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\begin{center}\section*{\textcolor{red}{Correction des exercices sur l'intégration}}\end{center}

\subsection{}
On souhaite calculer l'intégrale suivante : $I=\int_0^\pi t\cos(t)\text{ d}t$.\\
On pose $f:t\mapsto t\cos(t)$, définie sur $\mathbb{R}$.
\begin{enumerate}
\item %Démontrer que pour tout $t\in\mathbb{R}$, $f(t)=-2\sin(t)-f''(t).$
$f'(t)=1\times \cos(t)+t\times (-\sin(t))=\cos(t)-t\sin(t)$ donc $\\
f''(t)=-\sin(t)-\left[1\times \sin(t)+t\cos(t)\right]=-2\sin(t)-f(t)$.\\
Donc $f''(t)=-2\sin(t)-f(t)$. On en déduit que : $\boxed{\textcolor{red}{f(t)=-2\sin (t)-f''(t)}}$.

\item %En déduire la valeur de $I$
On en déduit une primitive $F$ de $f$ : $F(t)=2\cos(t)-f'(t)=2\cos(t)-\left[\cos(t)-t\sin(t)\right]=\cos(t)+t\sin(t)$.\\
Alors $I=F(\pi)-F(0)=\left(-1+\pi\sin(\pi)\right)-\left(1+0\right)=\boxed{\textcolor{red}{-2}}$

\end{enumerate}

\subsection{}
On souhaite calculer l'intégrale suivante : $I=\int_{-5}^1x\mathrm{e}^{x}\text{ d}x$
On pose $f:x\mapsto x\mathrm{e}^{x}$, définie sur $\mathbb{R}$.
\begin{enumerate}
\item %Pour tout réel $x$, calculer $f'(x)$ et l'exprimer en fonction de $f(x)$.
$f'(x)=\mathrm{e}^{x}+x\mathrm{e}^{x}=\mathrm{e}^{x}+f(x)$

\item %En déduire la valeur de $I$.
On en déduit $f(x)=f'(x)-\mathrm{e}^{x}$ ; une primitive de $f$ est alors $F$ avec $F(x)=f(x)-\mathrm{e}^{x}=x\mathrm{e}^{x}-\mathrm{e}^{x}=(x-1)\mathrm{e}^{x}$.\\
$I=F(1)-F(-5)=\left(\mathrm{e}-\mathrm{e}\right)-\left(-5\mathrm{e}^{-5}-\mathrm{e}^{-5}\right)=\boxed{\textcolor{red}{6\mathrm{e}^{-5}}}$

\end{enumerate}

\subsection{}
On considère les deux intégrales : $I=\int_0^\pi\cos^2(x)\text{ d}x\text{ et }J=\int_0^\pi\sin^2(x)\text{ d}x$

\begin{enumerate}
\item %Calculer $I+J$ et $I-J$.
Par linéarité, $I+J=\int_0^\pi\left(\cos^2 x+\sin^2 x\right)\text{ d}x=\int_0^1\text{ d}x=\pi-0=\pi$. (Car $\cos^2x+\sin^2x=1$)\\
De même : $I-J=\int_0^\pi\left(\cos^2 x-\sin^2 x\right)\text{ d}x=\int_0^\pi\cos(2x)\text{ d}x$. (Car $\cos^2x-\sin^2x=\cos(2x))$\\
Une primitive de $x\mapsto\cos(2x)$ est $x\mapsto\dfrac{1}{2}\sin(2x)$ donc $I-J=\dfrac{1}{2}\sin(2\pi)-\dfrac{1}{2}\sin(0)=0$.

