Calculate-lim-x-0-2x-1-2x-1-e-t-2-dt-1-1-e-t-2-dt-x-2-8e-Don-t-use-L-Hospital-s-rule-

Question Number 168696 by qaz last updated on 16/Apr/22

Calculate :: \lim_{x \to 0} \frac{\int_{2 x - 1}^{2 x + 1} e^{t^{2}} dt - \int_{- 1}^{1} e^{t^{2}} dt}{x^{2}} = 8 e, {Don}^{'} t use L^{'} {Hospital}^{'} s rule .

Answered by aleks041103 last updated on 17/Apr/22

l e t f (x) = \int_{2 x - 1}^{2 x + 1} e^{t^{2}} d t

\Rightarrow f (x) = f (0) + f^{'} (0) x + \frac{1}{2} f^{″} (0) x^{2} + \dots