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Solve Problems · Q17

Q.Evaluate the following integral:

(i) ∫0π/2sin⁡x dx\displaystyle\int_0^{\pi/2} \sin x\, dx
(ii) ∫01x dx\displaystyle\int_0^1 x\, dx
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Step 1 (i). ∫0π/2sin⁡x dx=[−cos⁡x]0π/2=−cos⁡π2−(−cos⁡0)=−0−(−1)=1\displaystyle\int_0^{\pi/2}\sin x\,dx = \Big[-\cos x\Big]_0^{\pi/2} = -\cos\dfrac{\pi}{2}-(-\cos 0) = -0-(-1) = 1. …

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