The value of integral ∑$$k=1n$$ ∫$$01$$ $$f(k$$−$$1+x)dx$$ is

Correct option is 3∫0n f(x)dx

Questions

The value of integral $\sum_{k = 1}^{n} \int_{0}^{1} f (k - 1 + x) dx$ is

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∫01 (x)dx

∫02 f(x)dx

∫0n f(x)dx

n∫01 f(x)dx

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detailed solution

Correct option is C

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

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The value of integral ∑k=1n ∫01 f(k−1+x)dx is

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The value of integral $\sum_{k = 1}^{n} \int_{0}^{1} f (k - 1 + x) dx$ is