Let $$p\left( x \right)$$ be a function defined on R such that $$p'\left( x \right) = p'\left( {1 - x} \right),$$ for all $$x \in \left[ {0,\,1} \right],p\left( 0 \right) = 1$$ and $$p\left( 1 \right) = 41.$$ Then
$$\int\limits_0^1 {p\left( x \right)dx} $$ equals-
A.
$$21$$
B.
$$41$$
C.
$$42$$
D.
$$\sqrt {41} $$
Answer :
$$21$$
Solution :
$$\eqalign{
& p'\left( x \right) = p'\left( {1 - x} \right)\,\, \Rightarrow p\left( x \right) = - p\left( {1 - x} \right) + c \cr
& {\text{At }}x = 0 \cr
& p\left( 0 \right) = - p\left( 1 \right) + c\,\, \Rightarrow 42 = c \cr
& {\text{Now }}p\left( x \right) = - p\left( {1 - x} \right) + 42 \cr
& \Rightarrow p\left( x \right) + p\left( {1 - x} \right) = 42 \cr
& \Rightarrow I = \int\limits_0^1 {p\left( x \right)dx} .....{\text{(i)}} \cr
& \Rightarrow I = \int\limits_0^1 {p\left( {1 - x} \right)dx} .....{\text{(ii)}} \cr} $$
on adding (i) and (ii), we get
$$2I = \int\limits_0^1 {\left( {42} \right)dx} \,\, \Rightarrow I = 21$$
Releted MCQ Question on Calculus >> Definite Integration
Releted Question 1
The value of the definite integral $$\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} \,dx$$ is-