if-the-F-x-1-x-1-x-2t-F-t-dt-what-the-F-1-value-using-the-Leibnitz-formula-

Question Number 138407 by tugu last updated on 13/Apr/21

i f t h e F (x) = \frac{1}{x} \underset{1}{\int^{x}} (2 t - F^{'} (t)) d t \Rightarrow w h a t t h e F^{'} (1) v a l u e u s i n g t h e L e i b n i t z f o r m u l a .

Answered by ajfour last updated on 13/Apr/21

F^{'} (x) = - \frac{F (x)}{x} + \frac{1}{x} [2 x - F^{'} (x)]

(x + 1) F^{'} (x) + F (x) = 2 x

x F^{'} (x) + F (x) = 2 x - F^{'} (x)

d [x F (x)] = [2 x - F^{'} (x)] d x

x F (x) = x^{2} - F (x) + c

F (x) = \frac{x^{2} + c}{x + 1}

x F (x) = (x^{2} - 1) - F (x) + F (1)

F (1) = 1 - x^{2} + x^{2} + c

F (1) = c + 1 = \frac{c + 1}{2}

\Rightarrow c = - 1

F (1) = 0

F (x) = \frac{x^{2} - 1}{x + 1} = x - 1

F^{'} (x) = 1 \Rightarrow F^{'} (1) = 1 ★

c h e c k

x (x - 1) = \int_{1}^{x} (2 t - 1) d t

x (x - 1) = x^{2} - 1 - (x - 1)

\Rightarrow x (x - 1) = x (x - 1) ✓

Commented by ajfour last updated on 13/Apr/21

a h! n o b o d y l i k e s t h i s . .

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