$\int \frac { x } { \sqrt { 1 - 9 x ^ { 2 } } } d x =$
(A) $- \frac { 1 } { 9 } \sqrt { 1 - 9 x ^ { 2 } } + C$
(B) $- \frac { 1 } { 18 } \ln \sqrt { 1 - 9 x ^ { 2 } } + C$
(C) $\frac { 1 } { 3 } \arcsin ( 3 x ) + C$
(D) $\frac { x } { 3 } \arcsin ( 3 x ) + C$

The function $f$ is defined by $f ( x ) = \sqrt { 25 - x ^ { 2 } }$ for $- 5 \leq x \leq 5$.
(a) Find $f ^ { \prime } ( x )$.
(b) Write an equation for the line tangent to the graph of $f$ at $x = - 3$.
(c) Let $g$ be the function defined by $g ( x ) = \left\{ \begin{array} { l } f ( x ) \text { for } - 5 \leq x \leq - 3 \\ x + 7 \text { for } - 3 < x \leq 5 . \end{array} \right.$
Is $g$ continuous at $x = - 3$ ? Use the definition of continuity to explain your answer.
(d) Find the value of $\int _ { 0 } ^ { 5 } x \sqrt { 25 - x ^ { 2 } } d x$.

Using the substitution $u = \sqrt { x }$, $\int _ { 1 } ^ { 4 } \frac { e ^ { \sqrt { x } } } { \sqrt { x } } d x$ is equal to which of the following?
(A) $2 \int _ { 1 } ^ { 16 } e ^ { u } d u$
(B) $2 \int _ { 1 } ^ { 4 } e ^ { u } d u$
(C) $2 \int _ { 1 } ^ { 2 } e ^ { u } d u$
(D) $\frac { 1 } { 2 } \int _ { 1 } ^ { 2 } e ^ { u } d u$
(E) $\int _ { 1 } ^ { 4 } e ^ { u } d u$

Let $f$ be a function such that $\int _ { 6 } ^ { 12 } f ( 2 x ) d x = 10$. Which of the following must be true?
(A) $\int _ { 12 } ^ { 24 } f ( t ) d t = 5$
(B) $\int _ { 12 } ^ { 24 } f ( t ) d t = 20$
(C) $\int _ { 6 } ^ { 12 } f ( t ) d t = 5$
(D) $\int _ { 6 } ^ { 12 } f ( t ) d t = 20$
(E) $\int _ { 3 } ^ { 6 } f ( t ) d t = 5$

Using the substitution $u = x ^ { 2 } - 3 , \int _ { - 1 } ^ { 4 } x \left( x ^ { 2 } - 3 \right) ^ { 5 } d x$ is equal to which of the following?
(A) $2 \int _ { - 2 } ^ { 13 } u ^ { 5 } d u$
(B) $\int _ { - 2 } ^ { 13 } u ^ { 5 } d u$
(C) $\frac { 1 } { 2 } \int _ { - 2 } ^ { 13 } u ^ { 5 } d u$
(D) $\int _ { - 1 } ^ { 4 } u ^ { 5 } d u$
(E) $\frac { 1 } { 2 } \int _ { - 1 } ^ { 4 } u ^ { 5 } d u$

Compute the following integral $$\int_{0}^{\frac{\pi}{2}} \frac{\mathrm{~d}x}{(\sqrt{\sin x} + \sqrt{\cos x})^{4}}.$$

There is a function $f ( x )$ that is differentiable on the set of all real numbers. For all real numbers $x$, $f ( 2 x ) = 2 f ( x ) f ^ { \prime } ( x )$, and $$f ( a ) = 0 , \quad \int _ { 2 a } ^ { 4 a } \frac { f ( x ) } { x } d x = k \quad ( a > 0,0 < k < 1 )$$ When this holds, what is the value of $\int _ { a } ^ { 2 a } \frac { \{ f ( x ) \} ^ { 2 } } { x ^ { 2 } } d x$ expressed in terms of $k$? [3 points]
(1) $\frac { k ^ { 2 } } { 4 }$
(2) $\frac { k ^ { 2 } } { 2 }$
(3) $k ^ { 2 }$
(4) $k$
(5) $2 k$

