Intégral et règles de calculs

#2bacsef

Sommaire

Propriété

Soient ff et gg deux fonctions continues sur un intervalle II, a,b,c∈Ia, b, c \in I, et α∈R\alpha \in \mathbb{R}. Alors :

  • ∫aaf(x) dx=0\displaystyle \int_{a}^{a} f(x)~dx = 0

  • ∫abf(x) dx=−∫baf(x) dx\displaystyle \int_{a}^{b} f(x)~dx = -\int_{b}^{a} f(x)~dx

  • Relation de Chasles :

∙∫acf(x) dx+∫cbf(x) dx=∫abf(x) dx\bullet\quad \begin{aligned} \int_{a}^{c} f(x)~dx + \int_{c}^{b} f(x)~dx = \int_{a}^{b} f(x)~dx \end{aligned}
  • Linéarité
∙∫ab(f(x)+g(x)) dx=∫abf(x) dx+∫abg(x) dx\bullet\quad \displaystyle \int_{a}^{b} \big(f(x) + g(x)\big)~dx = \int_{a}^{b} f(x)~dx + \int_{a}^{b} g(x)~dx
  • ∫abαf(x) dx=α∫abf(x) dx\displaystyle \int_{a}^{b} \alpha f(x)~dx = \alpha \int_{a}^{b} f(x)~dx

Application 2

  1. Calculer les intégrales suivantes : ∫03∣x−2∣dx \displaystyle\int_{0}^{3}|x-2|dx~ ;;  ∫−12∣x(x−1)∣dx~\displaystyle\int_{-1}^{2}|x(x-1)|dx

  2. On pose : I=∫0π4cos2(x)dxI=\displaystyle\int_{0}^{\dfrac\pi4}cos^2(x)dx et J=∫0π4sin2(x)dxJ=\displaystyle\int_{0}^{\dfrac\pi4}sin^2(x)dx

    a) Calculer I+JI+J et I−JI-J

    b) En déduire II et JJ

1/

∙\bullet \quad Calcul de ∫03∣x−2∣dx\displaystyle\int_{0}^{3}|x-2|dx on a :

x023x−2−0+\begin{array}{|c|ccccc|} \hline x & 0 & & 2 & & 3 \\ \hline x - 2 & & - & 0 & + &\\ \hline \end{array}
∫03∣x−2∣ dx=∫02∣x−2∣ dx+∫23∣x−2∣ dx=−∫02(x−2) dx+∫23(x−2) dx\begin{align*} \int_{0}^{3}|x-2|\,dx&= \int_{0}^{2}|x-2|\,dx + \int_{2}^{3}|x-2|\,dx \\ &= -\int_{0}^{2}(x-2)\,dx + \int_{2}^{3}(x-2)\,dx \end{align*}

on a :

  • ∫02(x−2) dx=[x22−2x]02=−2\displaystyle\int_{0}^{2}(x-2)\,dx=\left[\dfrac{x^2}2-2x\right]_0^2=-2
  • ∫23(x−2) dx=[x22−2x]23=12\displaystyle\int_{2}^{3}(x-2)\,dx=\left[\dfrac{x^2}2-2x\right]_2^3=\dfrac12

Donc : ∫03∣x−2∣ dx=52\boxed{\displaystyle\int_{0}^{3}|x-2|\,dx=\dfrac52}

∙\bullet \quad Calcul de ∫−12∣x(x−1)∣ dx\displaystyle\int_{-1}^{2}|x(x-1)|\,dx :

On a x(x−1)=0  ⟺  x=0x(x-1) = 0 \iff x = 0 ou x=1x = 1.

x−1012x(x−1)+0−0+\begin{array}{c|ccccccc} x & -1 & & 0 & & 1 & & 2 \\ \hline x(x-1) & & + & 0 & - & 0 & + & \end{array}

Donc :

∫−12∣x(x−1)∣ dx={+∫−10x(x−1) dx −∫01x(x−1) dx +∫12x(x−1) dx\begin{align*} \int_{-1}^{2}|x(x-1)|\,dx &= \begin{cases} +\displaystyle\int_{-1}^{0}x(x-1)\,dx \\~\\ \displaystyle- \int_{0}^{1}x(x-1)\,dx \\~\\ \displaystyle+ \int_{1}^{2}x(x-1)\,dx \end{cases} \end{align*}
∫−12∣x(x−1)∣ dx=∫−10x(x−1) dx−∫01x(x−1) dx+∫12x(x−1) dx\begin{align*} \int_{-1}^{2}|x(x-1)|\,dx &=\int_{-1}^{0}x(x-1)\,dx - \int_{0}^{1}x(x-1)\,dx + \int_{1}^{2}x(x-1)\,dx \end{align*}

puisque : x(x−1)=x2−xx(x-1)=x^2-x alors :

∙∫−10x(x−1) dx=[x33−x22]−10=56\bullet\displaystyle\int_{-1}^{0}x(x-1)\,dx=\left[\dfrac{x^3}3-\dfrac{x^2}2\right]_{-1}^0=\dfrac56

∙∫01x(x−1) dx=[x33−x22]01=−16\bullet\displaystyle\int_{0}^{1}x(x-1)\,dx=\left[\dfrac{x^3}3-\dfrac{x^2}2\right]_{0}^1=-\dfrac16

∙∫12x(x−1) dx=[x33−x22]12=56\bullet\displaystyle\int_{1}^{2}x(x-1)\,dx=\left[\dfrac{x^3}3-\dfrac{x^2}2\right]_{1}^2=\dfrac56

Donc :

∫−12∣x(x−1)∣ dx=56+16+56\int_{-1}^{2}|x(x-1)|\,dx =\dfrac56+\dfrac16+\dfrac56

et donc :

∫−12∣x(x−1)∣ dx=116\boxed{\int_{-1}^{2}|x(x-1)|\,dx=\dfrac{11}6}

2/

a/\quad Calcul de I+JI + J et I−JI - J :

I+J=∫0π4(cos⁡2(x)+sin⁡2(x)) dx=∫0π41 dx=[x]0π4=π4\begin{align*} I + J &= \int_{0}^{\dfrac{\pi}{4}} (\cos^2(x) + \sin^2(x)) \, dx \\ &= \int_{0}^{\dfrac{\pi}{4}} 1 \, dx = \left[ x \right]_0^{\dfrac{\pi}{4}} = \dfrac{\pi}{4} \end{align*}
I−J=∫0π4(cos⁡2(x)−sin⁡2(x)) dx=∫0π4cos⁡(2x) dx=[sin⁡(2x)2]0π4=12−0=12\begin{align*} I - J &= \int_{0}^{\dfrac{\pi}{4}} (\cos^2(x) - \sin^2(x)) \, dx \\ &= \int_{0}^{\dfrac{\pi}{4}} \cos(2x) \, dx = \left[ \dfrac{\sin(2x)}{2} \right]_0^{\dfrac{\pi}{4}} \\ &= \dfrac{1}{2} - 0 = \dfrac{1}{2} \end{align*}

b/\quad Détermination de II et JJ :

{I+J=π4I−J=12  ⟹  {I=π+28J=π−28\begin{cases} I + J = \dfrac{\pi}{4} \\ I - J = \dfrac{1}{2} \end{cases} \implies \begin{cases} I = \dfrac{\pi + 2}{8} \\ J = \dfrac{\pi - 2}{8} \end{cases}

Ainsi, les valeurs sont :

I=π+28,J=π−28I = \dfrac{\pi + 2}{8}, \quad J = \dfrac{\pi - 2}{8}