Let $n \in \mathbb{N}^*$, $h = \frac{1}{n+1}$, and $x_i = ih$ for all $i \in \{0, \ldots, n+1\}$. Show that for any function $v \in \mathcal { C } ^ { 4 } ( [ 0,1 ] , \mathbb { R } )$, there exists a constant $C \geq 0$, independent of $n$, such that
$$\forall i \in \{ 1 , \ldots , n \} , \left| v ^ { \prime \prime } \left( x _ { i } \right) - \frac { 1 } { h ^ { 2 } } \left( v \left( x _ { i + 1 } \right) + v \left( x _ { i - 1 } \right) - 2 v \left( x _ { i } \right) \right) \right| \leq C h ^ { 2 }$$

In the case $I = [0,1]$ and $\forall x \in I, w(x) = 1$, we seek to approximate $\int_0^1 f(x)\,\mathrm{d}x$ when $f$ is a continuous function from $[0,1]$ to $\mathbb{R}$.
Determine the order of the quadrature formula $I_0(f) = f(1/2)$ and represent graphically the associated error $e(f)$.

In the case $I = [0,1]$ and $\forall x \in I, w(x) = 1$, we seek to approximate $\int_0^1 f(x)\,\mathrm{d}x$ when $f$ is a continuous function from $[0,1]$ to $\mathbb{R}$.
Determine the coefficients $\lambda_0, \lambda_1, \lambda_2$ so that the formula $I_2(f) = \lambda_0 f(0) + \lambda_1 f(1/2) + \lambda_2 f(1)$ is exact on $\mathbb{R}_2[X]$. Is this quadrature formula of order 2?

Let $n \in \mathbb{N}$. We consider $n+1$ distinct points in $I$, denoted $x_0 < x_1 < \cdots < x_n$, and the Lagrange basis $(L_0, \ldots, L_n)$ associated with these points.
Suppose that, for all $k \in \mathbb{N}$, the map $x \mapsto x^k w(x)$ is integrable on $I$. Show that the quadrature formula $I_n(f) = \sum_{j=0}^n \lambda_j f(x_j)$ is exact on $\mathbb{R}_n[X]$ if and only if $$\forall j \in \llbracket 0, n \rrbracket, \quad \lambda_j = \int_I L_j(x) w(x)\,\mathrm{d}x.$$

We consider the case $I = [0,1]$ and $\forall x \in I, w(x) = 1$. Determine the Lagrange basis associated with the points $(0, 1/2, 1)$ and thus recover the coefficients of the quadrature formula $I_2(f)$ from question 3.

We assume that $I = [a,b]$ with $a < b$, $\forall x \in I, w(x) = 1$ (general weight $w$ in the formula for $e(f)$), and that $f$ is of class $\mathcal{C}^{m+1}$ on $I$, where $m$ is the order of the quadrature formula $I_n(f) = \sum_{j=0}^n \lambda_j f(x_j)$.
For every natural number $m$, consider the function $\varphi_m : \mathbb{R}^2 \rightarrow \mathbb{R}$ defined by $$\forall (x,t) \in \mathbb{R}^2, \quad \varphi_m(x,t) = \begin{cases} (x-t)^m & \text{if } x \geqslant t, \\ 0 & \text{if } x < t. \end{cases}$$
Using the Taylor formula with integral remainder, show that $e(f) = e(R_m)$, where $R_m$ is defined by $$\forall x \in [a,b], \quad R_m(x) = \frac{1}{m!} \int_a^b \varphi_m(x,t) f^{(m+1)}(t)\,\mathrm{d}t.$$

We assume that $I = [a,b]$ with $a < b$, and that $f$ is of class $\mathcal{C}^{m+1}$ on $I$, where $m \geqslant 1$ is the order of the quadrature formula $I_n(f) = \sum_{j=0}^n \lambda_j f(x_j)$.
For every natural number $m$, consider the function $\varphi_m : \mathbb{R}^2 \rightarrow \mathbb{R}$ defined by $$\forall (x,t) \in \mathbb{R}^2, \quad \varphi_m(x,t) = \begin{cases} (x-t)^m & \text{if } x \geqslant t, \\ 0 & \text{if } x < t. \end{cases}$$
Deduce that, if $m \geqslant 1$, $$e(f) = \frac{1}{m!} \int_a^b K_m(t) f^{(m+1)}(t)\,\mathrm{d}t$$ where the function $K_m : [a,b] \rightarrow \mathbb{R}$ is defined by $$\forall t \in [a,b], \quad K_m(t) = e\left(x \mapsto \varphi_m(x,t)\right) = \int_a^b \varphi_m(x,t) w(x)\,\mathrm{d}x - \sum_{j=0}^n \lambda_j \varphi_m(x_j, t).$$ You may use the following admitted result: for every continuous function $g : [a,b]^2 \rightarrow \mathbb{R}$, we have $$\int_a^b \left(\int_a^b g(x,t)\,\mathrm{d}t\right)\mathrm{d}x = \int_a^b \left(\int_a^b g(x,t)\,\mathrm{d}x\right)\mathrm{d}t.$$

