f(x) = (∫_0 ^1 f(x)dx)x^2 + (∫_0 ^2 f(x)dx)x + (∫


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Question Number 119214 by abdul88 last updated on 23/Oct/20

$f (x) = (\underset{0}{\int^{1}} f (x) d x) x^{2} + (\underset{0}{\int^{2}} f (x) d x) x + (\underset{0}{\int^{3}} f (x) d x) + 1$ $t h e v a l e u f (4) = ?$
	Answered by bemath last updated on 23/Oct/20
	$let ∫_0 ^1 f(x) dx = p , ∫_0 ^2 f(x) dx = q , ∫_0 ^3 f(x) dx= r ⇔ f(x) = px^2 +qx+r+1 ⇒ p =(1/3)p+(1/2)q+r+1 ⇒4p=3q+6r+6 ⇒q = (8/3)p+2q+2r+2 ⇒−q=(8/3)p+2r+2 ⇒r = 9p+(9/2)q+3r+3 ⇒−2r=9p+(9/2)q+3 we get { ((p=(1/3))),((q=−((80)/(63)))),((r=−(1/7))) :} ∵ f(x) = (x^2 /3)−((80x)/(63)) + (6/7) ⇒f(4) = ((16)/3)−((320)/(63))+(6/7)=((10)/9)$
	$l e t \underset{0}{\int^{1}} f (x) d x = p, \underset{0}{\int^{2}} f (x) d x = q, \underset{0}{\int^{3}} f (x) d x = r$ $\Leftrightarrow f (x) = {p x}^{2} + q x + r + 1$ $\Rightarrow p = \frac{1}{3} p + \frac{1}{2} q + r + 1 \Rightarrow 4 p = 3 q + 6 r + 6$ $\Rightarrow q = \frac{8}{3} p + 2 q + 2 r + 2 \Rightarrow - q = \frac{8}{3} p + 2 r + 2$ $\Rightarrow r = 9 p + \frac{9}{2} q + 3 r + 3 \Rightarrow - 2 r = 9 p + \frac{9}{2} q + 3$ $w e g e t {\begin{cases} p = \frac{1}{3} \\ q = - \frac{80}{63} \\ r = - \frac{1}{7} \end{cases}$ $∵ f (x) = \frac{x^{2}}{3} - \frac{80 x}{63} + \frac{6}{7}$ $\Rightarrow f (4) = \frac{16}{3} - \frac{320}{63} + \frac{6}{7} = \frac{10}{9}$
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