Question

$\int _ { 3 } ^ { 6 } \frac { 1 } { x + 1 } d x$ is equal to

• A. $[ \log ( x + 1 ) ] _ { 3 } ^ { 6 }$
• B. $\left[ \log ( t + 1 ] _ { 3 } ^ { 6 } \right.$
• C. Both (1) and (2)
• D. None of these

Answer 1

Answer: (C)

Both (1) and (2)

Answer 2

$I = \int _ { 3 } ^ { 6 } \frac { 1 } { x + 1 } d x = [ \log ( x + 1 ) ] _ { 3 } ^ { 6 }$, $I = \int _ { 3 } ^ { 6 } \frac { 1 } { t + 1 } d t = [ \log ( t + 1 ) ] _ { 3 } ^ { 6 }$

(2) $\int _ { a } ^ { b } f ( x ) d x = - \int _ { b } ^ { a } f ( x ) d x$ i.e., by the interchange in the limits of definite integral, the sign of the integral is changed.