We define on $[ 0,1 ]$ the function $P _ { n }$ by:
$$\forall x \in [ 0,1 ] , \quad P _ { n } ( x ) = \frac { 1 } { n ! } \frac { \mathrm { d } ^ { n } \left( x ^ { n } ( 1 - x ) ^ { n } \right) } { \mathrm { d } x ^ { n } } .$$
Let $n \in \mathbb { N } ^ { * }$. Show that for all $y \in ] 0,1 [$,
$$\int _ { 0 } ^ { 1 } \frac { P _ { n } ( x ) } { 1 - x y } \mathrm {~d} x = ( - y ) ^ { n } \int _ { 0 } ^ { 1 } \frac { x ^ { n } ( 1 - x ) ^ { n } } { ( 1 - x y ) ^ { n + 1 } } \mathrm {~d} x$$

We define on $[ 0,1 ]$ the function $P _ { n }$ by:
$$\forall x \in [ 0,1 ] , \quad P _ { n } ( x ) = \frac { 1 } { n ! } \frac { \mathrm { d } ^ { n } \left( x ^ { n } ( 1 - x ) ^ { n } \right) } { \mathrm { d } x ^ { n } } .$$
We set $I _ { n } = \int _ { 0 } ^ { 1 } \int _ { 0 } ^ { 1 } \frac { ( 1 - y ) ^ { n } P _ { n } ( x ) } { 1 - x y } \mathrm {~d} x \mathrm {~d} y$.
Deduce that
$$I _ { n } = ( - 1 ) ^ { n } \int _ { 0 } ^ { 1 } \int _ { 0 } ^ { 1 } \frac { x ^ { n } ( 1 - x ) ^ { n } y ^ { n } ( 1 - y ) ^ { n } } { ( 1 - x y ) ^ { n + 1 } } \mathrm {~d} x \mathrm {~d} y$$

The value of $$\int_{0}^{1} 4x^3 \left\{\frac{d^2}{dx^2}\left(1 - x^2\right)^5\right\} dx$$ is

If $\displaystyle\int f(x)\,dx = \psi(x)$, then $\displaystyle\int x^{5}f(x^{3})\,dx$ is equal to
(1) $\frac{1}{3}x^{3}\psi(x^{3}) - 3\displaystyle\int x^{3}\psi(x^{3})\,dx + C$
(2) $\frac{1}{3}\left[x^{3}\psi(x^{3}) - \displaystyle\int x^{2}\psi(x^{3})\,dx\right] + C$
(3) $\frac{1}{3}x^{3}\psi(x^{3}) - \displaystyle\int x^{2}\psi(x^{3})\,dx + C$
(4) $\frac{1}{3}\left[x^{3}\psi(x^{3}) - \displaystyle\int x^{3}\psi(x^{3})\,dx\right] + C$

If $\int f(x)\, dx = \psi(x)$, then $\int x^5 f\left(x^3\right) dx$, is equal to
(1) $\frac{1}{3}x^3 \psi\left(x^3\right) - \int x^2 \psi\left(x^3\right) dx + c$
(2) $\frac{1}{3}\left[x^3 \psi\left(x^3\right) - \int x^3 \psi\left(x^3\right) dx\right] + c$
(3) $\frac{1}{3}\left[x^3 \psi\left(x^3\right) - \int x^2 \psi\left(x^3\right) dx\right] + c$
(4) $\frac{1}{3}x^3 \psi\left(x^3\right) - 3\int x^3 \psi\left(x^3\right) dx + c$

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

For $\alpha , \beta , \gamma , \delta \in \mathbb { N }$, if $\int \left( \frac { x^2 e^x + e^{2x} } { x } \log _ { e } x \right) dx = \frac { 1 } { \alpha } \frac { x^{\beta} e^x } { 1 } - \frac { 1 } { \gamma } \frac { e ^ { \delta x } } { x } + C$, where $e = \sum _ { n = 0 } ^ { \infty } \frac { 1 } { n ! }$ and $C$ is constant of integration, then $\alpha + 2\beta + 3\gamma - 4\delta$ is equal to
(1) 1
(2) 4
(3) $-4$
(4) $-8$

