/(t^3 −3t^2 +3t)), f^((2019)) (1)=?


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Question Number 138819 by qaz last updated on 18/Apr/21

$(1) :: f (x) = \int_{0}^{1} ∣ x - 2 t ∣ d t, \int_{0}^{3} f (x) d x = ?$ $- - - - - - - - - - - - - - - - - - -$ $(2) :: f (x) = x^{2} \cdot \int_{1}^{x} \frac{d t}{t^{3} - 3 t^{2} + 3 t}, f^{(2019)} (1) = ?$
	Answered by TheSupreme last updated on 18/Apr/21

	$\int_{0}^{1} ∣ x - 2 t ∣ d t =$ $x < 0 \to \int_{0}^{1} ∣ x - 2 t ∣= \int_{0}^{1} - x + 2 t d t = - x t + t^{2} = 1 - x$ $x > 2 \to \int_{0}^{1} ∣ x - 2 t ∣= \int_{0}^{1} x - 2 t d t = x t - t^{2} = x - 1$ $0 < x < 2$ $\int_{0}^{1} ∣ x - 2 t ∣ d t = \int_{0}^{\frac{x}{2}} x - 2 t d t + \int_{\frac{x}{2}}^{1} - x + 2 t d t =$ $= x t - t^{2} ∣_{0}^{\frac{x}{2}} + (- x t + t^{2}) ∣_{\frac{x}{2}}^{1} = \frac{x^{2}}{2} - \frac{x^{2}}{4} - x + 1 + \frac{x^{2}}{2} - \frac{x^{2}}{4} = \frac{x^{2}}{2} - x + 1$
	Answered by mathmax by abdo last updated on 18/Apr/21

	$f (x) = \int_{0}^{1} ∣ x - 2 t ∣ dt =_{2 t = u} \frac{1}{2} \int_{0}^{2} ∣ x - u ∣ du$ $\Rightarrow 2 f (x) = \int_{0}^{x} ∣ x - u ∣ du + \int_{x}^{2} ∣ x - u ∣ du = \int_{0}^{x} (x - u) du + \int_{x}^{2} (u - x) du$ $= {[xu - \frac{u^{2}}{2}]}_{u = 0}^{x} + {[\frac{u^{2}}{2} - xu]}_{x}^{2} = x^{2} - \frac{x^{2}}{2} + 2 - 2 x - \frac{x^{2}}{2} + x^{2}$ $= {2 x}^{2} - x^{2} + 2 - 2 x = x^{2} - 2 x + 2 \Rightarrow f (x) = \frac{x^{2}}{2} - x + 1 \Rightarrow$ $\int_{0}^{3} f (x) dx = \int_{0}^{3} (\frac{x^{2}}{2} - x + 1) dx = {[\frac{x^{3}}{6} - \frac{x^{2}}{2} + x]}_{0}^{3} = \frac{27}{6} - \frac{9}{2} + 3 = . . .$
	Answered by mathmax by abdo last updated on 18/Apr/21

	$f (x) = x^{2} \int_{1}^{x} \frac{dt}{t (t^{2} - 3 t + 3)} let decompose F (t) = \frac{1}{t (t^{2} - 3 t + 3)}$ $\Rightarrow F (t) = \frac{a}{t} + \frac{bt + c}{t^{2} - 3 t + 3} we have a = \frac{1}{3}$ $\lim_{t \to + \infty} tF (t) = 0 = a + b \Rightarrow b = - \frac{1}{3}$ $F (1) = 1 = a + b + c \Rightarrow c = 1 - 0 = 1 \Rightarrow F (x) = \frac{1}{3 t} + \frac{- \frac{t}{3} + 1}{t^{2} - 3 t + 3} \Rightarrow$ $\int_{1}^{x} F (t) dt = \int_{1}^{x} \frac{dt}{3 t} - \frac{1}{6} \int_{1}^{x} \frac{2 t - 3 - 3}{t^{2} - 3 t + 3} dt$ $= \frac{1}{3} \log ∣ x ∣ - \frac{1}{6} {[\log (t^{2} - 3 t + 3)]}_{1}^{x} + \frac{3}{6} \int_{1}^{x} \frac{dt}{t^{2} - 3 t + 3}$ $= \frac{1}{3} \log ∣ x ∣ - \frac{1}{6} \log (x^{2} - 3 x + 3) + \frac{3}{6} \int_{1}^{x} \frac{dt}{t^{2} - 3 t + 3}$ $\int_{1}^{x} \frac{dt}{t^{2} - 3 t + 3} = \int_{1}^{x} \frac{dt}{t^{2} - 2 \frac{3}{2} t + \frac{9}{4} + 3 - \frac{9}{4}}$ $= \int_{1}^{x} \frac{dt}{{(t - \frac{3}{2})}^{2} + \frac{3}{4}} =_{t - \frac{3}{2} = \frac{\sqrt{3}}{2} y} \frac{4}{3} \int_{- \frac{1}{\sqrt{3}}}^{\frac{2 x - 3}{\sqrt{3}}} \frac{1}{y^{2} + 1} . \frac{\sqrt{3}}{2} dy$ $= \frac{2}{\sqrt{3}} {[arctany]}_{- \frac{1}{\sqrt{3}}}^{\frac{2 x - 3}{\sqrt{3}}} = \frac{2}{\sqrt{3}} (\arctan (\frac{2 x - 3}{\sqrt{3}}) + \arctan (\frac{1}{\sqrt{3}}))$ $\Rightarrow \int_{1}^{x} F (x) dx = \frac{1}{3} \log ∣ x ∣ - \frac{1}{6} \log (x^{2} - 3 x + 3)$ $+ \frac{1}{\sqrt{3}} (\arctan (\frac{2 x - 3}{\sqrt{3}}) + \frac{π}{6}) \Rightarrow$ $f (x) = \frac{1}{3} x^{2} \log ∣ x ∣ - \frac{x^{2}}{6} \log (x^{2} - 3 x + 3) + \frac{x^{2}}{\sqrt{3}} (\arctan (\frac{2 x - 3}{\sqrt{3}}) + \frac{π}{6})$ $rest to calculate f^{(n)} (x) . . . . be continued . . . .$
	Commented by qaz last updated on 19/Apr/21

	${i t}^{'} s j u s t a f i l l - i n - t h e - b l a n k s e x e r c i s e,$ $i d o n t t h i n k o f t h a t s o c o m p l i c a t e .$
	Answered by ajfour last updated on 19/Apr/21
	$f(x)=2∫_0 ^( 1) (√((t−(x/2))^2 ))dx =(t−(x/2))(√((t−(x/2))^2 )) ∣_0 ^1 =(1−(x/2))(√((1−(x/2))^2 ))+(x/2)(√(x^2 /4)) ∫_0 ^( 3) f(x)dx=∫_0 ^( 2) {(1−(x/2))^2 dx −∫_2 ^( 3) {(1−(x/2))^2 dx+∫_0 ^( 3) (x^2 /4)dx =(2/3)((x/2)−1)^3 ∣_0 ^2 −(2/3)((x/2)−1)^3 ∣_2 ^3 +(x^3 /(12))∣_0 ^3 =(2/3)−(1/(12))+(9/4)=((17)/6)$
