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 Post subject: Integral equals to zero
PostPosted: Sat Jan 16, 2016 1:55 am 
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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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PostPosted: Sat Jan 16, 2016 1:57 am 

Joined: Tue Dec 29, 2015 1:56 pm
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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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PostPosted: Sat Jan 16, 2016 1:57 am 
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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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