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

$$\mathrm{Calculate}::\underset{\mathrm{x}\to 0}{\lim }\frac{{\int }_{2\mathrm{x}-1}^{2\mathrm{x}+1}{\mathrm{e}}^{{\mathrm{t}}^2}\mathrm{dt}-{\int }_{-1}^1{\mathrm{e}}^{{\mathrm{t}}^2}\mathrm{dt}}{{\mathrm{x}}^2}=8\mathrm{e},{\mathrm{Don}}^{\prime }\mathrm{t}\mathrm{use}{\mathrm{L}}^{\prime }{\mathrm{Hospital}}^{\prime }\mathrm{s}\mathrm{rule}.$$

Answered by aleks041103 last updated on 17/Apr/22

$$letf\left(x\right)={\int }_{2x-1}^{2x+1}{e}^{t^2}dt$$$$\Rightarrow f\left(x\right)=f\left(0\right)+f^{\prime }\left(0\right)x+\frac{1}{2}{f}^{″}\left(0\right)x^2+...$$

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