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 $\_\_\_\_$

Let $\int x ^ { 3 } \sin x \mathrm {~d} x = g ( x ) + C$, where $C$ is the constant of integration. If $8 \left( g \left( \frac { \pi } { 2 } \right) + g ^ { \prime } \left( \frac { \pi } { 2 } \right) \right) = \alpha \pi ^ { 3 } + \beta \pi ^ { 2 } + \gamma , \alpha , \beta , \gamma \in Z$, then $\alpha + \beta - \gamma$ equals :
(1) 48
(2) 55
(3) 62
(4) 47

Let $\mathrm { f } : \mathbf { R } \rightarrow \mathbf { R }$ be a twice differentiable function such that $f ( 2 ) = 1$. If $\mathrm { F } ( x ) = x f ( x )$ for all $x \in \mathbf { R }$, $\int _ { 0 } ^ { 2 } x \mathrm {~F} ^ { \prime } ( x ) \mathrm { d } x = 6$ and $\int _ { 0 } ^ { 2 } x ^ { 2 } \mathrm {~F} ^ { \prime \prime } ( x ) \mathrm { d } x = 40$, then $\mathrm { F } ^ { \prime } ( 2 ) + \int _ { 0 } ^ { 2 } \mathrm {~F} ( x ) \mathrm { d } x$ is equal to :
(1) 11
(2) 13
(3) 15
(4) 9

Let $a$ be a positive real number. Let P denote the point of intersection of the following two curves
$$\begin{aligned} & C _ { 1 } : y = \frac { 3 } { x } \\ & C _ { 2 } : y = \frac { a } { x ^ { 2 } } , \end{aligned}$$
and let $\ell$ denote the tangent to $C _ { 2 }$ at P. Then we are to find the area $S$ of the region bounded by $C _ { 1 }$ and $\ell$.
Since the coordinates of P are $\left( \frac { a } { \mathbf { N } } , \frac { \mathbf { O } } { a } \right)$, the equation of $\ell$ is
$$y = - \frac { \mathbf { P Q } } { a ^ { 2 } } x + \frac { \mathbf { R S } } { a }$$
When we set
$$p = \frac { a } { \mathbf { T } } , \quad q = \frac { a } { \mathbf { U } } \quad ( p < q )$$
$S$ is obtained by calculating
$$S = [ \mathbf { V } ] _ { p } ^ { q }$$
where $\mathbf{V}$ is the appropriate expression from among (0) $\sim$ (5) below.
(0) $\frac { 18 } { a ^ { 2 } } x ^ { 2 } - \frac { 27 } { a } x + 3 \log | x |$
(1) $\frac { 9 } { a ^ { 2 } } x ^ { 2 } - \frac { 9 } { a } x + 3 \log | x |$
(2) $- \frac { 27 } { a ^ { 2 } } x ^ { 2 } + \frac { 18 } { a } x - 3 \log | x |$
(3) $- \frac { 27 } { a ^ { 2 } } x ^ { 2 } + \frac { 27 } { a } x - 3 \log | x |$
(4) $\frac { 27 } { a ^ { 2 } } x ^ { 2 } - \frac { 27 } { a } x + 3 \log | x |$
(5) $- \frac { 18 } { a ^ { 2 } } x ^ { 2 } + \frac { 27 } { a } x - 3 \log | x |$
Hence we obtain
$$S = \frac { \mathbf { W } } { \mathbf { X } } - 3 \log \mathbf { Y }$$

For a positive integer $n$ and a real number $a$, consider the function
$$f_n(a) = \int_0^{\pi} (\cos x + a\sin 2nx)^2\, dx$$
(1) When we transform $f_n(a)$ into
$$f_n(a) = \int_0^{\pi} \left\{\frac{1 + \cos \mathbf{L}\, x}{2} + a^2 \frac{1 - \cos \mathbf{M}\, nx}{2} + a(\sin(2n+1)x + \sin(2n-1)x)\right\} dx$$
and calculate the definite integral on the right side, we obtain
$$f_n(a) = \frac{\pi}{\mathbf{N}}\, a^2 + \frac{\mathbf{O}\, n}{\mathbf{P}\, n^2 - \mathbf{Q}}\, a + \frac{\pi}{\mathbf{R}}.$$
(2) Let $a_n$ denote the value of $a$ at which $f_n(a)$ is minimalized, and set $S_N = \sum_{n=1}^{N} \frac{a_n}{n}$.
Then
$$\begin{aligned} S_N &= -\frac{\mathbf{S}}{\pi} \sum_{n=1}^{N} \left(\frac{1}{2n - \mathbf{T}} - \frac{1}{2n + \mathbf{U}}\right) \\ &= -\frac{\mathbf{S}}{\pi} \left(\mathbf{V} - \frac{1}{\mathbf{W}N + \mathbf{U}}\right) \end{aligned}$$
Hence we obtain
$$\sum_{n=1}^{\infty} \frac{a_n}{n} = \lim_{N \to \infty} S_N = -\frac{\mathbf{Y}}{\pi}.$$