\item %En déduire les valeurs de $I$ et $J$.
On a : $\begin{cases}I+J=\pi\\I-J=0\end{cases}$.\\
En additionnant, on trouve $2I=\pi$ donc $I=\dfrac{\pi}{2}$.\\
Comme $I-J=0$, on a $J=I=\dfrac{\pi}{2}$. D'où $\boxed{\textcolor{red}{I=J=\dfrac{\pi}{2}}}$
\end{enumerate}

\subsection{}
Soit l'intégrale $I=\int_0^1\dfrac{1}{1+u^2}\text{ d}u$.
\begin{enumerate}
\item %Est-il possible, avec vos connaissances de Terminale, de calculer une primitive de la fonction \linebreak $f:u\mapsto \dfrac{1}{1+u^2}$ ?
Il n' y a aucune formule qui permette de trouver une primitive de cette fonction dans le tableau des primitives en Terminale.\\

\item Démontrons que, pour tout réel $u\geqslant 0$, $1-u^2\leqslant \dfrac{1}{1+u^2}\leqslant 1$.
\begin{enumerate}[$\bullet$]
\item $\forall u\geqslant 0$, $1+u^2\geqslant 1$ donc $\dfrac{1}{1+u^2}\leqslant 1$.

\item $\forall u\geqslant 0$, $\dfrac{1}{1+u^2}-\left(1-u^2\right)=\dfrac{1-\left(1+u^2\right)\left(1-u^2\right)}{1+u^2}=\dfrac{1-\left(1-u^4\right)}{1++u^2}=\dfrac{u^4}{1+u^2}\geqslant 0$ donc $1-u^2\leqslant \dfrac{1}{1+u^2}$.
\end{enumerate}

\item %En dédire un encadrement de $I$
Par \textbf{\textcolor{red}{conservation de l'ordre}}, on a : $\int_0^1\left(1-u^2\right)\text{ d}u\leqslant I\leqslant \int_0^11\text{ d}u$.\\
$\int_0^1\left(1-u^2\right)\text{ d}u=\left[u-\dfrac{u^3}{3}\right]_0^1=1-\dfrac{1}{3}=\dfrac{2}{3}$.\\
$\int_0^11\text{ d}u=1$.\\
Par conséquent : $\dfrac{2}{3}\leqslant I\leqslant 1$

\item %À l'aide de la calculatrice, calculer une valeur approchée de $4I$. Pouvez-vous en déduire une conjecture sur la valeur de $I$ ?
$4I\approx \numprint{3,141592654}$ qui ressemble au nombre $\pi$. ON peut donc conjecturer que $\boxed{\textcolor{red}{I=\dfrac{\pi}{4}}}$
\bigskip

\textbf{\textcolor{red}{Remarque}} : en fait, une primitive de $x\mapsto\dfrac{1}{1+x^2}$ est la fonction $\arctan$ (fonction réciproque de la fonction $\tan$ donc $I=\arctan(1)-\arctan(0)=\dfrac{\pi}{4}-0=\dfrac{\pi}{4}$ car $\tan\left(\dfrac{\pi}{4}=1\right)$
\end{enumerate}

\subsection{}
Soit $F:x\mapsto \int_0^x\mathrm{e}^{-t^2}\text{ d}t$ une fonction.
\begin{enumerate}
\item% Sur quel ensemble cette fonction est-elle définie ?
On pose $f(t)=\mathrm{e}^{-t^2}$ ; $f$ est continue sur $\mathbb{R}$ donc $F$ est définie sr $\mathbb{R}$.

\item 
\begin{enumerate}
\item  $F$ est dérivable sur $\mathbb{R}^+$ car $f$ est continue, donc $F$ est une primitive de $f$. (la primitive qui s'annule en 0)

\item $F'=f$ et $f\geqslant 0$ donc $F'(x)=f(x)$ ; or $f(x)\geqslant 0$ donc $F$ est croissante sur $\mathbb{R}^+$ (et même sur $\mathbb{R}$).
\end{enumerate}

\item Soit $H:x\mapsto\int_x^{2x}\mathrm{e}^{-t^2}\text{ d}t$.
\begin{enumerate}
\item 
\end{enumerate}
%Démontrer que $H(x)=F(2x)-F(x)$.
D'après la relation de Chasles, $H(x)=\int_x^0\mathrm{e}^{-t^2}\text{ d}t+\int_0^{2x}\mathrm{e}^{-t^2}\text{ d}x=-\int_0^x\mathrm{e}^{-t^2}\text{ d}t+\int_0^{2x1}\mathrm{e}^{-t^2}\text{ d}x\\
=-F(x)+F(2x)=\boxed{\textcolor{red}{F(2x)-F(x)}}$.

