1-calculate-f-x-x-1-x-2-1-e-xt-arctan-t-dt-2-find-lim-x-0-f-x-

Question Number 74345 by mathmax by abdo last updated on 22/Nov/19

1) c a l c u l a t e f (x) = \int_{x + 1}^{x^{2} + 1} e^{- x t} a r c t a n (t) d t

2) f i n d {l i m}_{x \to 0} f (x)

Commented by mathmax by abdo last updated on 24/Nov/19

1) f (x) = \int_{x + 1}^{x^{2} + 1} e^{- x t} a r c t a n (t) d t \Rightarrow f (x) =_{x t = u} \frac{1}{x} \int_{x^{2} + x}^{x^{3} + x} e^{- u} a r c t a n (\frac{u}{x}) d u

\Rightarrow x f (x) = \int_{x^{2} + x}^{x^{3} + x} e^{- u} a r c t a n (\frac{u}{x}) d x =_{b y p a r t s} {[- e^{- u} a r c t a n (\frac{u}{x})]}_{x^{2} + x}^{x^{3} + x}

+ \int_{x^{2} + x}^{x^{3} + x} e^{- u} \times \frac{1}{x (1 + \frac{u^{2}}{x^{2}})} d u

= e^{- (x^{2} + x)} a r c t a n (x + 1) - e^{- (x^{3} + x)} a r c t a n (x^{2} + 1)

+ x \int_{x^{2} + x}^{x^{3} + x} \frac{e^{- u}}{x^{2} + u^{2}} d u \Rightarrow

f (x) = \frac{1}{x} {e^{- (x^{2} + x)} a r c t a n (x + 1) - e^{- (x^{3} + x)} a r c t a n (x^{2} + 1)}

+ \int_{x^{2} + x}^{x^{3} + x} \frac{e^{- u}}{x^{2} + u^{2}} d u \dots b e c o n t i n u e d \dots

Commented by mathmax by abdo last updated on 24/Nov/19

2) \exists c_{x} \in] x + 1, x^{2} + 1 [/ f (x) = {a r c t a n c}_{x} \int_{x + 1}^{x^{2} + 1} e^{- x t} d t

=_{x t = u} a r c t a n (c_{x}) \int_{x^{2} + x}^{x^{3} + x} e^{- u} \frac{d u}{x} = \frac{1}{x} a r c t a n (c_{x}) {[- e^{- u}]}_{x^{2} + x}^{x^{3} + x}

= a r c t a n (c_{x}) \times \frac{e^{- (x^{2} + x)} - e^{- (x^{3} + x)}}{x}

w e h a v e {l i m}_{x \to 0} c_{x} = \frac{π}{4} a l s o w e h a v e e^{- (x^{2} + x)} \sim 1 - (x^{2} + x)

a n d e^{- (x^{3} + x)} \sim 1 - (x^{3} + x) \Rightarrow

e^{- (x^{2} + x)} - e^{- (x^{3} + x)} = 1 - x^{2} - x - 1 + x^{3} + x = x^{3} - x^{2} \Rightarrow

\frac{e^{- (x^{2} + x)} - e^{- (x^{3} + x)}}{x} \sim x^{2} - x \to 0 (x \to 0) \Rightarrow {l i m}_{x \to 0} f (x) = 0