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AP EAMCET201823 Apr 2018Evening ShiftMathematicsDefinite IntegrationActual

_ n 1 n ^5 ( 6 n )+ ^5 ( 2 6 n )+ ^5 ( 3 6 n )+ . .+ ^5 ( 2 ) =

Options

  1. A8 15
  2. B8 5
  3. C32 5
  4. D16 5

Correct answer

D. 16 5

Step-by-step solution

aligned _ n 1 n ^5 ( 6 n )+ ^5 ( 2 6 n ) . & + ^5 ( 3 6 n ) & .+ + ^5 ( 2 ) aligned aligned = _ n 1 n ^5 ( 6 n )+ ^5 ( 2 6 n ) . & + ^5 ( 3 6 n ) & .+ + ^5 ( 3 n 6 n ) = & _ n 1 n _ r=1 ^ 3 n ^5 ( r 6 n )= ₀^3 ^5 ( 6 x ) d x aligned Let 6 x=t For upper limit at x=3, t= / 2 and lower limit at x=0, t=0 and d x= 6 d t So, aligned ₀^3 ^5 ( 6 x ) d t & = 6 ₀^ / 2 ^5(t) d t & = 6 4 2 5 3 1 = 16 5 aligned

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