\item %En déduire que $H$ est dérivable sur $\mathbb{R}^+$ puis démontrer que $H'(x)=\mathrm{e}^{-4x^2}\left(2-\mathrm{e}^{3x^2}\right)$.
$H$ est dérivable comme somme et composée de fonction dérivables.\\
$H'(x)=2F'(2x)-F'(x)=2f(2x)-f(x)=2\mathrm{e}^{-4x^2}-\mathrm{e}^{-x^2}=2\mathrm{e}^{-4x^2}--\mathrm{e}^{-4x^2+3x^2}=\boxed{\textcolor{red}{\mathrm{e}^{-4x^2}\left(2-\mathrm{e}^{3x^2}\right)}}$

\item %En déduire les variations de $H$
$\mathrm{e}^{-4x^2}>0$ donc $H'(x)$ est du signe de $2-\mathrm{e}^{3x^2}$.\\
$2-\mathrm{e}^{3x^2}=0\Leftrightarrow \mathrm{e}^{3x^2}=2$.\\
On applique la fonction logarithme népérien (voir prochain chapitre): $\ln\left(\mathrm{e}^{3x^2}\right)=\ln(2)$ donc $3x^2=\ln (2)$ car $\ln\left(\mathrm{e}^{x}\right)=x$. Comme $x\geqslant 0$, on trouve $x=\sqrt{\dfrac{\ln(2)}{3}}$.\\
$2-\mathrm{e}^{3x^2}>0\Leftrightarrow \mathrm{e}^{3x^2}<2\Leftrightarrow 3x^2<\ln(2)\Leftrightarrow x<\sqrt{\dfrac{\ln(2)}{3}}$.\\
On en déduit le tableau de variation :
\begin{center}

\begin{variations}
x&0&&\sqrt{\dfrac{\ln(2)}{3}}&&\pI\\
\hline
H'(x)&&+&\z&-&\\
\hline
\m{H(x)}&0&\c&\h{H\left(\sqrt{\dfrac{\ln(2)}{3}}\right)}&\d&\\
\hline
\end{variations}
\end{center}

\end{enumerate}

\subsection{\textcolor{blue}{Méthode des rectangles}}
On souhaite déterminer une valeur approchée de $\displaystyle{\int_0^3} \dfrac{1}{\sqrt{x^2+1}}\,\mathrm{d}x$.
%\begin{enumerate}
% \item %Démontrer que la fonction à intégrer est décroissante sur $I = [0\,;3]$.
% $x\mapsto x^2+1$ est croissante donc $x\mapsto \sqrt{x^2+1}$ est croissante et $x\mapsto \dfrac{1}{\sqrt{x^2+1}} $ est alors décroissante.
% 
% \item Pour $n \in \mathbb{N}^{*}$, on découpe $I$ en $n$ intervalles
%   de même amplitude, que l'on note $I_k$, pour $k=0, \ldots, n-1$ et
%   sur chaque intervalle, on construit le rectangle  \og
%   inférieur \fg{}.
%  \begin{enumerate}
%   \item Faire un schéma.
%   \item Quelle est l'amplitude des intervalles $I_k$ ?
%   \item On note $I_k=[x_k\,; x_{k+1}]$. Pour tout
%     $k=0, \ldots, n$,exprimer $x_k$ en fonction de $k$ et
%     $n$.
%   \item Pour tout $k= 0, \ldots, n-1$, quelle est l'aire du 
%     rectangle construit sur l'intervalle $I_k$ ?
%   \item Compléter l'algorithme permettant de calculer la somme des aires des rectangles inférieurs :
%   
%\textbf{Algorithme}
%
%     %\titreAlgo{Méthode Rectangles} 
%\begin{minipage}{11cm}
%\fbox{\begin{tabular}{ll}\\
%1.&Liste des variables utilisées\\
%2.&$k$, $n$ entiers\\
%3.&Sinf : réel\\
%4.&Entrée\\
%5.&Saisir $n$\\
%6;&Traitement\\
%7.&Donner à Sin la valeur 0\\
%8.&Pour k variant de \dots à \dots faire\\
%9.&Donner à Sinf la valeur \dots\\
%10.&FinPour\\
%11.&Sortie\\
%12.&Afficher Sinf\\
%13.&Fin de l'algortithme
%\end{tabular}
%}
%\end{minipage} 
%
% \end{enumerate}
%  \item Reprendre la question 2) avec les rectangles supérieurs et compléter l'algorithme pour qu'il affiche aussi la valeur de \texttt{Ssup}.
%  
%  \item Modifier cet algorithme pour qu'il s'arrête lorsque la différence entre \texttt{Ssup} et \texttt{Sinf} devient inférieure à $10^{-2}$. On fera aussi afficher la valeur de $n$ correspondant à la subdivision de $I$ permettant d'obtenir cette précision.
%\end{enumerate}

