Mathematics

Evaluate $$I = \displaystyle \int_{\pi /6}^{\pi /3}\sin x\:dx$$


ANSWER

$$\displaystyle \frac{1-\sqrt{3}}{2}$$


SOLUTION
Given : $$I = \displaystyle \int_{\pi /6}^{\pi /3}\sin x\:dx$$

Integeration of $$\sin x dx$$ is $$-cos x dx + c$$

$$I = -cos x dx$$

Substuting the upper and lower limit values we get,

$$I = -cos\dfrac{\pi}{3}+cos\dfrac{\pi}{6}$$

$$I = \dfrac{-\sqrt{3}}{2} + \dfrac{1}{2}$$

$$I = \dfrac{1-\sqrt{3}}{2}$$
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Single Correct Medium Published on 17th 09, 2020
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