We consider $n$ a non-zero natural integer. We consider the function $f_n$ defined on the interval $[0; 1]$ by:
$$f_n(x) = x^n e^{1-x}$$
We admit that the function $f_n$ is differentiable on $[0; 1]$ and we denote $f_n'$ its derivative function.

Part A

In this part we study the case where $n = 1$. We thus study the function $f_1$ defined on $[0; 1]$ by:
$$f_1(x) = x e^{1-x}$$

1. Show that $f_1'(x)$ is strictly positive for all real $x$ in $[0; 1[$.
2. Deduce the table of variations of the function $f_1$ on the interval $[0; 1]$.
3. Deduce that the equation $f_1(x) = 0.1$ admits a unique solution in the interval $[0; 1]$.

Part B

We consider the sequence $(u_n)$ defined for all non-zero natural integers $n$ by
$$u_n = \int_0^1 f_n(x) \, dx \quad \text{that is} \quad u_n = \int_0^1 x^n e^{1-x} \, dx$$
We admit that $u_1 = e - 2$.

1. a. Justify that for all $x \in [0; 1]$ and for all non-zero natural integers $n$, $$0 \leq x^{n+1} \leq x^n$$ b. Deduce that for all non-zero natural integers $n$, $$0 \leq u_{n+1} \leq u_n.$$ c. Show that the sequence $(u_n)$ is convergent.
2. a. Using integration by parts, prove that for all non-zero natural integers $n$ we have: $$u_{n+1} = (n+1)u_n - 1$$ b. Consider the Python script below defining the function suite(): \begin{verbatim} from math import exp def suite(): u = ... for n in range (1, ...): u = ... return \end{verbatim} Copy and complete the Python script above so that the function suite() returns the value of $\int_0^1 x^8 e^{1-x} \, dx$.
3. a. Prove that for all non-zero natural integers $n$ we have: $$u_n \leq \frac{e}{n+1}$$ b. Deduce the limit of the sequence $(u_n)$.

We equip the plane with an orthonormal coordinate system. For every natural integer $n$, we consider the function $f_n$ defined on $[0; +\infty[$ by: $$f_0(x) = \mathrm{e}^{-x} \text{ and, for } n \geqslant 1,\ f_n(x) = x^n \mathrm{e}^{-x}.$$ For every natural integer $n$, we denote $\mathscr{C}_n$ the representative curve of the function $f_n$.
Parts A and B are independent.
Part A: Study of the functions $f_n$ for $n \geqslant 1$
We consider a natural integer $n \geqslant 1$.

1. a. We admit that the function $f_n$ is differentiable on $[0; +\infty[$. Show that for all $x \geqslant 0$, $$f_n'(x) = (n - x)x^{n-1}\mathrm{e}^{-x}.$$ b. Justify all elements of the table below:
   

   $x$	0		$n$		$+\infty$
   $f_n'(x)$		+	0	-
   			$\left(\frac{n}{\mathrm{e}}\right)^n$
   $f_n$
   	0				0

   
   
2. Justify by calculation that the point $\mathrm{A}\left(1; \mathrm{e}^{-1}\right)$ belongs to the curve $\mathscr{C}_n$.

Part B: Study of the integrals $\int_0^1 f_n(x)\,\mathrm{d}x$ for $n \geqslant 0$
In this part, we study the functions $f_n$ on $[0; 1]$ and we consider the sequence $(I_n)$ defined for every natural integer $n$ by: $$I_n = \int_0^1 f_n(x)\,\mathrm{d}x = \int_0^1 x^n \mathrm{e}^{-x}\,\mathrm{d}x.$$

