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0 ∫ 1 (1+e -x^2 )dx=?

01(1+e-x^2)dx=?
 

Grade:12

1 Answers

jagdish singh singh
173 Points
5 years ago
\hspace{-0.8 cm}$ Given $\bf{\int_{0}^{1}(1+e^{-x^2})dx = 1+\int_{0}^{1}e^{-x^2}dx\;,}$ Here we can not find\\\\ antiderivative of that Integrant in terms of elementry function.\\\\ So Here we can only find Range of $\bf{I.\;\; }$\\\\So $\bf{e^{-x}<e^{-x^2}<1\Rightarrow \int_{0}^{1}e^{-x}dx <\int_{0}^{1}e^{-x^2}dx<\int_{0}^{1}1dx}$\\\\ So we get $\bf{\left(1-\frac{1}{e}\right)<\int_{0}^{1}e^{-x^2}dx<1\Rightarrow \left(2-\frac{1}{e}\right)<1+\int_{0}^{1}e^{-x^2}dx<2}$

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