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प्रश्न
If f' (x) = x − \[\frac{1}{x^2}\] and f (1) \[\frac{1}{2}, find f(x)\]
बेरीज
उत्तर
\[f'\left( x \right) = x - \frac{1}{x^2}\]
\[ f'\left( x \right) = x - x^{- 2} \]
\[\int f'\left( x \right)dx = \int\left( x - x^{- 2} \right)dx\]
\[ f\left( x \right) = \frac{x^2}{2} - \frac{x^{- 2 + 1}}{- 2 + 1} + C\]
\[ = \frac{x^2}{2} + \frac{1}{x} + C\]
\[f\left( 1 \right) = \frac{1}{2} \left( Given \right)\]
\[ \Rightarrow \frac{1^2}{2} + \frac{1}{1} + C = \frac{1}{2}\]
\[ \Rightarrow C = - 1\]
\[ \therefore f\left( x \right) = \frac{x^2}{2} + \frac{1}{x} - 1\]
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