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प्रश्न
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
उत्तर
\[\int_1^2 \frac{1}{x^2} e^\frac{- 1}{x} d x\]
\[Let \frac{- 1}{x} = t, then \frac{1}{x^2} dx = dt\]
\[\text{When, }x \to 1 ; t \to - 1\]
\[\text{And }x \to 2 ; t \to \frac{- 1}{2}\]
Therefore the integral becomes
\[ \int_{- 1}^\frac{- 1}{2} e^t d t\]
\[ = \left[ e^t \right]_{- 1}^\frac{- 1}{2} \]
\[ = e^\frac{- 1}{2} - e^{- 1} \]
\[ = \frac{\sqrt{e} - 1}{e}\]
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