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Question Number 111975 by 175mohamed last updated on 05/Sep/20

	Answered by mathmax by abdo last updated on 05/Sep/20
	$f(0)=d=k f(−2)=0 ⇒−8a+4b−2c+d=0 (1) f(−4) =−16 ⇒−64a+16b−4c+d =−16 (2) f^′ (x)=3ax^2 +2bx +c ⇒f^((2)) (x)=6ax +2b f^((2)) (−2)=0 ⇒−12a+2b =0 ⇒−6a+b=0 ⇒b=6a (1)⇒−8a+24a−2c +d =0 ⇒16a−2c +d =0 (2)⇒−64a +96a−4c +d =−16 ⇒32a−4c +d =−16 ⇒ { ((16a−2c +d =0)),((16a−2c +(d/2)=−8 ⇒(d/2) =8 ⇒d =16 =k)) :} ⇒∫_(−16) ^k f^(−1) (x)dx =∫_(−16) ^(16) f^(−1) (x)dx changementf^(−1) (x)=t give x =f(t) ⇒ ∫_(−16) ^(16) f^(−1) (x)dx =∫_(f^(−1) (−16)) ^(f^(−1) (16)) t f^′ (t)dt =∫_(−4) ^0 t f^′ (t)dt =∫_(−4) ^0 t{3at^2 +2bt+c} dt =∫_(−4) ^0 (3at^3 +2bt^2 +ct)dt =∫_(−4) ^0 (3at^3 +12at^2 +ct)dt ( b=6a) we have 2c =16a +d =16a +16 ⇒c =8a+8 ⇒ ∫_(−16) ^(16) f^(−1) (x)dx =∫_(−4) ^0 (3at^3 +12at^2 +(8a+8)t)dt =[((3a)/4)t^4 +4at^3 +(4a+4)t^2 ]_(−4) ^0 =−(((3a)/4)(−4)^4 +4a(−4)^3 +(4a+4)(−4)^2 ) =−(3a(64) −256a +64a +64) =−64$
	$f (0) = d = k$ $f (- 2) = 0 \Rightarrow - 8 a + 4 b - 2 c + d = 0 (1)$ $f (- 4) = - 16 \Rightarrow - 64 a + 16 b - 4 c + d = - 16 (2)$ $f^{^{'}} (x) = {3 ax}^{2} + 2 bx + c \Rightarrow f^{(2)} (x) = 6 ax + 2 b$ $f^{(2)} (- 2) = 0 \Rightarrow - 12 a + 2 b = 0 \Rightarrow - 6 a + b = 0 \Rightarrow b = 6 a$ $(1) \Rightarrow - 8 a + 24 a - 2 c + d = 0 \Rightarrow 16 a - 2 c + d = 0$ $(2) \Rightarrow - 64 a + 96 a - 4 c + d = - 16 \Rightarrow 32 a - 4 c + d = - 16$ $\Rightarrow {\begin{cases} 16 a - 2 c + d = 0 \\ 16 a - 2 c + \frac{d}{2} = - 8 \Rightarrow \frac{d}{2} = 8 \Rightarrow d = 16 = k \end{cases}$ $\Rightarrow \int_{- 16}^{k} f^{- 1} (x) dx = \int_{- 16}^{16} f^{- 1} (x) dx$ ${changementf}^{- 1} (x) = t give x = f (t) \Rightarrow$ $\int_{- 16}^{16} f^{- 1} (x) dx = \int_{f^{- 1} (- 16)}^{f^{- 1} (16)} t f^{^{'}} (t) dt = \int_{- 4}^{0} t f^{^{'}} (t) dt$ $= \int_{- 4}^{0} t {{3 at}^{2} + 2 bt + c} dt = \int_{- 4}^{0} ({3 at}^{3} + {2 bt}^{2} + ct) dt$ $= \int_{- 4}^{0} ({3 at}^{3} + {12 at}^{2} + ct) dt (b = 6 a)$ $we have 2 c = 16 a + d = 16 a + 16 \Rightarrow c = 8 a + 8 \Rightarrow$ $\int_{- 16}^{16} f^{- 1} (x) dx = \int_{- 4}^{0} ({3 at}^{3} + {12 at}^{2} + (8 a + 8) t) dt$ $= {[\frac{3 a}{4} t^{4} + {4 at}^{3} + (4 a + 4) t^{2}]}_{- 4}^{0} = - (\frac{3 a}{4} {(- 4)}^{4} + 4 a {(- 4)}^{3} + (4 a + 4) {(- 4)}^{2})$ $= - (3 a (64) - 256 a + 64 a + 64)$ $= - 64$
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