Let for $x \in \mathbb{R}$, $f(x) = \frac{x + |x|}{2}$ and $g(x) = \begin{cases} x, & x < 0 \\ x^2, & x \geq 0 \end{cases}$. Then area bounded by the curve $y = f(g(x))$ and the lines $y = 0$, $2y - x = 15$ is equal to $\underline{\hspace{1cm}}$.

Let $y = p(x)$ be the parabola passing through the points $(-1, 0)$, $(0, 1)$ and $(1, 0)$. If the area of the region $\{(x, y) : (x+1)^2 + (y-1)^2 \leq 1,\; y \leq p(x)\}$ is $A$, then $12\pi - 4A$ is equal to $\_\_\_\_$.

If the area of the region $\left\{ ( x , y ) : \frac { \mathrm { a } } { x ^ { 2 } } \leq y \leq \frac { 1 } { x } , 1 \leq x \leq 2,0 < \mathrm { a } < 1 \right\}$ is $\left( \log _ { \mathrm { e } } 2 \right) - \frac { 1 } { 7 }$ then the value of $7 \mathrm { a } - 3$ is equal to:
(1) 0
(2) 2
(3) $- 1$
(4) 1

The area of the region in the first quadrant inside the circle $x ^ { 2 } + y ^ { 2 } = 8$ and outside the parabola $y ^ { 2 } = 2 x$ is equal to : (1) $\frac { \pi } { 2 } - \frac { 1 } { 3 }$ (2) $\pi - \frac { 1 } { 3 }$ (3) $\frac { \pi } { 2 } - \frac { 2 } { 3 }$ (4) $\pi - \frac { 2 } { 3 }$

The area (in square units) of the region enclosed by the ellipse $x ^ { 2 } + 3 y ^ { 2 } = 18$ in the first quadrant below the line $y = x$ is
(1) $\sqrt { 3 } \pi - \frac { 3 } { 4 }$
(2) $\sqrt { 3 } \pi + 1$
(3) $\sqrt { 3 } \pi$
(4) $\sqrt { 3 } \pi + \frac { 3 } { 4 }$

Let the area of the region $\left\{ ( x , y ) : x - 2 y + 4 \geq 0 , x + 2 y ^ { 2 } \geq 0 , x + 4 y ^ { 2 } \leq 8 , y \geq 0 \right\}$ be $\frac { m } { n }$, where $m$ and $n$ are coprime numbers. Then $\mathrm { m } + \mathrm { n }$ is equal to $\_\_\_\_$.

If the area of the region $( x , y ) : 0 \leq y \leq \min 2 x , 6 x - x ^ { 2 }$ is $A$, then $12 A$ is equal to $\_\_\_\_$ .

The area of the region enclosed by the parabola $(y-2)^2 = x - 1$, the line $x - 2y + 4 = 0$ and the positive coordinate axes is $\underline{\hspace{1cm}}$.

Let the area of the region enclosed by the curve $y = \min \{ \sin x , \cos x \}$ and the $x$ axis between $x = - \pi$ to $x = \pi$ be $A$. Then $A ^ { 2 }$ is equal to $\_\_\_\_$

Let $y = y ( x )$ be the solution of the differential equation $\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 2 x } { \left( 1 + x ^ { 2 } \right) ^ { 2 } } y = x \mathrm { e } ^ { \frac { 1 } { \left( 1 + x ^ { 2 } \right) } } ; y ( 0 ) = 0$. Then the area enclosed by the curve $f ( x ) = y ( x ) \mathrm { e } ^ { - \frac { 1 } { \left( 1 + x ^ { 2 } \right) } }$ and the line $y - x = 4$ is $\_\_\_\_$

The area of the region enclosed by the curves $y = \mathrm{e}^{x}$, $y = \left| \mathrm{e}^{x} - 1 \right|$ and $y$-axis is:
(1) $1 - \log_{e} 2$
(2) $\log_{e} 2$
(3) $1 + \log_{e} 2$
(4) $2 \log_{e} 2 - 1$

