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

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a
∫010 f(x)dx
b
∫01 f(x)dx
c
10∫01 f(x)dx
d
9∫01 f(x)dx

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detailed solution

Correct option is A

I=∑r=110 ∫01 f(r−1+x)dxIt=∫01 f(t−1+x)dx Put t−1+x=y⇒dx=dy It=∫t−1t f(y)dy⇒It=∫t−1t f(x)dxI1=∫01 f(x)dx,I2=∫12 f(x)dx I3=∫23 f(x)dxr...I10=∫910 f(x)dx I = I1+I2+....+I10=∫01 f(x)dx+∫12 f(x)dx+........+∫910 f(x)dx=∫010 f(x)dx


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