lf f(x) is continuous for all real values of x, then ∑n=110 ∫01 f(r−1+x)dx

 lf f(x) is continuous for all real values of x, then n=11001f(r1+x)dx

  1. A

    010f(x)dx

  2. B

    01f(x)dx

  3. C

    1001f(x)dx

  4. D

    901f(x)dx

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    Solution:

    I=r=11001f(r1+x)dxIt=01f(t1+x)dx

     Put t1+x=ydx=dy It=t1tf(y)dyIt=t1tf(x)dxI1=01f(x)dx,I2=12f(x)dx I3=23f(x)dxr...I10=910f(x)dx I = I1+I2+....+I10=01f(x)dx+12f(x)dx+........+910f(x)dx=010f(x)dx

     

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