$\int_{0}^{1} e^{2 I n x} d x =$ [MP PET 1990]

0

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\frac{1}{2}

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\frac{1}{3}

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$\frac{1}{4}$

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Solution

The correct option is C $\frac{1}{3}$
$\int_{0}^{1} e^{2 l o g x} d x = \int_{0}^{1} e^{l o g x^{2}} d x = \int_{0}^{1} x^{2} d x = {[\frac{x^{3}}{3}]}_{0}^{1} = \frac{1}{3}$

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