Question

Let f(x) be the real valued function not identically zero, such that $f\left( x+{{y}^{n}} \right)=f\left( x \right)+{{\left( f\left( y \right) \right)}^{n}}$ where $\forall x,y\in R$and $n\in N,\left( n\ne 1 \right)$and ${f}'\left( 0 \right)\ge 0$. We may get an explicit form of the function f(x). Then, $\int\limits_{0}^{1}{f\left( x \right)dx}$ is equal to:
(a) $\dfrac{1}{2n}$
(b) 2n
(c) $\dfrac{1}{2}$
(d) 2

Answer 1

First, before proceeding for this, we must know the formula for the derivative of the function f(x) where h is any constant is given by${f}'\left( x \right)=\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$. Then, by substituting the value of h as ${{y}^{n}}$in the derivative formula, we get${f}'\left( x \right)=\dfrac{f\left( x+{{y}^{n}} \right)-f\left( x \right)}{{{y}^{n}}}$. Then, by integrating both sides and solving the given integration, we get the desired value.

Complete step by step answer:
In this question, we are supposed to find the value of $\int\limits_{0}^{1}{f\left( x \right)dx}$ when f(x) be the real valued function not identically zero, such that $f\left( x+{{y}^{n}} \right)=f\left( x \right)+{{\left( f\left( y \right) \right)}^{n}}$ where $\forall x,y\in R$and $n\in N,\left( n\ne 1 \right)$and ${f}'\left( 0 \right)\ge 0$.
So, before proceeding for this, we must know the formula for the derivative of the function f(x) where h is any constant is given by:
${f}'\left( x \right)=\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$
Now, we are given in our question to get the value of $f\left( x+{{y}^{n}} \right)$.
So, by substituting the value of h as ${{y}^{n}}$in the derivative formula, we get:
${f}'\left( x \right)=\dfrac{f\left( x+{{y}^{n}} \right)-f\left( x \right)}{{{y}^{n}}}$
Then, we are given with the value as $f\left( x+{{y}^{n}} \right)=f\left( x \right)+{{\left( f\left( y \right) \right)}^{n}}$and by substituting it in above expression, we get:
$\begin{aligned} & {f}'\left( x \right)=\dfrac{f\left( x \right)+{{\left( f\left( y \right) \right)}^{n}}-f\left( x \right)}{{{y}^{n}}} \\\ & \Rightarrow {f}'\left( x \right)=\dfrac{{{\left( f\left( y \right) \right)}^{n}}}{{{y}^{n}}} \\\ \end{aligned}$
Now, by replacing the value of y as x, then we get the above expression as:
$\begin{aligned} & {f}'\left( x \right)=\dfrac{{{\left( f\left( x \right) \right)}^{n}}}{{{x}^{n}}} \\\ & \Rightarrow {f}'\left( x \right)={{\left( \dfrac{f\left( x \right)}{x} \right)}^{n}} \\\ \end{aligned}$
Now, by integrating both sides, we get:
\int{{{\left\\{ f\left( x \right) \right\\}}^{-n}}{f}'\left( x \right)dx=\int{{{x}^{-n}}dx}}
Then, by using substitution in left hand side as $t=f\left( x \right)$and then differentiating both sides, we get $dt={f}'\left( x \right)dx$ and by using substitution, we get:
$\int{{{t}^{-n}}dt=\int{{{x}^{-n}}dx}}$

& \dfrac{{{t}^{-n+1}}}{-n+1}=\dfrac{{{x}^{-n+1}}}{-n+1} \\\ & \Rightarrow {{t}^{-n+1}}={{x}^{-n+1}} \\\ \end{aligned}$$ Then, by cancelling the same powers from both sides and also substituting the value of t as supposed, we get: $\begin{aligned} & t=x \\\ & \Rightarrow f\left( x \right)=x \\\ \end{aligned}$ Now, we are asked in the question to find the value of $\int\limits_{0}^{1}{f\left( x \right)dx}$. So, by substituting the value of f(x) as x and by solving the integration, we get: $\begin{aligned} & \int\limits_{0}^{1}{xdx}=\left[ \dfrac{{{x}^{2}}}{2} \right]_{0}^{1} \\\ & \Rightarrow \dfrac{{{1}^{2}}}{2}-0 \\\ & \Rightarrow \dfrac{1}{2} \\\ \end{aligned}$ So, we get the value of integration as $\dfrac{1}{2}$. **So, the correct answer is “Option C”.** **Note:** Now, to solve these types of the questions we need to know some of the basics of the integration beforehand so that we can solve easily and appropriately. So, the required formulas are as: $$\begin{aligned} & \int{{{\left\\{ f\left( x \right) \right\\}}^{-n}}{f}'\left( x \right)dx=f\left( x \right)} \\\ & \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}} \\\ \end{aligned}$$