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
$$\lim_{n \to \infty} \sum_{k=1}^{n} ka_k = \frac{\mathbf{X}}{\mathbf{Y}} e^{\frac{\mathrm{JK}}{\mathrm{L}}} - 1.$$

Let $a > 0$. Consider the region of a plane bounded by the curve $y = \sqrt { x } e ^ { - x }$, the $x$-axis, and the straight line $x = a$ which passes through the point $\mathrm{ A }( a , 0 )$, and let $V$ be the volume of the solid obtained by rotating this region once about the $x$-axis.
(1) $V$ is expressed as a function in $a$ by
$$V = - \frac { \pi } { 4 } \left\{ ( \mathbf { N } a + \mathbf { O } ) e ^ { - \mathbf { P } a } - \mathbf { Q R } \right\} .$$
(2) Suppose that the point A starts at the origin and moves along the $x$-axis in the positive direction and that its speed at $t$ seconds is $4t$. Then the rate of change of $V$ at $t$ seconds is
$$\frac { d V } { d t } = \mathbf { R } \pi t ^ { \mathbf { S } } e ^ { - \mathbf { T } t ^ { \mathbf { U } } } .$$
This rate of change is maximized at
$$t = \frac { \sqrt { \mathbf { V } } } { 4 } ,$$
and the value of $V$ at this time is
$$V = - \frac { \pi } { 8 } \left( \mathbf { W } e ^ { - \frac { \mathbf { X } } { \mathbf { Y } } } - \mathbf { Z } \right) .$$

Consider the two functions
$$y = x \log a x , \tag{1}$$ $$y = 2 x - 3 , \tag{2}$$
where $a > 0$, and where $\log$ is the natural logarithm.
(1) Let us find $a$ such that the graph of (1) is tangent to the graph of (2).
The equation of the tangent to the graph of (1) at the point $( t , t \log a t )$ is $\mathbf { A }$ (for A, choose the correct answer from among choices (0) $\sim$ (3) below). (0) $y = ( \log a t + 1 ) x - t$
(1) $y = ( \log a t + a ) x - t$
(2) $y = ( a \log t + 1 ) x + t$
(3) $y = ( a \log t + a ) x + t$
Hence, the graph of (1) is tangent to the graph of (2) when $a = \frac { e } { \square \mathbf{B} }$, and the coordinates of the tangent point are ( $\mathbf { C }$, $\mathbf { D }$ ).
(2) When $a = \frac { e } { \mathbf{B} }$, function (1) is minimized at $x = \square e ^ { - \mathbf { F } }$, and in this case the minimum value is $- \mathbf { G } \cdot e ^ { - \mathbf { H } }$.
(3) When $a = \frac { e } { \mathbf{B} }$, let us find the area $S$ of the region bounded by the graphs of (1) and (2) and the $x$-axis.
For the following indefinite integral, we have
$$\int x \log a x \, d x = \square + C , \quad \text { where } C \text { is an integral constant }$$
(for I, choose the correct answer from among (0) $\sim$ (3) below). (0) $\frac { 1 } { 2 } x ^ { 2 } \log a x - \frac { 1 } { 2 } x ^ { 2 }$
(1) $2 x ^ { 2 } \log a x - 2 x ^ { 2 }$
(2) $\frac { 1 } { 2 } x ^ { 2 } \log a x - \frac { 1 } { 4 } x ^ { 2 }$
(3) $2 x ^ { 2 } \log a x - 4 x ^ { 2 }$
Hence we obtain
$$S = \frac { \mathbf { J } } { \mathbf { K } }$$

