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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