	$f (x) = 2 \int_{0}^{1} \sqrt{{(t - \frac{x}{2})}^{2}} d x$ $= (t - \frac{x}{2}) \sqrt{{(t - \frac{x}{2})}^{2}} ∣_{0}^{1}$ $= (1 - \frac{x}{2}) \sqrt{{(1 - \frac{x}{2})}^{2}} + \frac{x}{2} \sqrt{\frac{x^{2}}{4}}$ $\int_{0}^{3} f (x) d x = \int_{0}^{2} {{(1 - \frac{x}{2})}^{2} d x$ $- \int_{2}^{3} {{(1 - \frac{x}{2})}^{2} d x + \int_{0}^{3} \frac{x^{2}}{4} d x$ $= \frac{2}{3} {(\frac{x}{2} - 1)}^{3} ∣_{0}^{2} - \frac{2}{3} {(\frac{x}{2} - 1)}^{3} ∣_{2}^{3}$ $+ \frac{x^{3}}{12} ∣_{0}^{3}$ $= \frac{2}{3} - \frac{1}{12} + \frac{9}{4} = \frac{17}{6}$
	Answered by ajfour last updated on 19/Apr/21
	$(df/dx)=(x^2 /(x^3 −3x^2 +3x))+((2f)/x) x^2 (((df(x))/dx))−2xf(x)=(x^3 /(x^2 −3x+3)) d(((f(x))/x^2 ))=(1/(x(x^2 −3x+3))) (3/(3x(x^2 −3x+3)))=(1/(3x))+((Ax+B)/(x^2 −3x+3)) ⇒ x^2 −3x+3+3Ax^2 +3Bx=3 ⇒ A=−(1/3), B=1=(3/3) f(x)=x^2 {((ln x)/3)−(1/3)∫(((x−3)dx)/(x^2 −3x+3))+k} f^( 1) (x)=((2xln x)/3)+(x/3)−((x^2 (x−3))/(3(x^2 −3x+3))) +2x(((f(x))/x^2 )−((ln x)/3)) f^( 1) (x)=(x/3)−((x^2 (x−3))/(3(x^2 −3x+3)))+((2f(x))/x) xf^( 1) (x)=(x^2 /3)−((x^2 (x^2 −3x))/(3(x^2 −3x+3)))+2f(x) f^( 1) (x)+xf^( 2) (x)=((2x(x^2 −3x+3)−x^2 (2x−3))/((x^2 −3x+3)^2 ))+2f^( 1) (x) xf^( 2) (x)=−((3x^2 )/((x^2 −3x+3)^2 ))+f^( 1) (x) x(x^2 −3x+3)^2 f^( 2) (x)+3x^2 =(x^2 −3x+3)^2 f^( 1) (x) {(x^2 −3x+3)^2 +2x(2x−3)(x^2 −3x+3)}f^( 2) (x) +x(x^2 −3x+3)^2 f^( 3) (x)+6x =2(x^2 −3x+3)(2x−3)f^( 1) (x) +(x^2 −3x+3)^2 f^( 2) (x) ⇒ 2x(2x−3)(x^2 −3x+3)f^( 2) (x) +x(x^2 −3x+3)^2 f^( 3) (x)+6 =2(x^2 −3x+3)(2x−3)f^( 1) (x) ⇒ 2(2x−3)(3x^2 )=6(x^2 −3x+3) +x(x^2 −3x+3)^2 f^( 3) (x) f^( 3) (x)=((6x^2 (2x−3)−6(x^2 −3x+3))/(x(x^2 −3x+3)^2 )) =((6(2x^3 −4x^2 +3x−3))/(x(x^2 −3x+3)^2 )) let x=t+h f^( 3) (t)=((6(2t^3 +6ht^2 +6h^2 t+h^3 −4t^2 −8ht−4h^2 +3t+3h−3))/((t+h)(t^2 +2ht+h^2 −3t−3h+3)^2 )) let −2h^3 +6h^3 −6h^3 +h^3 −4h^2 +8h^2 −4h^2 −3h+3h−3=0 ⇒ −h^3 −3=0 h=−3^(1/3) ...$