Let $f : \mathbf { R } \rightarrow \mathbf { R }$ be a twice differentiable function such that $f ( x + y ) = f ( x ) f ( y )$ for all $x , y \in \mathbf { R }$. If $f ^ { \prime } ( 0 ) = 4 \mathrm { a }$ and $f$ satisfies $f ^ { \prime \prime } ( x ) - 3 \mathrm { a } f ^ { \prime } ( x ) - f ( x ) = 0 , \mathrm { a } > 0$, then the area of the region $\mathrm { R } = \{ ( x , y ) \mid 0 \leq y \leq f ( \mathrm { a } x ) , 0 \leq x \leq 2 \}$ is:
(1) $e ^ { 2 } - 1$
(2) $\mathrm { e } ^ { 2 } + 1$
(3) $e ^ { 4 } + 1$
(4) $e ^ { 4 } - 1$

Let the function, $f ( x ) = \left\{ \begin{array} { l l } - 3 a x ^ { 2 } - 2 , & x < 1 \\ a ^ { 2 } + b x , & x \geqslant 1 \end{array} \right.$ be differentiable for all $x \in \mathbf { R }$, where $\mathbf { a } > 1 , \mathbf { b } \in \mathbf { R }$. If the area of the region enclosed by $y = f ( x )$ and the line $y = - 20$ is $\alpha + \beta \sqrt { 3 } , \alpha , \beta \in Z$, then the value of $\alpha + \beta$ is $\_\_\_\_$

If the area of the larger portion bounded between the curves $x ^ { 2 } + y ^ { 2 } = 25$ and $y = | x - 1 |$ is $\frac { 1 } { 4 } ( b \pi + c ) , b , c \in N$, then $b + c$ is equal to

Let $0 < a < 1$. Let $S ( a )$ denote the sum of the areas of two regions, one region bounded by the curve $y = x e ^ { 2 x }$, the $x$-axis, and the straight line $x = a - 1$, and the other region bounded by the curve $y = x e ^ { 2 x }$, the $x$-axis, and the straight line $x = a$. We are to find the value of $a$ at which $S ( a )$ is minimized.
The indefinite integral of $x e ^ { 2 x }$ is to be determined, where $C$ is the constant of integration.
The value of $x e ^ { 2 x }$ is $x e ^ { 2 x } < 0$ for $x < 0$ and $x e ^ { 2 x } \geqq 0$ for $x \geqq 0$. Hence we have
$$S ( a ) = \frac { \mathbf { L M } } { \mathbf { N } } \left\{ \mathbf { O } + \left( \mathbf { P } a - \mathbf { Q } \right) e ^ { 2 ( a - 1 ) } + ( \mathbf { R } a - 1 ) e ^ { 2 a } \right\} .$$
Further, since
$$S ^ { \prime } ( a ) = ( a - \mathbf { S } ) e ^ { 2 ( a - 1 ) } + a e ^ { 2 a } ,$$
the value of $a$ at which $S ( a )$ is minimized is $a = \dfrac { \square \mathbf { T } } { e ^ { 2 } + \mathbf { U } }$, which satisfies $0 < a < 1$.

Let $a$ and $t$ be positive real numbers. Let $\ell$ be the tangent to the graph $C$ of $y = a x ^ { 3 }$ at a point $\mathrm { P } \left( t , a t ^ { 3 } \right)$, and let Q be the point at which $\ell$ intersects the curve $C$ again. Further, let $p$ be the line passing through the point P parallel to the $x$-axis; let $q$ be the line passing through the point Q parallel to the $y$-axis; and let R be the point of intersection of $p$ and $q$.
Also, let us denote by $S _ { 1 }$ the area of the region bounded by the curve $C$, the straight line $p$ and the straight line $q$, and denote by $S _ { 2 }$ the area of the region bounded by the curve $C$ and the tangent $\ell$. We are to find the value of $\frac { S _ { 1 } } { S _ { 2 } }$.
First, since the equation of the tangent $\ell$ is
$$y = \mathbf { A } a t ^ { \mathbf{B} } x - \mathbf { C } a t ^ { \mathbf{D} } \text {, }$$
the $x$-coordinate of Q is $- \mathbf { E } t$.
Hence, $S _ { 1 }$ is
$$S _ { 1 } = \frac { \mathbf { F G } } { \mathbf { H } } a t ^ { \mathbf { I } } .$$
Also, since $S _ { 2 }$ is obtained by subtracting $S _ { 1 }$ from the area of the triangle PQR, we have
$$S _ { 2 } = \frac { \mathbf { J K } } { \mathbf { L } } a t ^ { \mathbf { M } } .$$
Hence, the value of $\frac { S _ { 1 } } { S _ { 2 } }$ is always
$$\frac { S _ { 1 } } { S _ { 2 } } = \mathbf { N } ,$$
independent of the values of $a$ and $t$.