1. On the graph in APPENDIX, the curves $\mathscr{C}_0, \mathscr{C}_1, \mathscr{C}_2, \mathscr{C}_{10}$ and $\mathscr{C}_{100}$ are represented. a. Give a graphical interpretation of $I_n$. b. By reading this graph, what conjecture can be made about the limit of the sequence $(I_n)$?
2. Calculate $I_0$.
3. a. Let $n$ be a natural integer. Prove that for all $x \in [0; 1]$, $$0 \leqslant x^{n+1} \leqslant x^n.$$ b. Deduce that for every natural integer $n$, we have: $$0 \leqslant I_{n+1} \leqslant I_n.$$
4. Prove that the sequence $(I_n)$ is convergent, towards a limit greater than or equal to zero that we will denote $\ell$.
5. Using integration by parts, prove that for every natural integer $n$ we have: $$I_{n+1} = (n+1)I_n - \frac{1}{\mathrm{e}}.$$
6. a. Prove that if $\ell > 0$, the equality from question 5 leads to a contradiction. b. Prove that $\ell = 0$. You may use question 6.a.
7. The script of the \texttt{mystere} function is given below, written in Python language. The constant \texttt{e} has been imported. \begin{verbatim} def mystere(n): I = 1 - 1/e L = [I] for i in range(n): I = (i + 1)*I - 1/e L.append(I) return L \end{verbatim} What does \texttt{mystere(100)} return in the context of the exercise?

For every pair of natural integers $( p , q )$ and for every $\varepsilon \in ] 0,1 [$, we denote $$I _ { p , q } = \int _ { 0 } ^ { 1 } t ^ { p } ( \ln t ) ^ { q } \mathrm {~d} t \quad \text { and } \quad I _ { p , q } ^ { \varepsilon } = \int _ { \varepsilon } ^ { 1 } t ^ { p } ( \ln t ) ^ { q } \mathrm {~d} t$$
Show that $\left. \forall p \in \mathbb { N } , \forall q \in \mathbb { N } ^ { * } , \forall \varepsilon \in \right] 0,1 \left[ , \quad I _ { p , q } ^ { \varepsilon } = - \frac { q } { p + 1 } I _ { p , q - 1 } ^ { \varepsilon } - \frac { \varepsilon ^ { p + 1 } ( \ln \varepsilon ) ^ { q } } { p + 1 } \right.$.

For every pair of natural integers $( p , q )$ and for every $\varepsilon \in ] 0,1 [$, we denote $$I _ { p , q } = \int _ { 0 } ^ { 1 } t ^ { p } ( \ln t ) ^ { q } \mathrm {~d} t \quad \text { and } \quad I _ { p , q } ^ { \varepsilon } = \int _ { \varepsilon } ^ { 1 } t ^ { p } ( \ln t ) ^ { q } \mathrm {~d} t$$
Deduce that we have $\forall p \in \mathbb { N } , \forall q \in \mathbb { N } ^ { * } , \quad I _ { p , q } = - \frac { q } { p + 1 } I _ { p , q - 1 }$.

For every pair of natural integers $( p , q )$ and for every $\varepsilon \in ] 0,1 [$, we denote $$I _ { p , q } = \int _ { 0 } ^ { 1 } t ^ { p } ( \ln t ) ^ { q } \mathrm {~d} t \quad \text { and } \quad I _ { p , q } ^ { \varepsilon } = \int _ { \varepsilon } ^ { 1 } t ^ { p } ( \ln t ) ^ { q } \mathrm {~d} t$$
Deduce an expression for $I _ { p , q }$ as a function of the integers $p$ and $q$.

For every natural integer $k$ we set $$m_{k} = \frac{1}{2\pi} \int_{-2}^{2} x^{k} \sqrt{4 - x^{2}} \, \mathrm{d}x$$ Using integration by parts, show that, for every natural integer $k$, $$m_{2k+2} = \frac{2(2k+1)}{k+2} m_{2k}$$

For every natural integer $k$ we set $$m_{k} = \frac{1}{2\pi} \int_{-2}^{2} x^{k} \sqrt{4 - x^{2}} \, \mathrm{d}x$$ Using integration by parts, show that, for every natural integer $k$, $$m_{2k+2} = \frac{2(2k+1)}{k+2} m_{2k}$$