	$\frac{d f}{d x} = \frac{x^{2}}{x^{3} - 3 x^{2} + 3 x} + \frac{2 f}{x}$ $x^{2} (\frac{d f (x)}{d x}) - 2 x f (x) = \frac{x^{3}}{x^{2} - 3 x + 3}$ $d (\frac{f (x)}{x^{2}}) = \frac{1}{x (x^{2} - 3 x + 3)}$ $\frac{3}{3 x (x^{2} - 3 x + 3)} = \frac{1}{3 x} + \frac{A x + B}{x^{2} - 3 x + 3}$ $\Rightarrow x^{2} - 3 x + 3 + 3 {A x}^{2} + 3 B x = 3$ $\Rightarrow A = - \frac{1}{3}, B = 1 = \frac{3}{3}$ $f (x) = x^{2} {\frac{\ln x}{3} - \frac{1}{3} \int \frac{(x - 3) d x}{x^{2} - 3 x + 3} + k}$ $f^{1} (x) = \frac{2 x \ln x}{3} + \frac{x}{3} - \frac{x^{2} (x - 3)}{3 (x^{2} - 3 x + 3)}$ $+ 2 x (\frac{f (x)}{x^{2}} - \frac{\ln x}{3})$ $f^{1} (x) = \frac{x}{3} - \frac{x^{2} (x - 3)}{3 (x^{2} - 3 x + 3)} + \frac{2 f (x)}{x}$ ${x f}^{1} (x) = \frac{x^{2}}{3} - \frac{x^{2} (x^{2} - 3 x)}{3 (x^{2} - 3 x + 3)} + 2 f (x)$ $f^{1} (x) + {x f}^{2} (x) = \frac{2 x (x^{2} - 3 x + 3) - x^{2} (2 x - 3)}{{(x^{2} - 3 x + 3)}^{2}} + 2 f^{1} (x)$ ${x f}^{2} (x) = - \frac{3 x^{2}}{{(x^{2} - 3 x + 3)}^{2}} + f^{1} (x)$ $x {(x^{2} - 3 x + 3)}^{2} f^{2} (x) + 3 x^{2}$ $= {(x^{2} - 3 x + 3)}^{2} f^{1} (x)$ ${{(x^{2} - 3 x + 3)}^{2} + 2 x (2 x - 3) (x^{2} - 3 x + 3)} f^{2} (x)$ $+ x {(x^{2} - 3 x + 3)}^{2} f^{3} (x) + 6 x$ $= 2 (x^{2} - 3 x + 3) (2 x - 3) f^{1} (x)$ $+ {(x^{2} - 3 x + 3)}^{2} f^{2} (x)$ $\Rightarrow 2 x (2 x - 3) (x^{2} - 3 x + 3) f^{2} (x)$ $+ x {(x^{2} - 3 x + 3)}^{2} f^{3} (x) + 6$ $= 2 (x^{2} - 3 x + 3) (2 x - 3) f^{1} (x)$ $\Rightarrow$ $2 (2 x - 3) (3 x^{2}) = 6 (x^{2} - 3 x + 3)$ $+ x {(x^{2} - 3 x + 3)}^{2} f^{3} (x)$ $f^{3} (x) = \frac{6 x^{2} (2 x - 3) - 6 (x^{2} - 3 x + 3)}{x {(x^{2} - 3 x + 3)}^{2}}$ $= \frac{6 (2 x^{3} - 4 x^{2} + 3 x - 3)}{x {(x^{2} - 3 x + 3)}^{2}}$ $l e t x = t + h$ $f^{3} (t) = \frac{6 (2 t^{3} + 6 {h t}^{2} + 6 h^{2} t + h^{3} - 4 t^{2} - 8 h t - 4 h^{2} + 3 t + 3 h - 3)}{(t + h) {(t^{2} + 2 h t + h^{2} - 3 t - 3 h + 3)}^{2}}$ $l e t - 2 h^{3} + 6 h^{3} - 6 h^{3} + h^{3} - 4 h^{2} + 8 h^{2} - 4 h^{2} - 3 h + 3 h - 3 = 0$ $\Rightarrow - h^{3} - 3 = 0$ $h = - 3^{1 / 3}$ $. . .$
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