Let $a > 1$. We divide the region defined by the two inequalities
$$0 \leqq x \leqq \frac { \pi } { 6 } , \quad 0 \leqq y \leqq a \cos 3 x$$
into two sections by the straight line $y = 1$. Let us denote the area of the section where $y \geq 1$ by $S$ and the area of the section where $y \leq 1$ by $T$. We are to find the value of $a$ such that $T - S$ is maximized, and also find the maximum value of $T - S$.
Let $t$ denote the value of $x \left( 0 \leqq x \leqq \frac { \pi } { 6 } \right)$ satisfying the equation $a \cos 3 x = 1$. Then we have
$$\begin{aligned} S & = \frac { \sin 3 t } { \mathbf { A } \cos 3 t } - t \\ S + T & = \frac { 1 } { \mathbf { B } \cos 3 t } . \end{aligned}$$
When we set $f ( t ) = T - S$, we see that
$$f ^ { \prime } ( t ) = \frac { ( \mathbf { C } - \mathbf { D } \sin 3 t ) \sin 3 t } { \cos ^ { \mathbf { E } } 3 t } .$$
Hence $T - S$ is maximized at $t = \frac { \pi } { \mathbf { F G } }$. Thus, $T - S$ is maximized at $a = \frac { \mathbf { H } \sqrt { \mathbf { I } } } { \mathbf{J} }$, and the maximum value is $\frac { \pi } { \mathbf { K } }$.

Consider the function $f ( x ) = x \sin ^ { 2 } x$ on the interval $0 \leqq x \leqq \pi$. Let $\ell$ be the tangent to the curve $y = f ( x )$ that passes through the origin, where $\ell$ is not the $x$-axis. We are to find the area $S$ of the region bounded by the curve $y = f ( x )$ and the tangent $\ell$.
(1) For each of $\mathbf{A}$ $\sim$ $\mathbf{D}$ in the following sentences, choose the correct answer from among (0) $\sim$ (9) below.
When we denote the point of tangency of the curve $y = f ( x )$ and the tangent $\ell$ by $( t , f ( t ) )$, we have the equality $\mathbf{A}$, since $\ell$ passes through the origin. Further, since
$$f ^ { \prime } ( t ) = \mathbf { B } + 2 t \, \mathbf { C }$$
the $x$-coordinate of the point of tangency is $t = \mathbf { D }$.
(0) $f ( t ) = t f ^ { \prime } ( t )$ (1) $f ^ { \prime } ( t ) = t f ( t )$ (2) $\sin t$ (3) $\sin ^ { 2 } t$ (4) $\cos ^ { 2 } t$ (5) $\sin t \cos t$ (6) $\frac { \pi } { 2 }$ (7) $\frac { \pi } { 3 }$ (8) $\frac { \pi } { 4 }$ (9) $\frac { \pi } { 6 }$
(2) For each of $\mathbf { E }$ $\sim$ $\mathbf { G }$ in the following sentences, choose the correct answer from among (0) $\sim$ (9) below.
The antiderivative of the function $f ( x )$ is
$$\int f ( x ) d x = \mathbf { E } \left( 2 x ^ { 2 } - 2 x \mathbf { F } - \mathbf { G } \right) + C ,$$
where $C$ is the integral constant.
(0) $\frac { 1 } { 8 }$ (1) $\frac { 1 } { 4 }$ (2) $\frac { 1 } { 2 }$ (3) 2 (4) 4 (5) 8 (6) $\sin x$ (7) $\cos x$ (8) $\sin 2 x$ (9) $\cos 2 x$
(3) Thus, the area $S$ of the region bounded by the curve $y = f ( x )$ and the tangent $\ell$ is
$$S = \frac { \mathbf { H } } { \mathbf { I J } } \pi ^ { \mathbf { K } } - \frac { \mathbf { L } } { \mathbf{M} } .$$

