Question

$\int _ { 0 } ^ { 1 } f ( 1 - x ) d x$ has the same value as the integral

• A. $\int _ { 0 } ^ { 1 } f ( x ) d x$
• B. $\int _ { 0 } ^ { 1 } f ( - x ) d x$
• C. $\int _ { 0 } ^ { 1 } f ( x - 1 ) d x$
• D. $\int _ { - 1 } ^ { 1 } f ( x ) d x$

Answer 1

Answer: (A)

$\int _ { 0 } ^ { 1 } f ( x ) d x$

Answer 2

Put $1 - x = t \Rightarrow - d x = d t$ . Also as $x = 0$ to 1, $t = 1$to 0

Therefore, $\int _ { 0 } ^ { 1 } f ( 1 - x ) d x = \int _ { 1 } ^ { 0 } f ( t ) ( - d t ) = \int _ { 0 } ^ { 1 } f ( t ) d t = \int _ { 0 } ^ { 1 } f ( x ) d x$.