Solveeit Logo
Login

Question

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

The value of the integral ∑k=1n∫01f(k−1+x)dx\sum_{k = 1}^{n}{\int_{0}^{1}{f(k - 1 + x)dx}} is

A

∫01f(x)dx\int_{0}^{1}{f(x)dx}

B

∫02f(x)dx\int_{0}^{2}{f(x)dx}

C

∫0nf(x)dx\int_{0}^{n}{f(x)dx}

D

n∫01f(x)dxn\int_{0}^{1}{f(x)dx}

Answer

∫0nf(x)dx\int_{0}^{n}{f(x)dx}

Explanation

Solution

Let I=∫01f(k−1+x)dxI = \int_{0}^{1}{f(k - 1 + x)dx}

⇒ I=∫k−1kf(t)dt,I = \int_{k - 1}^{k}{f(t)dt,} where t=k−1+xt = k - 1 + x ⇒ I=∫k−1kf(x)dxI = \int_{k - 1}^{k}{f(x)dx}

∴∑k=1n∫k−1kf(x)dx=∫01f(x)dx+∫12f(x)dx+.....+∫n−1nf(x)dx\therefore\sum_{k = 1}^{n}{\int_{k - 1}^{k}{f(x)dx = \int_{0}^{1}{f(x)dx + \int_{1}^{2}{f(x)dx + ..... + \int_{n - 1}^{n}{f(x)dx}}}}}

=∫0nf(x)dx= \int_{0}^{n}{f(x)dx}.