Integral inequality (I)

August 20, 2021 / Leave a comment

Let f:[0, 1] \rightarrow \mathbb{R} be a differentiable and convex function such that f(0)=0 and f'(1)= 2. Prove that \int_{0}^{1} f(x) \, \mathrm{d}x < 1.

Symmetry

July 24, 2021 / Leave a comment

Let f, g : \mathbb{R} \rightarrow \mathbb{R} be continuous functions. If \mathcal{C}_f is symmetric around the line x=\frac{\alpha + \beta}{2} then prove that:

    \[\int_{\alpha}^{\beta} x g \left ( f(x) \right )\, \mathrm{d}x = \frac{\alpha+\beta}{2} \int_{\alpha}^{\beta}g \left ( f(x) \right )\, \mathrm{d}x\]

An integral from a geometric view

July 5, 2021 / Leave a comment

Evaluate the integral

    \[\mathcal{J} = \int_0^1 \sqrt{4-x^2} \, \mathrm{d}x\]

Solution

 

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A definite integral

July 5, 2021 / Leave a comment

Let f:\mathbb{R} \rightarrow \mathbb{R} be a continuous function such that

(1)   \begin{equation*} e^x f(x) + f \left ( e^{-x} \right ) = xe^x +e^{-x} \quad \text{for all} \;\; x \in \mathbb{R} \end{equation*}

Evaluate the integral \int_0^1 f(x)\, \mathrm{d}x.

Solution

First of all we note that the equation

(2)   \begin{equation*} e^{-x} = x \end{equation*}

has a unique root , lets call it a. Hence e^{-a} = a and thus e^{-2a}=a^2. We note that (1) is rewritten as

(3)   \begin{equation*} f(x) + e^{-x} f \left ( e^{-x} \right ) = x +e^{-2x} \quad \text{for all} \;\; x \in \mathbb{R} \end{equation*}

Integrating (3) from 0 to a we get

    \begin{align*} \int _0^a f(x)\, \mathrm{d}x + \int_0^a e^{-x} f \left ( e^{-x} \right ) \, \mathrm{d}x = \frac {a^2}{2} - \frac{1}{2} \left ( e^{-2a} -1 \right ) & \Leftrightarrow \\ \int_{0}^{a} f(x) \, \mathrm{d}x - \int_{1}^{e^{-a}} f(t) \, \mathrm{d}t = \frac{a^2}{2} - \frac{1}{2} \left ( e^{-2a} -1 \right ) &\Leftrightarrow \\ \int_{0}^{a} f(x) \, \mathrm{d}x - \int_{1}^{a} f(x) \, \mathrm{d}x = \frac{a^2}{2} - \frac{1}{2} \left ( e^{-2a} -1 \right ) &\Leftrightarrow \\ \int_{0}^{a} f(x) \, \mathrm{d}x + \int_{a}^{1} f(x) \, \mathrm{d}x = \frac{a^2}{2} - \frac{1}{2} \left ( e^{-2a} -1 \right ) & \Leftrightarrow \\ \int_{0}^{1} f(x) \, \mathrm{d}x = \frac{a^2}{2} - \frac{a^2}{2} + \frac{1}{2} &\Leftrightarrow \\ \int_{0}^{1} f(x) \, \mathrm{d}x = \frac{1}{2} \end{align*}

 

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Periodicity and integral

July 3, 2021 / Leave a comment

Let f:\mathbb{R} \rightarrow \mathbb{R} be a continuous and periodic function with period T \neq 0. If \kappa, \mu, \nu \in \mathbb{N} then prove that:

  1. \displaystyle \int_{0}^{\kappa \mathrm{T}} f(x) \, \mathrm{d}x = \kappa \int_0^{\mathrm{T}} f(x) \, \mathrm{d}x
  2. \displaystyle \int_{\alpha + \mu \mathrm{T}}^{\beta + \nu \mathrm{T}} f(x) \, \mathrm{d}x = \int_{\alpha}^{\beta} f(x) \, \mathrm{d}x + \left ( \nu - \mu \right ) \int_{0}^{\mathrm{T}} f(x) \, \mathrm{d}x where \alpha, \beta \in \mathbb{R}.

Solution

  1. We have successively:

        \begin{align*} \int_{0}^{\kappa T} f(x) \, \mathrm{d}x &= \sum_{n=0}^{\kappa-1} \int_{nT}^{(n+1) T} f(x) \, \mathrm{d}x \\ &= \sum_{n=0}^{\kappa-1} \int_{0}^{T} f \left ( u + nT \right )\, \mathrm{d}u \\ &= \sum_{n=0}^{\kappa-1} \int_{0}^{T} f(x) \, \mathrm{d}x \\ &=\kappa \int_{0}^{T} f(x) \, \mathrm{d}x \end{align*}

  2. We have successively:

        \begin{align*} \int_{\alpha + \mu \mathrm{T}}^{\beta + \nu \mathrm{T}} f(x) \, \mathrm{d}x - \int_{\alpha}^{\beta} f(x) \, \mathrm{d}x &= \int_{\alpha + \mu \mathrm{T}}^{\beta + \nu \mathrm{T}} f(x) \, \mathrm{d}x - \\ & - \left ( \int_{\alpha}^{\alpha + \mu T} f(x) \, \mathrm{d}x + \int_{\alpha +\mu T}^{\beta } f(x) \, \mathrm{d}x \right )\\ &=\int_{\beta }^{\beta + \nu \mathrm{T}} f(x) \, \mathrm{d}x - \int_{\alpha}^{\alpha + \mu T} f(x) \, \mathrm{d}x \\ &=\nu \int_{0}^{\mathrm{T}} f(x) \, \mathrm{d}x - \mu \int_{0}^{\mathrm{T}} f(x) \, \mathrm{d}t \\ &= \left ( \nu - \mu \right ) \int_{0}^{\mathrm{T}} f(x) \, \mathrm{d}x \end{align*}

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