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 Post subject: Integral equals to zeroPosted: Sat Jan 16, 2016 1:55 am
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Let $\displaystyle{f:\mathbb{R}\longrightarrow \mathbb{R}}$ be a continuous function such that

$$\int_{0}^{x}f(t)\,\mathrm{d}t>\int_{x}^{1}f(t)\,\mathrm{d}t\,\,,\forall\,x\in\left(-\infty,0\right)\cup\left(0,1\right)\cup\left(1,+\infty\right)$$.

Prove that $\displaystyle{\int_{0}^{1}f(t)\,\mathrm{d}t=0}$ .

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 Post subject: Re: Integral equals to zeroPosted: Sat Jan 16, 2016 1:57 am

Joined: Tue Dec 29, 2015 1:56 pm
Posts: 1
Papapetros Vaggelis wrote:
Let $\displaystyle{f:\mathbb{R}\longrightarrow \mathbb{R}}$ be a continuous function such that

$$\int_{0}^{x}f(t)\,\mathrm{d}t>\int_{x}^{1}f(t)\,\mathrm{d}t\,\,,\forall\,x\in\left(-\infty,0\right)\cup\left(0,1\right)\cup\left(1,+\infty\right)$$.

Prove that $\displaystyle{\int_{0}^{1}f(t)\,\mathrm{d}t=0}$ .

The functions $\displaystyle \int\limits_0^x {f\left( t \right)dt} ,\int\limits_x^1 {f\left( t \right)dt} ,x \in R$ are differentiable cause of the continuouty of f hence they are continuous too.
That means
$\displaystyle \mathop {\lim }\limits_{x \to 1} \int\limits_0^x {f\left( t \right)dt} = \int\limits_0^1 {f\left( t \right)dt} ,\mathop {\lim }\limits_{x \to 1} \int\limits_x^1 {f\left( t \right)dt} = \int\limits_1^1 {f\left( t \right)dt} = 0$
and
$\displaystyle \mathop {\lim }\limits_{x \to 0} \int\limits_0^x {f\left( t \right)dt} = \int\limits_0^0 {f\left( t \right)dt} = 0,\mathop {\lim }\limits_{x \to 0} \int\limits_x^1 {f\left( t \right)dt} = \int\limits_0^1 {f\left( t \right)dt}$

Taking limits to inequality gives
$\displaystyle \mathop {\lim }\limits_{x \to 0} \int\limits_0^x {f\left( t \right)dt} \ge \mathop {\lim }\limits_{x \to 0} \int\limits_x^1 {f\left( t \right)dt} \Rightarrow 0 \ge \int\limits_0^1 {f\left( t \right)dt}$ and
$\displaystyle \mathop {\lim }\limits_{x \to 1} \int\limits_0^x {f\left( t \right)dt} \ge \mathop {\lim }\limits_{x \to 1} \int\limits_x^1 {f\left( t \right)dt} \Rightarrow \int\limits_0^1 {f\left( t \right)dt} \ge 0$
Finally $\displaystyle \int\limits_0^1 {f\left( t \right)dt} = 0$

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 Post subject: Re: Integral equals to zeroPosted: Sat Jan 16, 2016 1:57 am
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Thank you mathxl for your solution.

Here is another one.

We define $\displaystyle{g:\mathbb{R}\longrightarrow \mathbb{R}}$ by $\displaystyle{g(x)=\int_{0}^{x}f(t)\,\mathrm{d}t-\int_{x}^{1}f(t)\,\mathrm{d}t}$.

The function $\displaystyle{g}$ is well defined and continuous at $\displaystyle{\mathbb{R}}$ and according to the intial relation, we have that

$\displaystyle{g(x)>0\,,\forall\,x\in\left(-\infty,0\right)\cup\left(0,1\right)\cup\left(1,+\infty\right)}$ .

We observe that $\displaystyle{g(0)=-\int_{0}^{1}f(t)\,\mathrm{d}t}$ and $\displaystyle{g(1)=\int_{0}^{1}f(t)\,\mathrm{d}t}$ .

Suppose that $\displaystyle{\int_{0}^{1}f(t)\,\mathrm{d}t\neq 0}$.

Then, $\displaystyle{g(0)\cdot g(1)=-\left(\int_{0}^{1}f(t)\,\mathrm{d}t\right)^2<0}$

and since $\displaystyle{g}$ is continuous at $\displaystyle{\left[0,1\right]}$,

it follows that there exists $\displaystyle{x_0\in\left(0,1\right)}$ such that $\displaystyle{g(x_0)=0}$ ($\displaystyle{\rm{Bolzano's Theorem}}$) ,

a contradiction, because $\displaystyle{g(x)>0\,\forall\,x\in\left(0,1\right)}$ .

So, $\displaystyle{\int_{0}^{1}f(t)\,\mathrm{d}t=0}$.

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