On the coordinate plane, there is an annular region formed by the intersection of the exterior of the circle $x ^ { 2 } + y ^ { 2 } = 3$ and the interior of the circle $x ^ { 2 } + y ^ { 2 } = 4$ . A person wants to use a straight scanning rod of length 1 to scan a certain region $R$ above the $x$-axis of this annular region. He designs the scanning rod with black and white ends moving respectively on the semicircles $C _ { 1 } : x ^ { 2 } + y ^ { 2 } = 3 ( y \geq 0 )$ and $C _ { 2 } : x ^ { 2 } + y ^ { 2 } = 4 ( y \geq 0 )$ . Initially, the black end of the scanning rod is at point $A ( \sqrt { 3 } , 0 )$ and the white end is at point $B$ on $C _ { 2 }$ . Then the black and white ends move counterclockwise along $C _ { 1 }$ and $C _ { 2 }$ respectively until the white end reaches point $B ^ { \prime } ( - 2,0 )$ on $C _ { 2 }$ , at which point scanning stops.
(Continuing from Question 19) Let $\Omega$ denote the region swept by the scanning rod in the first quadrant. Find the areas of $\Omega$ and $R$ respectively. (Non-multiple choice question, 6 points)

Let $a , b$ be real numbers, and let $O$ be the origin of the coordinate plane. It is known that the graph of the quadratic function $f ( x ) = a x ^ { 2 }$ and the circle $\Omega : x ^ { 2 } + y ^ { 2 } - 3 y + b = 0$ both pass through point $P \left( 1 , \frac { 1 } { 2 } \right)$, and let point $C$ be the center of $\Omega$.
Find the area of the region bounded by the graph of $y = f ( x )$ above and the lower semicircular arc of $\Omega$.

On the coordinate plane, let $\Gamma$ be the graph of the cubic function $f(x) = x^{3} - 9x^{2} + 15x - 4$. The tangent line $L$ to $\Gamma$ at point $P(1, 3)$ was found in question 16. Continuing from 16, find the area of the bounded region enclosed by $\Gamma$ and $L$.

Let $f(x) = 3ax^{2} + (1 - a)$ be a real coefficient polynomial function, where $-\frac{1}{2} \leq a \leq 1$. On the coordinate plane, let $\Gamma$ be the region enclosed by $y = f(x)$ and the $x$-axis for $-1 \leq x \leq 1$.
Prove that for all $a \in \left[-\frac{1}{2}, 1\right]$, the area of $\Gamma$ is always 2. (Non-multiple choice question, 2 points)

A piece of paper in the shape of rectangle $ABCD$ has a point E marked on side $DC$ and a point F marked on side AB. When this paper is folded along line EF, AF and EC intersect perpendicularly as shown in the figure.
Given that the area of the figure obtained after the folding operation is $\mathbf { 18 }$ square units less than the area before the folding operation, what is the length |AD| in units?
A) 3
B) 4
C) 6
D) 8
E) 9

In the rectangular coordinate plane, the shaded region between the curve $y = x ^ { 2 }$, the x-axis, and the line $x = 3$ is shown.
This shaded region is divided into three equal-area sub-regions by the lines $x = a$ and $x = b$.
Accordingly, what is the product $\mathbf { a } \cdot \mathbf { b }$?
A) $5 \sqrt { 2 }$
B) $4 \sqrt { 3 }$
C) $6 \sqrt { 3 }$
D) $3 \sqrt [ 3 ] { 6 }$
E) $2 \sqrt [ 3 ] { 9 }$

The graph of a one-to-one and onto function f defined on the interval [2, 6] is given in the figure.
Given that the area of the shaded region is 13 square units,
$$\int _ { 2 } ^ { 6 } f ^ { - 1 } ( x ) d x$$
What is the value of the integral?
A) 18
B) 19
C) 20
D) 21
E) 22