Question

If $p(x)$ be a function defined on $\mathbb{R}$ such that $p'(x) = p'(1 - x)$ , $\forall x \in [0,1]$ , $p(0) = 1$ and $p(1) = 41$ . Then, $\int_0^1 {p(x)dx = } $
A) $\sqrt {41} $
B) $21$
C) $41$
D) $42$

Answer 1

Hint:Suppose we write $\int {f(x)dx = F(x) + C} $ , such integrals are known as Indefinite Integrals where $C$ is called the integration constant. If it is given as $\int_a^b {f(x)dx} $ then, it would be termed as a Definite Integral, which can be easily evaluated by: a) The definite integral as the limit of a sum. b) $\int_a^b {f(x)dx = F(b) - F(a)} $ , where $F$ is the antiderivative of $f(x)$ .

Formula Used:
$\int_a^b {f(x)dx} = \int_a^b {f(b + a - x)dx} $

Complete step by step Solution:
Given, $p(x)$ a function such that,
$p'(x) = p'(1 - x)$
If we integrate this equation on both sides, with respect to $x$ we will be able to get the function $p(x)$
$\int {p'(x)dx = \int {p'(1 - x)} } dx$
     $p(x) = - p(1 - x) + C$
We have to find the value of $C$.
Here, we have the initial conditions
$p(0) = 1$ and $p(1) = 41$
If we substitute $x = 0$ , in equation (1.1) we get
$p(0) = - p(1) + C$
$\therefore $ $C = 42$
Now, we have to find the value of ${\rm I} = \int_0^1 {p(x)dx} $ . We are familiar with the formula $\int_a^b {f(x)dx} = \int_a^b {f(b + a - x)dx} $
Using this we can write
${\rm I} = \int_0^1 {p(1 - x)dx} $
We get,
$2{\rm I} = \int_0^1 {p(x)dx + \int_0^1 {p(1 - x)dx} } $
$2{\rm I} = \int_0^1 {Cdx} $
${\rm I} = \int_0^1 {21dx} $
${\rm I} = \left[ {21} \right]_0^1$
${\rm I} = 21$

Therefore, the correct option is B.

Note:Geometrically, the equation $\int {f(x)dx = F(x) + C} = y$ represents a family of curves. The process of integration and differentiation are inverse of one another. If we have two functions that differ by a constant then, they have the same derivative. The integral of the sum of two functions is equal to the sum of the integrals of the functions.