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# ISI MStat PSB 2007 Problem 4 | Application of Newton Leibniz theorem This is a very beautiful sample problem from ISI MStat PSB 2007 Problem 4 based on use of Newton Leibniz theorem . Let's give it a try !!

## Problem- ISI MStat PSB 2007 Problem 4

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a bounded continuous function. Define $g:[0, \infty) \rightarrow \mathbb{R}$ by,
$g(x)=\int_{-x}^{x}(2 x t+1) f(t) dt$
Show that g is differentiable on $(0, \infty)$ and find the derivative of g.

### Prerequisites

Riemann integrability

Continuity

Newton Leibniz theorem

## Solution :

As $f: \mathbb{R} \rightarrow \mathbb{R}$ be a bounded continuous function hence the function

$|\Phi(t)|=|(2xt+1)f(t)|=|2xt+1||f(t)|<(|2xt|+1)M<(2|x|^2+1)M$ , which is finite for a particular x so it's a riemann integrable function on t.

Now, by fundamental theorem we have g(x)=F(x)-F(-x) , where F is antiderivative of $\Phi(t)$ .

Hence from above we can say that g(x) is differentiable function over x .
Now by Leibniz integral rule we have $g'(x)=(2x^2+1)f(x)+f(-x)(1-2x^2) + \int_{-x}^{x} (2t)f(t) dt$.

## Food For Thought

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function. Now, we define $g(x)$ such that $g(x)=f(x) \int_{0}^{x} f(t) d t$
Prove that if g is a non increasing function, then f is identically equal to 0.

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This is a very beautiful sample problem from ISI MStat PSB 2007 Problem 4 based on use of Newton Leibniz theorem . Let's give it a try !!

## Problem- ISI MStat PSB 2007 Problem 4

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a bounded continuous function. Define $g:[0, \infty) \rightarrow \mathbb{R}$ by,
$g(x)=\int_{-x}^{x}(2 x t+1) f(t) dt$
Show that g is differentiable on $(0, \infty)$ and find the derivative of g.

### Prerequisites

Riemann integrability

Continuity

Newton Leibniz theorem

## Solution :

As $f: \mathbb{R} \rightarrow \mathbb{R}$ be a bounded continuous function hence the function

$|\Phi(t)|=|(2xt+1)f(t)|=|2xt+1||f(t)|<(|2xt|+1)M<(2|x|^2+1)M$ , which is finite for a particular x so it's a riemann integrable function on t.

Now, by fundamental theorem we have g(x)=F(x)-F(-x) , where F is antiderivative of $\Phi(t)$ .

Hence from above we can say that g(x) is differentiable function over x .
Now by Leibniz integral rule we have $g'(x)=(2x^2+1)f(x)+f(-x)(1-2x^2) + \int_{-x}^{x} (2t)f(t) dt$.

## Food For Thought

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function. Now, we define $g(x)$ such that $g(x)=f(x) \int_{0}^{x} f(t) d t$
Prove that if g is a non increasing function, then f is identically equal to 0.

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