We assume that $I = [0,1]$, $\forall x \in I, w(x) = 1$, and we consider the quadrature formula $$I_1(g) = \frac{g(0) + g(1)}{2},$$ which is of order $m = 1$.
Calculate the associated Peano kernel $t \mapsto K_1(t)$ and show that, for every function $g$ of class $\mathcal{C}^2$ from $[0,1]$ to $\mathbb{R}$, we have the following bound on the associated quadrature error: $$|e(g)| \leqslant \frac{1}{12} \sup_{x \in [0,1]} |g''(x)|.$$

We consider the case of an arbitrary segment $I = [a,b]$ (with $a < b$), subdivided into $n+1$ equidistant points $a_0, \ldots, a_n$: $$\forall i \in \llbracket 0, n \rrbracket, \quad a_i = a + ih,$$ where $h = \frac{b-a}{n}$ is the step of the subdivision. The trapezoidal rule is $$T_n(f) = \frac{b-a}{n} \sum_{i=0}^{n-1} \frac{f(a_i) + f(a_{i+1})}{2}.$$
Represent graphically $T_n(f)$.

We consider the trapezoidal rule on $I = [a,b]$: $$T_n(f) = \frac{b-a}{n} \sum_{i=0}^{n-1} \frac{f(a_i) + f(a_{i+1})}{2},$$ where $a_i = a + ih$ and $h = \frac{b-a}{n}$, with associated error $e_n(f) = \int_a^b f(x)\,\mathrm{d}x - T_n(f)$.
Suppose that $f$ is a function of class $\mathcal{C}^2$ from $[a,b]$ to $\mathbb{R}$. Show that $$e_n(f) = \frac{b-a}{n} \sum_{i=0}^{n-1} e(g_i)$$ where $e$ is the error associated with the quadrature formula $I_1$ studied in question 11 and the $g_i : [0,1] \rightarrow \mathbb{R}$ are functions to be specified.

We consider the trapezoidal rule on $I = [a,b]$ with associated error $e_n(f) = \int_a^b f(x)\,\mathrm{d}x - T_n(f)$, where $f$ is of class $\mathcal{C}^2$.
Deduce the error bound $$\left|e_n(f)\right| \leqslant \frac{(b-a)^3}{12n^2} \sup_{x \in [a,b]} |f''(x)|.$$

Consider a quadrature formula $I_n(f) = \sum_{j=0}^n \lambda_j f(x_j)$ where $n \in \mathbb{N}$, $\lambda_0, \ldots, \lambda_n \in \mathbb{R}$ and $x_0 < x_1 < \cdots < x_n$ are $n+1$ distinct points in $I$. We assume that the coefficients $(\lambda_j)_{0 \leqslant j \leqslant n}$ are chosen as $$\forall j \in \llbracket 0, n \rrbracket, \quad \lambda_j = \int_I L_j(x) w(x)\,\mathrm{d}x,$$ where $(L_0, \ldots, L_n)$ is the Lagrange basis associated with the points $(x_0, \ldots, x_n)$. Thus, the formula $I_n(f)$ is of order $m \geqslant n$.
By reasoning with the polynomial $\prod_{i=0}^n (X - x_i)$, show that $m \leqslant 2n+1$.

Consider a quadrature formula $I_n(f) = \sum_{j=0}^n \lambda_j f(x_j)$ where the coefficients $(\lambda_j)_{0 \leqslant j \leqslant n}$ are chosen as $$\forall j \in \llbracket 0, n \rrbracket, \quad \lambda_j = \int_I L_j(x) w(x)\,\mathrm{d}x,$$ where $(L_0, \ldots, L_n)$ is the Lagrange basis associated with the points $(x_0, \ldots, x_n)$. Let $(p_n)_{n \in \mathbb{N}}$ be the sequence of orthogonal polynomials associated with the weight $w$.
Show that $m = 2n+1$ if and only if the $x_i$ are the roots of $p_{n+1}$.

We consider the case where $I = [-1,1]$ and $w(x) = 1$. Let $(p_n)_{n \in \mathbb{N}}$ be the sequence of orthogonal polynomials associated with the weight $w$.
Deduce explicitly a quadrature formula of order 5 (you will determine the points $x_j$ and the coefficients $\lambda_j$).

124- What is the value of $\displaystyle\int_1^4 \sqrt{\left(\frac{1}{2}x^2 - \frac{1}{x}\right)^2 + 1}\,dx$?
$4\ (1$ $5\ (2$ $6\ (3$ $7\ (4$

Let $f : ( 0,2 ) \rightarrow R$ be defined as $f ( x ) = \log _ { 2 } \left( 1 + \tan \left( \frac { \pi x } { 4 } \right) \right)$. Then, $\lim _ { n \rightarrow \infty } \frac { 2 } { n } \left( f \left( \frac { 1 } { n } \right) + f \left( \frac { 2 } { n } \right) + \ldots + f ( 1 ) \right)$ is equal to $\_\_\_\_$.

