Q. If $f(x)$ is continuous for all real values of $x,$ then $\sum_{r = 1}^{10}{\int_{0}^{1}{(r - 1 + x)dx}}$ is equal to
A) $\int_{0}^{10}{f(x)dx}$
B) $\int_{0}^{1}{f(x)dx}$
C) $10\int_{0}^{1}{f(x)dx}$
D) $9\int_{0}^{1}{f(x)dx}$
Explanation
(a)
$I = \sum_{r = 1}^{10}{\int_{0}^{1}{f(r - 1 + x)dx}}$
$I_{t} = \int_{0}^{1}{f(t - 1 + x)dx}$
Put $t - 1 + x = y\ \Rightarrow dx = dy$
$I_{t} = \int_{t - 1}^{t}{f(y)dy\ \Rightarrow \ I_{t} = \int_{t - 1}^{t}{f(x)dx}}$
$\therefore I_{1} = \int_{0}^{1}{f(x)dx},\ I_{2} = \int_{1}^{2}{f(x)dx},$
$I_{3} = \int_{2}^{3}{f(x)dx},\ldots,\ I_{10} = \int_{9}^{10}{f(x)dx}$
So, $I = I_{1} + I_{2} + ... + I_{10} = \int_{9}^{10}{f(x)dx}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.