The function $f ( x ) = 3 x ^ { 2 } - a x$ satisfies
$$\lim _ { n \rightarrow \infty } \frac { 1 } { n } \sum _ { k = 1 } ^ { n } f \left( \frac { 3 k } { n } \right) = f ( 1 )$$
Find the value of the constant $a$. [4 points]

What is the value of $\int _ { 0 } ^ { e } \frac { 5 } { x + e } d x$? [3 points]
(1) $\ln 2$
(2) $2 \ln 2$
(3) $3 \ln 2$
(4) $4 \ln 2$
(5) $5 \ln 2$

A continuous function $f ( x )$ defined on $x > 0$ satisfies $$2 f ( x ) + \frac { 1 } { x ^ { 2 } } f \left( \frac { 1 } { x } \right) = \frac { 1 } { x } + \frac { 1 } { x ^ { 2 } }$$ for all positive $x$. What is the value of $\int _ { \frac { 1 } { 2 } } ^ { 2 } f ( x ) d x$? [4 points]
(1) $\frac { \ln 2 } { 3 } + \frac { 1 } { 2 }$
(2) $\frac { 2 \ln 2 } { 3 } + \frac { 1 } { 2 }$
(3) $\frac { \ln 2 } { 3 } + 1$
(4) $\frac { 2 \ln 2 } { 3 } + 1$
(5) $\frac { 2 \ln 2 } { 3 } + \frac { 3 } { 2 }$

What is the value of $\int _ { e } ^ { e ^ { 2 } } \frac { \ln x - 1 } { x ^ { 2 } } d x$? [3 points]
(1) $\frac { e - 2 } { e ^ { 2 } }$
(2) $\frac { e - 1 } { e ^ { 2 } }$
(3) $\frac { 1 } { e }$
(4) $\frac { e + 1 } { e ^ { 2 } }$
(5) $\frac { e + 2 } { e ^ { 2 } }$

Two functions $f(x)$ and $g(x)$ are defined and differentiable on the set of all positive real numbers. $g(x)$ is the inverse function of $f(x)$, and $g'(x)$ is continuous on the set of all positive real numbers. For all positive numbers $a$, $$\int_1^a \frac{1}{g'(f(x))f(x)}\,dx = 2\ln a + \ln(a+1) - \ln 2$$ and $f(1) = 8$. Find the value of $f(2)$. [3 points]
(1) 36
(2) 40
(3) 44
(4) 48
(5) 52

Throughout this part, $\lambda$ is a real number belonging to the interval $]0,1[$ and $f, g, h$ are functions in $C^{0}(\mathbb{R}, \mathbb{R}_{+})$ that are integrable and satisfy the following inequality $$\forall x \in \mathbb{R}, \forall y \in \mathbb{R}, \quad h(\lambda x + (1-\lambda) y) \geq f(x)^{\lambda} g(y)^{1-\lambda}.$$ In questions 3), 4) and 5) we additionally assume that $f$ and $g$ are strictly positive, that is, for all real $x$, $f(x) > 0$ and $g(x) > 0$.
Show that the image set of the application $w$ defined on $]0,1[$ by $$\forall t \in ]0,1[, \quad w(t) = \lambda u(t) + (1-\lambda) v(t),$$ is equal to $\mathbb{R}$. Then prove that $w$ defines a change of variable from $]0,1[$ to $\mathbb{R}$. Using this and $\int_{-\infty}^{+\infty} h(w)\,dw$, show that $f$, $g$ and $h$ satisfy the "P-L" inequality $$\int_{-\infty}^{+\infty} h(x)\,dx \geq \left(\int_{-\infty}^{+\infty} f(x)\,dx\right)^{\lambda} \left(\int_{-\infty}^{+\infty} g(x)\,dx\right)^{1-\lambda}.$$

