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 Author: Papapetros Vaggelis [ Sat Jan 16, 2016 1:55 am ] Post subject: Integral equals to zero 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 $$\displaystyle{\int_{0}^{1}f(t)\,\mathrm{d}t=0}$$ .

 Author: mathxl [ Sat Jan 16, 2016 1:57 am ] Post subject: Re: Integral equals to zero 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 $$\displaystyle \int\limits_0^1 {f\left( t \right)dt} = 0$$

 Author: Papapetros Vaggelis [ Sat Jan 16, 2016 1:57 am ] Post subject: Re: Integral equals to zero 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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