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6) Without finding the value of the integral prove that dz D C is the arc of -l= 2 in the first quadrant. (327 + In R FC is I

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(ن) we have f(2) 2²1 and t is the are of 121=2 in the first quadreest Thus Length of the curve is = 25X2 2 A - Length (L) = xhere f(a) = enz 72 and e is 121= R where R71 so ¿z Reit Where d E 20,21] Thin length of the curve e is 25R . L=QAR Now, Me ma

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