Let $f \in \mathcal { C } ^ { 0 } ( [ a , b ] )$ such that $f ( a ) \neq 0$ and $\varphi \in \mathcal { C } ^ { 1 } ( [ a , b ] )$. For every parameter $t \in \mathbb { R }$, we denote $$F ( t ) = \int _ { a } ^ { b } e ^ { - t \varphi ( x ) } f ( x ) d x$$ Case where the phase $\varphi$ has no critical point in $[ a , b ]$. We assume that $\varphi ^ { \prime } ( x ) > 0$ for all $x \in [ a , b ]$.
(a) Show that $\Phi : x \mapsto \varphi ( x ) - \varphi ( a )$ is a bijection from $[ a , b ]$ onto an interval of the form $[ 0 , \beta ]$, and that it is of class $\mathcal { C } ^ { 1 }$.
(b) Show that $$F ( t ) \underset { t \rightarrow + \infty } { \sim } \frac { e ^ { - t \varphi ( a ) } f ( a ) } { \varphi ^ { \prime } ( a ) t }$$ Hint. One can reduce to the case treated in question 1a) using a change of variable.

Let $f \in \mathcal { C } ^ { 0 } ( [ a , b ] )$ such that $f ( a ) \neq 0$ and $\varphi \in \mathcal { C } ^ { 1 } ( [ a , b ] )$. For every parameter $t \in \mathbb { R }$, we denote $$F ( t ) = \int _ { a } ^ { b } e ^ { - t \varphi ( x ) } f ( x ) d x$$ Case where the phase $\varphi$ has a critical point at $a$. We now assume that $\varphi \in \mathcal { C } ^ { 2 } ( [ a , b ] )$, $\varphi ^ { \prime } ( a ) = 0 , \varphi ^ { \prime \prime } ( a ) > 0$, and $\varphi ^ { \prime } ( x ) > 0$ for all $\left. \left. x \in \right] a , b \right]$.
(a) Show that the formula $\psi ( x ) = \sqrt { \varphi ( x ) - \varphi ( a ) }$ defines a function of class $\mathcal { C } ^ { 1 }$ on $[ a , b ]$. Calculate $\psi ^ { \prime } ( a )$.
(b) Show that $\psi$ is a bijection from $[ a , b ]$ onto an interval of the form $[ 0 , \beta ]$.
(c) Show that $$F ( t ) \underset { t \rightarrow + \infty } { \sim } \sqrt { \frac { \pi } { 2 \varphi ^ { \prime \prime } ( a ) } } \frac { e ^ { - t \varphi ( a ) } f ( a ) } { \sqrt { t } } .$$ Hint. One can reduce to the case treated in question 1b) using a change of variable.

Let $\varepsilon$ and $r$ be fixed such that $0 < \varepsilon < r$. With the change of variables $q = r\cos\theta$, establish that $$\int_\varepsilon^r \frac{\mathrm{d}q}{q^2\sqrt{r^2-q^2}} = \frac{\sqrt{r^2-\varepsilon^2}}{r^2\varepsilon}$$

For $( x , y )$ in $\left( \mathbb { R } ^ { + * } \right) ^ { 2 }$, we define $\beta ( x , y ) = \int _ { 0 } ^ { 1 } t ^ { x - 1 } ( 1 - t ) ^ { y - 1 } \mathrm {~d} t$. Throughout the rest of this question we assume $x > 1$ and $y > 1$.
Show that $\beta ( x , y ) = \int _ { 0 } ^ { + \infty } \frac { u ^ { x - 1 } } { ( 1 + u ) ^ { x + y } } \mathrm {~d} u$.
One may use the change of variable $t = \frac { u } { 1 + u }$.

We recall that $\Gamma ( y ) = e ^ { - y } y ^ { y } \int _ { - 1 } ^ { + \infty } e ^ { - y \phi ( s ) } d s$ where $\phi ( s ) = s - \ln(1+s)$, and that $\phi_-^{-1} : ]0,+\infty[ \to ]-1,0[$ and $\phi_+^{-1} : ]0,+\infty[ \to ]0,+\infty[$ are the inverse bijections of the restrictions of $\phi$.
Show that $$\Gamma ( y ) = e ^ { - y } y ^ { y } \int _ { 0 } ^ { \infty } e ^ { - y q } \left( \left( \phi _ { + } ^ { - 1 } \right) ^ { \prime } ( q ) - \left( \phi _ { - } ^ { - 1 } \right) ^ { \prime } ( q ) \right) d q$$

