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AP EAMCET20227 Jul 2022Evening ShiftMathematicsDefinite IntegrationActual

For n N , if I_n= n x x d x= 2 n-1 (n-1) x+I_ n-2 and ₀^ n x x d x= k 2 , then k=

Options

  1. A(-1)^n-1
  2. B1-(-1)^n
  3. C(-1)^n
  4. D(-1)^ n+1

Correct answer

B. 1-(-1)^n

Step-by-step solution

aligned & Consider I_n= ₀^ n x x d x, n N & I_n= [ 2 n-1 (n-1) x ]₀^n+I_ n-2 =I_ n-2 aligned aligned & I₁=I₃=I₅=I₇ . & and I₂+I₄=I₆=I₈ & Now, I₁= ₀^ x x d x= & and I₂= ₀^ 2 x d x=0 & I₁=I₃=I₅ . & = =2 . ( 2 )= (1-(-1)^n ) 2 , n odd & and I₂=I₄=I₆ . .=0 & = (1-(-1)^n ) 2 , n is even. & aligned

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