Question

If $\underset{0}{\overset{100}{\int }}f(x)dx=a$, then $\sum _{r=1}^{100}\left(\underset{0}{\overset{1}{\int }}f(r-1+x)dx\right)=$

• A. $100a$
• B. $a$
• C. 0
• D. $10a$

Answer 1

Answer: (B)

$\sum _{r=1}^{100}\left(\underset{0}{\overset{1}{\int }}f(r-1+x)dx\right)$
$=\underset{0}{\overset{1}{\int }}f(x)dx+\underset{0}{\overset{1}{\int }}f(1+x)dx+\underset{0}{\overset{1}{\int }}f(2+x)dx+\dots \dots +\underset{0}{\overset{1}{\int }}f(99+x)dx$
$=\underset{0}{\overset{1}{\int }}f(x)dx+\underset{1}{\overset{2}{\int }}f(x)dx+\dots ..+\underset{99}{\overset{100}{\int }}f(x)dx$
( using shifting property )
$=\underset{0}{\overset{100}{\int }}f(x)dx=a$