Consider the following integral $I _ { n } ( \alpha )$ for $\alpha \geq 1$ and $n > 0$.
$$I _ { n } ( \alpha ) = \int _ { \frac { 1 } { n } } ^ { n } \frac { f ( \alpha x ) - f ( x ) } { x } \mathrm {~d} x$$
Assume that a real-valued function $f ( x )$ is continuous and differentiable on $x \geq 0$, its derivative is continuous, and $\lim _ { x \rightarrow \infty } f ( x ) = 0$. Answer the following questions.
(1) Define $J _ { n } ( \alpha ) = \frac { \mathrm { d } I _ { n } ( \alpha ) } { \mathrm { d } \alpha }$. Show that $J _ { n } ( \alpha ) = \frac { 1 } { \alpha } \left( f ( \alpha n ) - f \left( \frac { \alpha } { n } \right) \right)$.
You can use the fact that the integration and the differentiation commute in this context.
(2) Define $I ( \alpha ) = \lim _ { n \rightarrow \infty } I _ { n } ( \alpha )$. Show that $\lim _ { n \rightarrow \infty } J _ { n } ( \beta )$ exists for any $\beta \in [ 1 , \alpha ]$ and it uniformly converges on $[ 1 , \alpha ]$, and show that
$$I ( \alpha ) = \int _ { 1 } ^ { \alpha } \left( \lim _ { n \rightarrow \infty } J _ { n } ( \beta ) \right) \mathrm { d } \beta$$
(3) Obtain $I ( \alpha )$.
(4) Calculate the following integral. Note that $p > q > 0$.
$$\int _ { 0 } ^ { \infty } \frac { e ^ { - p x } \cos ( p x ) - e ^ { - q x } \cos ( q x ) } { x } \mathrm {~d} x$$

For the function $f$ whose graph is given above, $$\int_{1}^{3} \frac{x \cdot f'(x) - f(x)}{x^{2}}\, dx$$ What is the value of the integral?
A) $\frac{7}{2}$
B) $\frac{3}{2}$
C) $\frac{2}{3}$
D) $\frac{1}{3}$
E) $\frac{5}{4}$

$$\int _ { 1 } ^ { e } \ln ^ { 3 } x \, d x = 6 - 2 e$$
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) $11 e - 28$

f is a differentiable function on the set of real numbers and
$$\begin{aligned} & \int _ { 0 } ^ { 3 } f ( x ) d x = 2 \\ & \int _ { 0 } ^ { 3 } x f ^ { \prime } ( x ) d x = 1 \end{aligned}$$
Given this, what is the value of $\mathbf { f } \boldsymbol { ( } \mathbf { 3 } \boldsymbol { ) }$?
A) 0
B) 1
C) 2
D) 3
E) 4

For a function f defined on the set of real numbers and twice differentiable,
$$\begin{aligned} & f ( 1 ) = f ( 2 ) = 2 \\ & f ^ { \prime } ( 1 ) = f ^ { \prime } ( 2 ) = - 1 \end{aligned}$$
the following equalities are given.
Accordingly, what is the value of the integral $\int _ { 1 } ^ { 2 } x \cdot f ^ { \prime \prime } ( x ) d x$?
A) $- 1$
B) $- 2$
C) $- 3$
D) $\frac { - 1 } { 2 }$
E) $\frac { - 2 } { 3 }$

$\int _ { 1/2 } ^ { e } x \ln ( 2 x ) \, d x$\ What is the value of the integral?\ A) $\frac { e ^ { 2 } } { 2 }$\ B) $\frac { e ^ { 2 } - 1 } { 4 }$\ C) $\frac { e ^ { 2 } + 1 } { 16 }$\ D) 1\ E) 2

In a mathematics class where the topic of integration by parts is being taught, Teacher Ebru writes on the board
$$\int u d v = u v - \int v d u$$
rule. Then, in solving a problem, Mehmet chooses functions $f ( x )$ and $g ( x )$ in place of u and v respectively and applies this rule to obtain
$$\int f ( x ) g ^ { \prime } ( x ) d x = \frac { f ( x ) } { x } - \int \frac { 2 } { x ^ { 2 } } d x$$
equality.
Given that $\mathbf { f } ( \mathbf { 1 } ) = \mathbf { 2 }$, what is the value of $\mathbf { f } ( \mathbf { e } )$?
A) 2 B) 4 C) 6 D) 8 E) 10

The functions $f$ and $g$ are defined and differentiable on the set of real numbers and satisfy
$$\begin{aligned} & \int_{1}^{2} f^{\prime}(3x)\, dx = 4 \\ & \int f(2x)\, dx = g(x) + C, \quad (C \text{ constant}) \end{aligned}$$
If $f(3) = 5$, what is the value of the derivative $g^{\prime}(3)$?
A) 1 B) 5 C) 9 D) 13 E) 17