e-i-2-2-x-2-sinx-1-x-2-4-dx-lim-x-2-2-x-log-x-8-dx-x-2-

Question Number 58940 by Tony Lin last updated on 01/May/19

e^{i \int_{- 2}^{2} (x^{2} s i n x + \sqrt{1 - \frac{x^{2}}{4}}) d x} + \lim_{x \to 2} \frac{\int_{2}^{x} l o g (x + 8) d x}{x - 2} = ?

Commented by maxmathsup by imad last updated on 01/May/19

l e t A = \int_{- 2}^{2} (x^{2} s i n x + \sqrt{1 - \frac{x^{2}}{4}}) d x \Rightarrow A = \int_{- 2}^{2} x^{2} s i n x d x + \int_{- 2}^{2} \sqrt{1 - \frac{x^{2}}{4}} d x

x \to x^{2} s i n (x) i s o d d f u n c t i o n \Rightarrow \int_{- 2}^{2} x^{2} s i n (x) d x = 0

\int_{- 2}^{2} \sqrt{1 - \frac{x^{2}}{4}} d x = 2 \int_{0}^{2} \sqrt{1 - \frac{x^{2}}{4}} d x =_{\frac{x}{2} = s i n (t)} 2 \int_{0}^{\frac{π}{2}} \sqrt{1 - {s i n}^{2} t} (2 c o s t) d t

= 4 \int_{0}^{\frac{π}{2}} {c o s}^{2} t d t = 4 \int_{0}^{\frac{π}{2}} \frac{1 + c o s (2 t)}{2} d t = 2 \int_{0}^{\frac{π}{2}} (1 + c o s (2 t)) d t

= π + {[s i n (2 t)]}_{0}^{\frac{π}{2}} = π \Rightarrow e^{i \int_{- 2}^{2} (x^{2} s i n x + \sqrt{1 - \frac{x^{2}}{4}}) d x} = e^{i π} = - 1

\int_{2}^{x} l n (t + 8) d t =_{t + 8 = u} \int_{10}^{x + 8} l n (u) d u = {[u l n (u) - u]}_{10}^{x + 8}

= (x + 8) l n (x + 8) - (x + 8) - 10 l n (10) + 10 \Rightarrow

{l i m}_{x \to 2} \frac{\int_{2}^{x} l n (x + 8) d x}{x - 2} = {l i m}_{x \to 2} \frac{(x + 8) l n (x + 8) - (x + 8) - 10 l n (10) + 10}{x - 2}

= {l i m}_{x \to 2} \frac{u^{^{'}} (x)}{v^{^{'}} (x)} (u s i n g h o s p i t a l t h e o r e m)

u (x) = (x + 8) l n (x + 8) - (x + 8) - 10 l n (10) + 10 \Rightarrow

u^{^{'}} (x) = l n (x + 8) + 1 - 1 = l n (x + 8) \Rightarrow u^{^{'}} (2) = l n (10)

v (x) = x - 2 \Rightarrow v^{^{'}} (x) = 1 \Rightarrow {l i m}_{x \to 2} \frac{\int_{2}^{x} l n (x + 8)}{x - 2} d x = l n (10) \Rightarrow

e^{i \int_{- 2}^{2} (\dots .)} d x + {l i m}_{x \to 2} \frac{\int_{2}^{x} (\dots .) d x}{x - 2} = - 1 + l n (10)

h e r e i h a v e t a k e n l n n o t l o g b e c a u s e l o g (x) = \frac{l n (x)}{l n (10)} \dots .

Commented by maxmathsup by imad last updated on 01/May/19

i f w e t a k e l o g (d d c i m a l l o g .) t h e r e s u l t w i l l b e - 1 + 1 = 0

Commented by Tony Lin last updated on 02/May/19

t h a n k s, s i r X D

Answered by tanmay last updated on 02/May/19

l o g (x + 8) = \frac{d F}{d x}

\int_{2}^{x} \frac{d F}{d x} \times d x

= F (x) - F (2)

n o w \lim_{x \to 2} \frac{F (x) - F (2)}{x - 2}

h = x - 2

\lim_{h \to 0} \frac{F (h + 2) - F (2)}{h}

= F^{^{'}} (2)

= l o g (2 + 8) = l o g 10

Commented by Tony Lin last updated on 02/May/19

w e l l d o n e, s i r