Question

If the integral is $\int\limits_{0}^{x}{{{t}^{2}}f\left( t \right)dt}=x+\int\limits_{x}^{1}{f\left( t \right)dt},$ then the value of $f\left( \dfrac{1}{2} \right)=$
$\left( \text{a} \right)\text{ }\dfrac{1}{5}$
$\left( \text{b} \right)\text{ }\dfrac{4}{5}$
$\left( \text{c} \right)\text{ }\dfrac{3}{5}$
$\left( \text{d} \right)\text{ }\dfrac{2}{5}$

Answer 1

To solve the given question, we will first find out the value of f(x). To find the value of f(x), we will differentiate the equation given in the question with respect to x. Then, we will apply Leibniz’s rule of differentiation which says that the derivative of $\int\limits_{a\left( x \right)}^{b\left( x \right)}{f\left( x,t \right)dt}$ is given by the formula
$\dfrac{d}{dx}\left( \int\limits_{a\left( x \right)}^{b\left( x \right)}{f\left( x,t \right)dt} \right)=\left[ f\left( x,b\left( x \right) \right).\dfrac{d}{dx}b\left( x \right) \right]-\left[ f\left( x,a\left( x \right) \right).\dfrac{d}{dx}a\left( x \right) \right]+\int\limits_{a\left( x \right)}^{b\left( x \right)}{\dfrac{\partial }{\partial x}f\left( x,t \right)dt}$
Using this, we will find the value of f(x). After finding the value of f(x), we will put $x=\dfrac{1}{2}$ to get the value of $f\left( \dfrac{1}{2} \right).$

Complete step-by-step answer:
The question demands that we have to find the value of $f\left( \dfrac{1}{2} \right)$ but we do not know the function f(x). So, the first thing we are going to do is to find the function f(x). Now, the equation given in the question is as shown.
$\int\limits_{0}^{x}{{{t}^{2}}f\left( t \right)dt}=x+\int\limits_{x}^{1}{f\left( t \right)dt}$
Now, we will differentiate the above equation with respect to x. Thus, we will get,
$\dfrac{d}{dx}\left[ \int\limits_{0}^{x}{{{t}^{2}}f\left( t \right)dt} \right]=\dfrac{d}{dx}\left( x \right)+\dfrac{d}{dx}\left[ \int\limits_{x}^{1}{f\left( t \right)dt} \right].....\left( i \right)$
Now, we cannot differentiate these functions directly because we are not able to evaluate the values of the integrals. Thus, we will use Leibniz’s Rule of derivative which says that the derivative of the integral $\int\limits_{a\left( x \right)}^{b\left( x \right)}{g\left( x,t \right)dt}$ is calculated by the formula shown below.
$\dfrac{d}{dx}\left( \int\limits_{a\left( x \right)}^{b\left( x \right)}{g\left( x,t \right)dt} \right)=\left[ g\left( x,b\left( x \right) \right).\dfrac{d}{dx}b\left( x \right) \right]-\left[ g\left( x,a\left( x \right) \right).\dfrac{d}{dx}a\left( x \right) \right]+\int\limits_{a\left( x \right)}^{b\left( x \right)}{\dfrac{\partial }{\partial x}g\left( x,t \right)dt}$
Now, we will calculate the value of $\dfrac{d}{dx}\left[ \int\limits_{0}^{x}{{{t}^{2}}f\left( t \right)dt} \right]$ using Leibniz’s rule of differentiation. Thus, we will get,
$\dfrac{d}{dx}\left[ \int\limits_{0}^{x}{{{t}^{2}}f\left( t \right)dt} \right]=\left[ {{x}^{2}}f\left( x \right).\dfrac{d}{dx}\left( x \right) \right]-\left[ {{x}^{2}}f\left( 0 \right).\dfrac{d}{dx}\left( 0 \right) \right]+\int\limits_{0}^{x}{\dfrac{\partial }{\partial x}\left( 0 \right)dt}$
The differentiation of x is 1 and the differentiation of zero is 0. Thus, we will get,
$\Rightarrow \dfrac{d}{dx}\left[ \int\limits_{0}^{x}{{{t}^{2}}f\left( t \right)dt} \right]={{x}^{2}}f\left( x \right)........\left( ii \right)$
Now, we will calculate the value of $\dfrac{d}{dx}\left[ \int\limits_{x}^{1}{f\left( t \right)dt} \right]$ using Leibniz’s rule of differentiation. Thus, we will get,
$\dfrac{d}{dx}\left[ \int\limits_{x}^{1}{f\left( t \right)dt} \right]=\left[ f\left( 1 \right).\dfrac{d}{dx}\left( 1 \right) \right]-\left[ f\left( x \right).\dfrac{d}{dx}\left( x \right) \right]+\int\limits_{x}^{1}{\dfrac{\partial }{\partial x}\left( 0 \right)dt}$
$\Rightarrow \dfrac{d}{dx}\left[ \int\limits_{x}^{1}{f\left( t \right)dt} \right]=-f\left( x \right)......\left( iii \right)$
From (i), (ii) and (iii), we will get,
$\Rightarrow {{x}^{2}}f\left( x \right)=\dfrac{d}{dx}\left( x \right)-f\left( x \right)$
$\Rightarrow {{x}^{2}}f\left( x \right)=1-f\left( x \right)$
$\Rightarrow {{x}^{2}}f\left( x \right)+f\left( x \right)=1$
$\Rightarrow f\left( x \right)\left[ {{x}^{2}}+1 \right]=1$
$\Rightarrow f\left( x \right)=\dfrac{1}{1+{{x}^{2}}}$
Now, we have to find out the function f(x). Now, we will put $x=\dfrac{1}{2}$ to get the value of $f\left( \dfrac{1}{2} \right).$ Thus,
$\Rightarrow f\left( \dfrac{1}{2} \right)=\dfrac{1}{1+{{\left( \dfrac{1}{2} \right)}^{2}}}=\dfrac{1}{1+\dfrac{1}{4}}=\dfrac{1}{\left( \dfrac{5}{4} \right)}$
$\Rightarrow f\left( \dfrac{1}{2} \right)=\dfrac{4}{5}$
Hence, option (b) is the right answer.

Note: The condition to apply Leibnitz theorem of differentiation is that $-\infty < a\left( x \right),b\left( x \right) < \infty$ Also, the limit should be only in one variable. Thus, if the limit is of the form a(x, y) and b(x, y) then we cannot apply this rule.