Using the change of variable $t = \frac { \ln x } { 2 \pi }$, demonstrate that $$\forall p \in \mathbb { N } \quad I _ { p } = \frac { e ^ { - p ^ { 2 } \pi ^ { 2 } } } { 2 \pi } \int _ { 0 } ^ { + \infty } x ^ { p - 1 } \exp \left( - \frac { \ln ^ { 2 } x } { 4 \pi ^ { 2 } } \right) \sin \left( \frac { \ln x } { 2 \pi } \right) \mathrm { d } x$$

Let $a < b$ be two real numbers and $f : [a,b] \rightarrow \mathbb{R}$ be an infinitely differentiable function. Let us call (H) the following hypothesis: there exists a unique point $x_0 \in [a,b]$ where $f$ attains its maximum, we have $a < x_0 < b$, and $f''(x_0) \neq 0$.
Under hypothesis (H), show that for all $\delta > 0$ such that $\delta < \min(x_0 - a, b - x_0)$, we have the asymptotic equivalence, as $t \rightarrow +\infty$, $$\int_a^b e^{tf(x)} \mathrm{d}x \sim \int_{x_0 - \delta}^{x_0 + \delta} e^{tf(x)} \mathrm{d}x.$$

Let $a < b$ be two real numbers and $f : [ a , b ] \rightarrow \mathbb { R }$ be an infinitely differentiable function. Let us call (H) the following hypothesis: there exists a unique point $x _ { 0 } \in [ a , b ]$ where $f$ attains its maximum, we have $a < x _ { 0 } < b$, and $f ^ { \prime \prime } \left( x _ { 0 } \right) \neq 0$.
Under hypothesis $( \mathrm { H } )$, show that for all $\delta > 0$ such that $\delta < \min \left( x _ { 0 } - a , b - x _ { 0 } \right)$, we have the asymptotic equivalence, as $t \rightarrow + \infty$, $$\int _ { a } ^ { b } e ^ { t f ( x ) } \mathrm { d } x \sim \int _ { x _ { 0 } - \delta } ^ { x _ { 0 } + \delta } e ^ { t f ( x ) } \mathrm { d } x$$

Let $a < b$ be two real numbers and $f : [a,b] \rightarrow \mathbb{R}$ be an infinitely differentiable function. Let us call (H) the following hypothesis: there exists a unique point $x_0 \in [a,b]$ where $f$ attains its maximum, we have $a < x_0 < b$, and $f''(x_0) \neq 0$.
We admit the identity $\int_{-\infty}^{+\infty} \exp(-x^2) \mathrm{d}x = \sqrt{\pi}$.
Under hypothesis (H), show the asymptotic equivalence, as $t \rightarrow +\infty$, $$\int_a^b e^{tf(x)} \mathrm{d}x \sim e^{tf(x_0)} \sqrt{\frac{2\pi}{t|f''(x_0)|}}$$

Let $a < b$ be two real numbers and $f : [ a , b ] \rightarrow \mathbb { R }$ be an infinitely differentiable function. Let us call (H) the following hypothesis: there exists a unique point $x _ { 0 } \in [ a , b ]$ where $f$ attains its maximum, we have $a < x _ { 0 } < b$, and $f ^ { \prime \prime } \left( x _ { 0 } \right) \neq 0$.
Under hypothesis (H), show the asymptotic equivalence, as $t \rightarrow + \infty$, $$\int _ { a } ^ { b } e ^ { t f ( x ) } \mathrm { d } x \sim e ^ { t f \left( x _ { 0 } \right) } \sqrt { \frac { 2 \pi } { t \left| f ^ { \prime \prime } \left( x _ { 0 } \right) \right| } }$$

If $a$ and $b$ are two real numbers, we denote $K _ { a , b }$ the function defined for all real $t$ by $K _ { a , b } ( t ) = \begin{cases} \frac { \mathrm { e } ^ { \mathrm { i } t b } - \mathrm { e } ^ { \mathrm { i } t a } } { 2 \mathrm { i } t } & \text { if } t \neq 0 , \\ \frac { b - a } { 2 } & \text { if } t = 0 . \end{cases}$ Show that $\int _ { - N } ^ { N } K _ { a , b } ( t ) \mathrm { d } t = \int _ { N a } ^ { N b } \operatorname { sinc } ( s ) \mathrm { d } s$.

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 the change of variable $x = 2\sin t$, calculate $m_{0}$.