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

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

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

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

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$n \int_{0}^{1} f (x) dx$

$\int_{0}^{1} (x) dx$

$\int_{0}^{n} f (x) dx$

$\int_{0}^{2} f (x) dx$

answer is C.

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Detailed Solution

$\begin{array}{ll} Let I = \int_{0}^{1} f (k - 1 + x) dx \\ \Rightarrow I = \int_{k - 1}^{k} f (t) dt, where t = k - 1 + x \\ \Rightarrow I = \int_{k - 1}^{k} f (x) dx \\ ∴ \sum_{k = 1}^{n} \int_{k - 1}^{k} f (x) dx \\ = \int_{0}^{1} f (x) dx + \int_{1}^{2} f (x) dx + \dots + \int_{n - 1}^{n} f (x) dx = \int_{0}^{n} f (x) dx \end{array}$