Integral question! Sqrt(6x - x^2)?

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(integral from 3 to 4.5 of) sqrt(6x - x^2) dx I have no idea how to start this. I've tried U-sub, trig sub, etc.. Nothing is working.

 

Answer No.1

Anonymous

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A u-substitution will NOT work. For this integral, we must complete the square under the radical sign as follows: ∫ √(6x - x²) dx = ∫ √(-9 + 9 + 6x - x²) dx = ∫ √[9 - (x - 3)²] dx. From here, let: x - 3 = 3sin(u) ==> x = 3sin(u) + 3 ==> u = arcsin[(x - 3)/3] dx = 3cos(u) du. Then, our limits of integration become: Lower limit = arcsin[(3 - 3)/3] = arcsin(0) = 0 Upper limit = arcsin[(4.5 - 3)/3 = arcsin(1/2) = π/6. Applying these substitutions yields: ∫ √[9 - (x - 3)²] dx (from 3 to 4.5) = ∫ √[9 - 9sin²(u)] [3cos(u) du] (from 0 to π/6) = 9 ∫ √[1 - sin²(u)] cos(u) du (from 0 to π/6) = 9 ∫ cos(u) * cos(u) du (from 0 to π/6) 0 on (0, π/6) = 9 ∫ cos²(u) du (from 0 to π/6) = 9/2 ∫ [1 + cos(2u)] du (from 0 to π/6) = [(9/2)u + (9/4)sin(2u)] (evaluated from 0 to π/6) = [(9/2)(π/6) + (9/4)(√3/2)] - [(9/2)(0) + (9/2)(0)] = (9/8)√3 + (3/4)π ≈ 4.30475. get my doctor prescribe viagra canada pharmacies online prescriptions order viagra online india canadian pharmacies that ship to the us best viagra pills women viagra without doctor does walgreens sell generic viagra viagra without doctor how much does viagra cost per pill 2012

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