GATE CSE 2024 | Set 2 | Question: 6

Question

​​​​​Let $f(x)$ be a continuous function from $\mathbb{R}$ to $\mathbb{R}$ such that

\[
f(x)=1-f(2-x)
\]
Which one of the following options is the CORRECT value of $\int_{0}^{2} f(x) d x$ ?

• A. $0$
• B. $1$
• C. $2$
• D. $-1$

Answer 1

$ \underline{\text{Given:}}\ \ \ f(x) = 1 - f(2-x) \qquad \to (1) $

$ \underline{\text{To find:}}\ \ \int_{0}^{2} f(x) \,dx  $

$ \underline{\large \text{Method 1:}} $

Notice that the continuous function $f(x) = 1/2$ from $\mathbb{R}$ to $\mathbb{R}$ satisfies eqn. $(1)$

So, let $f(x) = 1/2$.

Now,

$
\begin{align*}
\large \int_{0}^{2} f(x) \,dx  &= \large \int_{0}^{2} \tfrac{1}{2} \cdot \,dx = \large \tfrac{x}{2}\Big|_0^2 = 1 \\
\end{align*}
$

$\bf \therefore Ans = B.$

---

$ \underline{\large \text{A more formal method:}} $

Let $u=a-x$, we have $du=-dx$, then

$
\begin{align*}
\large \int_{0}^{a}f(a-x)dx &= \large \int_{u=a}^{u=0}-f(u)du = \large -\int_a^0f(u)du = \large \int_0^af(u)du
\end{align*}
$

So, $ \large  \int_{0}^{a} f(a-x) \,dx = \int_{0}^{a} f(u) \,dx = \int_{0}^{a} f(x) \,dx. \qquad \small \text{(} \because \text{variable of integration is a dummy variable)}$

Now,

$
\begin{align*}
\large \int_{0}^{2} f(x) \,dx &= \large \int_{0}^{2} f(2-x) \,dx = \large \int_{0}^{2} 1-f(x) \,dx = \large 2 - \int_{0}^{2} f(x) \,dx \\
\large 2 \cdot \int_{0}^{2} f(x) \,dx &= \large 2 \\
\large \int_{0}^{2} f(x) \,dx &= \large 1 \\
\blacksquare \\
\end{align*}
$ 

Answer 2

Given ,   f(x) = 1 -  f(2-x)

      or,     f(2-x) = 1 - f(x)   ....(1)

By property of definite integrals: $\int_{a}^{b}$f(x) dx = $\int_{a}^{b}$ f(a+b-x) dx

Let, I = $\int_{0}^{2}$f(x) dx =$\int_{0}^{2}$ f(0+2-x) dx  ... {using property}

       I = $\int_{0}^{2}$ f(2-x) dx   

       I  =  $\int_{0}^{2}$ (1 - f(x)) dx   ...{using eq 1}

       I  =  $\int_{0}^{2}$ 1dx  - $\int_{0}^{2}$ f(x) dx

       I  =   2 - I

     2 I = 2

 $\therefore$ ,  I = 1

 i.e.  $\int_{0}^{2}$ f(x) dx = 1 ; option (b) 1