For $I(x) = \int \frac{\sec^2 x - 2022}{\sin^{2022} x} dx$, if $I\left(\frac{\pi}{4}\right) = 2^{1011}$, then
(1) $3^{1010} I\left(\frac{\pi}{3}\right) - I\left(\frac{\pi}{6}\right) = 0$
(2) $3^{1010} I\left(\frac{\pi}{6}\right) - I\left(\frac{\pi}{3}\right) = 0$
(3) $3^{1011} I\left(\frac{\pi}{3}\right) - I\left(\frac{\pi}{6}\right) = 0$
(4) $3^{1011} I\left(\frac{\pi}{6}\right) - I\left(\frac{\pi}{3}\right) = 0$

The value of $\dfrac { e ^ { - \frac { \pi } { 4 } } + \int _ { 0 } ^ { \frac { \pi } { 4 } } e ^ { - x } \tan ^ { 50 } x \, d x } { \int _ { 0 } ^ { \frac { \pi } { 4 } } e ^ { - x } \left( \tan ^ { 49 } x + \tan ^ { 51 } x \right) d x }$
(1) 51
(2) 50
(3) 25
(4) 49

If $\int \operatorname { cosec } ^ { 5 } x \, d x = \alpha \cot x \operatorname { cosec } x \left( \operatorname { cosec } ^ { 2 } x + \frac { 3 } { 2 } \right) + \beta \log _ { e } \left| \tan \frac { x } { 2 } \right| + C$ where $\alpha , \beta \in \mathbb { R }$ and C is the constant of integration, then the value of $8 ( \alpha + \beta )$ equals $\_\_\_\_$

A sequence $\{a_n\}$ is defined as
$$a_n = \int_0^{\frac{1}{4}} x^n e^{-x}\, dx \quad (n = 1, 2, 3, \cdots).$$
Then
$$a_1 = -\frac{\mathbf{H}}{\mathbf{I}} e^{\frac{\mathbf{JK}}{\mathbf{L}}} + 1.$$
Also $a_{n+1}$ can be expressed in terms of $a_n$ as
$$a_{n+1} = -\left(\frac{\mathbf{M}}{\mathbf{N}}\right)^{n+1} e^{\frac{\mathrm{JK}}{\mathrm{L}}} + (n + \mathbf{O})a_n \quad (n = 1, 2, 3, \cdots).$$
When this is transformed into
$$na_n = a_{n+1} - a_n + \left(\frac{\mathrm{M}}{\mathrm{N}}\right)^{n+1} e^{\frac{\mathrm{JK}}{\mathrm{L}}},$$
we have
$$\sum_{k=1}^{n} ka_k = a_{n+1} - a_1 + \frac{\mathbf{P}}{\mathbf{PQ}} e^{\frac{\mathrm{JK}}{\mathrm{L}}} \left\{1 - \left(\frac{\mathbf{S}}{\mathbf{I}}\right)^n\right\}.$$
Since, for $0 \leqq x$, the range of values of $e^{-x}$ is $0 < e^{-x} \leqq \mathbf{U}$, it follows that
$$0 < a_n < \int_0^{\frac{1}{4}} \square\, x^n\, dx = \frac{1}{\square^{n+1}(n+1)}.$$
Thus, since
$$\lim_{n \to \infty} a_n = \mathbf{W},$$
we obtain
$$\lim_{n \to \infty} \sum_{k=1}^{n} ka_k = \frac{\mathbf{X}}{\mathbf{Y}} e^{\frac{\mathrm{JK}}{\mathrm{L}}} - 1.$$

$$\int _ { 1 } ^ { e } \ln ^ { 3 } x \, d x = 6 - 2 e$$
Given this, what is the value of the integral $\int _ { 1 } ^ { e } \ln ^ { 4 } x \, d x$?
A) $7 e - 16$
B) $8 e - 18$
C) $9 e - 24$
D) $10 e - 26$
E) $11 e - 28$