The value of integral ∑k=1n ∫01 f(k−1+x)dx is
∫01 (x)dx
∫02 f(x)dx
∫0n f(x)dx
n∫01 f(x)dx
Let I=∫01 f(k−1+x)dx⇒ I=∫k−1k f(t)dt, where t=k−1+x⇒ I=∫k−1k f(x)dx∴ ∑k=1n ∫k−1k f(x)dx= ∫01 f(x)dx+∫12 f(x)dx+…+∫n−1n f(x)dx=∫0n f(x)dx