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2 ∫ 1 1 X 2 E − 1 / X D X

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प्रश्न

\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]

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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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अध्याय 19: Definite Integrals - Revision Exercise [पृष्ठ १२१]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Revision Exercise | Q 18 | पृष्ठ १२१

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