\begin{enumerate}
 \item Soit $f : x \mapsto \dfrac{1}{\sqrt{x^2+1}}$ définie sur $I = [0 \,; 3]$. On a successivement :
 \begin{enumerate}
  \item[•] $x \mapsto x^2+1$ est croissante sur $I$ ;
  \item[•] $x \mapsto \sqrt{x^2+1}$ est croissante sur $I$ car la fonction racine carrée est croissante sur $[0 \,; +\infty[$ ;
  \item[•] $x \mapsto \dfrac{1}{\sqrt{x^2+1}}$ est décroissante sur $I$ car la fonction inverse est décroissante sur $[0 \,; +\infty[$.
 \end{enumerate}
 
 \item \begin{enumerate}
  \item Ici, avec $n = 5$, on obtient :
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 \item L'amplitude de chaque intervalle est $\dfrac{3}{n}$.
 \item Pour tout $k = 0, \ldots, n$, on a $x_k = \dfrac{3k}{n}$.
 \item Le rectangle construit sur l'intervalle $I_k$ a pour largeur $\dfrac{3}{n}$ et pour longueur \\
$f\left( x_{k+1} \right) = \dfrac{1}{\sqrt{1+\left(\dfrac{3(k+1)}{n}\right)^2}}=\dfrac{1}{\sqrt{\dfrac{n^2+9(k+1)^2}{n^2}}}=\dfrac{1}{\dfrac{\sqrt{n^2+9(k+1)^2}}{n}}= \dfrac{n}{\sqrt{n^2+9(k+1)^2}}$, son aire est donc $\mathscr{A}_k =\dfrac{3}{n}\times  \dfrac{n}{\sqrt{n^2+9(k+1)^2}}= \boxed{\textcolor{red}{\dfrac{3}{\sqrt{n^2+9(k+1)^2}}}}$.
 \item La ligne 8 se complète en \og \textbf{Pour k variant de 0 à n$-$1} \fg{}.\\
 La ligne 9 se complète en \og Donner à Sinf la valeur Sinf + $\dfrac{3}{\sqrt{n^2+9(k+1)^2}}$ \fg{}.
 \end{enumerate}
 
\newpage


 \item On commence par le schéma :
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 L'amplitude des intervalles est la même : les rectangles ont donc la même largeur. Seule leur longueur change puisqu'elle est maintenant de $f\left( x_k \right) = \dfrac{n}{\sqrt{n^2+9k^2}}$ et leur aire est ainsi $\mathscr{A}_k = \dfrac{3}{\sqrt{n^2+9k^2}}$.\\
 Dans l'algorithme, seule la ligne 9 est à modifiée en \og Donner à Ssup la valeur Ssup + $\dfrac{3}{\sqrt{n^2+9k^2}}$ \fg{}.
 
 \item \hfill\\
 Liste des variables utilisées :\\
 k, n : entiers\\
 Ssup, Sinf : réels\\
 Entrée\\
 Saisir n\\
 Traitement\\
 Faire\\
 Donner à Sinf la valeur 0\\
 Donner à Ssup la valeur 0\\
 Pour k variant de 0 à n-1, faire\\
 Donner à Sinf la valeur Sinf + $\dfrac{3}{\sqrt{n^2+9(k+1)^2}}$\\
 Donner à Ssup la valeur Ssup + $\dfrac{3}{\sqrt{n^2+9k^2}}$\\
 FinPour\\
 Donner à n la valeur n+1\\
 TantQue(Ssup-Sinf>1/100)\\
 Afficher Sinf\\
 Afficher Ssup\\
 Affiche n\\
\end{enumerate}


\label{fin}
\end{document}  