7329769 - 1. Curves r=2sin and r=2sin2, 0 2 Area of the...

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Curves = r 2sinθ and = r 2sin2θ , 0 θ π2 Area of the region outside the first curve and inside the second curve: The point of intersection of both curves: = => = 2sinθ 2sin2θ θ π3 For given case, 0 θ π3 Required area is: = A 0π32sinθ2sin2θrdrdθ = - 420π3sin22θ sin2θdθ = - - - 20π31 cos4θ2 1 cos2θ2dθ = - 0π3cos2θ cos4θdθ = - sin2θ2 sin4θ40π3 = A 338 2. Convergence of series: = ∞ n 1 15n = + + +…+ Sn 15 110 115 15n And, + - = ( + )+ ( + )+…+ ( + )> ( + )+ ( + )+…+ ( + ) Sn p Sn 15 n 1 15 n 2 15 n p 15 n p 15 n p 15 n p Now, + = = p5n p 110 when p n i.e. - > S2n Sn 110 Thus for < ε 110 , we cannot have - < S2n Sn ε for all n N . Hence series is divergent . 3. Series: = ∞ + + n 1 n 12nn 2 = + + an n 12nn 2 Now, →∞ + + = limn n 12nn 2 1 which is not equal to 0. Hence, series is divergent. Sum = DNE 4. Series: = ∞ + n 1 4n 15n By ratio test, →∞ + < limn an 1an 1 for convergence. Now,
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7329769 - 1. Curves r=2sin and r=2sin2, 0 2 Area of the...

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