Given, f(x)=min[x−1],[x−2],…,[x−10]
⇒f(x)=[x]−10
Now finding ∫010f(x)⋅dx=−10−9−8−…..−1
=−210⋅11=−55
Finding∫010(f(x)2)dx=102+92+82+…+12
=610⋅11⋅21=385
And ∫010∣f(x)∣=10+9+8+….+1
=210⋅11=55
So, ∫010f(x)dx+∫010(f(x))2dx+∫010∣f(x)∣dx
=−55+385+55=385