# Consider the following function f ( x ) = integral x 0 1 - t 2/ 1 + t 2 d t . Find the...

## Question:

Consider the following function

{eq}f(x)= \int^x_0 \frac{1-t^2}{1+t^2} \ dt {/eq}.

Find the interval in which the function {eq}f {/eq} is concave down. (Remember that a function is concave down if its double derivative is negative)

## Concavity:

As per Newton-Leibnitz formula for differentiation to differentiate {eq}f(x)=\int_{p(x)}^{q(x)}g(t)dt {/eq} w.r.t. {eq}x {/eq}, the derivative is given by:-

{eq}\frac{d}{dx}\int_{p(x)}^{q(x)}g(t)dt=q'(x)g(q(x))-p'(x)g(p(x)) {/eq}. Formulae helpful to set-up the concavity for the given function is:-

{eq}\frac{d}{dx}(\frac{m}{n})=\frac{m'\cdot n-n'\cdot m}{n^2},\ \frac{d}{dx}x^n=nx^{n-1} {/eq}.

## Answer and Explanation:

As per given info:-

{eq}f(x)= \int^x_0 \frac{1-t^2}{1+t^2} \ dt {/eq}, Now as we know the fact that:-

- As per Newton-Leibnitz formula for differentiation to differentiate {eq}f(x)=\int_{p(x)}^{q(x)}g(t)dt {/eq} w.r.t. {eq}x {/eq}, the derivative is given by:-

{eq}\frac{d}{dx}\int_{p(x)}^{q(x)}g(t)dt=q'(x)g(q(x))-p'(x)g(p(x)) {/eq}.

- Formulae helpful to set-up the concavity for the given function {eq}f {/eq} is:-

{eq}\frac{d}{dx}(\frac{m}{n})=\frac{m'\cdot n-n'\cdot m}{n^2},\ \frac{d}{dx}x^n=nx^{n-1} {/eq}, Hence:-

{eq}f'(x)=\frac{d}{dx}\int_0^x\frac{1-t^2}{1+t^2}dt=\frac{1-x^2}{1+x^2} {/eq}.

{eq}f''(x)=\frac{-2x(1+x^2)-2x(1-x^2)}{(1+x^2)^2}=\frac{-4x}{(1+x^2)^2}<0\ \forall\ x\epsilon R^+ {/eq}

{eq}\Rightarrow f {/eq} is concave downwards for {eq}x\epsilon R^+ {/eq}.

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