Mathematics

Evaluate the given integral.
$$\displaystyle\int{{e}^{\log{\sqrt{x}}}dx}$$


SOLUTION
$$I=\displaystyle\int{{e}^{\log{\sqrt{x}}}dx}$$

$$=\displaystyle\int{\sqrt{x}\,dx}$$

$$=\displaystyle\int{{x}^{\frac{1}{2}}\,dx}$$

$$=\dfrac{{x}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}+c$$

$$=\dfrac{{x}^{\frac{3}{2}}}{\dfrac{3}{2}}+c$$

$$=\dfrac{2}{3}{x}^{\frac{3}{2}}+c$$
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Subjective Medium Published on 17th 09, 2020
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