$\lim_{n \rightarrow \infty} \frac{3}{n}\left\{4 + \left(2 + \frac{1}{n}\right)^{2} + \left(2 + \frac{2}{n}\right)^{2} + \ldots + \left(3 - \frac{1}{n}\right)^{2}\right\}$ is equal to
(1) 12
(2) $\frac{19}{3}$
(3) 0
(4) 19

It is given that $f ( x ) = - 2 x ^ { 2 } + 10$
Consider the following three curves:
(1) $y = f ( x )$
(2) $y = f ( x + 1 )$
(3) the curve $y = f ( x + 1 )$ reflected in the line $y = 6$
The trapezium rule is used to estimate the area under each of these three curves between $x = 0$ and $x = 1$.
State whether the trapezium rule gives an overestimate or underestimate for each of these areas.

	(1)	(2)	(3)
A	underestimate	underestimate	underestimate
B	underestimate	underestimate	overestimate
C	underestimate	overestimate	underestimate
D	underestimate	overestimate	overestimate
E	overestimate	underestimate	underestimate
F	overestimate	underestimate	overestimate
G	overestimate	overestimate	underestimate
H	overestimate	overestimate	overestimate

The function f is such that $0 < f(x) < 1$ for $0 \leq x \leq 1$. The trapezium rule with $n$ equal intervals is used to estimate $\int_0^1 f(x) \, dx$ and produces an underestimate.
Using the same number of equal intervals, for which one of the following does the trapezium rule produce an overestimate?

A student approximates the integral $\int _ { a } ^ { b } \sin ^ { 2 } x \mathrm {~d} x$ using the trapezium rule with 4 strips. The resulting approximation is an overestimate.
Which of the following is/are necessarily true?
I If the student approximates $\int _ { - b } ^ { - a } \sin ^ { 2 } x \mathrm {~d} x$ in the same way, the result will be an overestimate.
II If the student approximates $\int _ { a } ^ { b } \cos ^ { 2 } x \mathrm {~d} x$ in the same way, the result will be an underestimate.

Which one of $\mathbf { A } - \mathbf { F }$ correctly completes the following statement? Given that $a < b$, and $\mathrm { f } ( x ) > 0$ for all $x$ with $a < x < b$, the trapezium rule produces an overestimate for $\int _ { a } ^ { b } \mathrm { f } ( x ) \mathrm { d } x \ldots$
A ... if $\mathrm { f } ^ { \prime } ( x ) > 0$ and $\mathrm { f } ^ { \prime \prime } ( x ) < 0$ for all $x$ with $a < x < b$
B ... only if $\mathrm { f } ^ { \prime } ( x ) > 0$ and $\mathrm { f } ^ { \prime \prime } ( x ) < 0$ for all $x$ with $a < x < b$
C ... if and only if $\mathrm { f } ^ { \prime } ( x ) > 0$ and $\mathrm { f } ^ { \prime \prime } ( x ) < 0$ for all $x$ with $a < x < b$
D ... if $\mathrm { f } ^ { \prime } ( x ) < 0$ and $\mathrm { f } ^ { \prime \prime } ( x ) > 0$ for all $x$ with $a < x < b$
E $\ldots$ only if $\mathrm { f } ^ { \prime } ( x ) < 0$ and $\mathrm { f } ^ { \prime \prime } ( x ) > 0$ for all $x$ with $a < x < b$ F ... if and only if $\mathrm { f } ^ { \prime } ( x ) < 0$ and $\mathrm { f } ^ { \prime \prime } ( x ) > 0$ for all $x$ with $a < x < b$

Use the trapezium rule with 3 strips to estimate
$$\int _ { \frac { 1 } { 2 } } ^ { 2 } 2 \log _ { 10 } x \mathrm {~d} x$$
A $\log _ { 10 } \frac { \sqrt { 6 } } { 2 }$
B $\log _ { 10 } \frac { 3 } { 2 }$
C $\log _ { 10 } \frac { 9 } { 4 }$
D $\log _ { 10 } 3$
E $\quad \log _ { 10 } \frac { 81 } { 16 }$
F $\log _ { 10 } \frac { \sqrt { 23 } } { 2 }$

The trapezium rule with 4 strips is used to estimate the integral:
$$\int _ { - 2 } ^ { 2 } \sqrt { 4 - x ^ { 2 } } d x$$
What is the positive difference between the estimate and the exact value of the integral?

$$\begin{aligned} & f : [ 1,3 ] \rightarrow [ 2,10 ] \\ & f ( x ) = 1 + x ^ { 2 } \end{aligned}$$
The interval $[ 1,3 ]$ is divided into two subintervals of equal length, and the right endpoints of these subintervals are marked as $x _ { 1 }$ and $x _ { 2 }$. Then, two rectangles are drawn with each subinterval as the base and heights $f \left( x _ { 1 } \right), f \left( x _ { 2 } \right)$ respectively.
If the sum of the areas of these rectangles is A and the area of the region between the function f and the x-axis is B, what is the difference A - B in square units?
A) $\frac { 11 } { 2 }$
B) $\frac { 13 } { 3 }$
C) $\frac { 15 } { 4 }$
D) $\frac { 19 } { 6 }$
E) $\frac { 23 } { 6 }$