Area bounded by region of single curve Questions in English

(A) The area $A$ is given by the integral of the absolute value of the function over the interval $[0, \pi]$.
$A = \int_{0}^{\pi} |\cos x| \, dx$
Since $\cos x \ge 0$ for $x \in [0, \pi/2]$ and $\cos x \le 0$ for $x \in [\pi/2, \pi]$,we split the integral:
$A = \int_{0}^{\pi/2} \cos x \, dx + \int_{\pi/2}^{\pi} (-\cos x) \, dx$
$A = [\sin x]_{0}^{\pi/2} - [\sin x]_{\pi/2}^{\pi}$
$A = (\sin(\pi/2) - \sin(0)) - (\sin(\pi) - \sin(\pi/2))$
$A = (1 - 0) - (0 - 1)$
$A = 1 + 1 = 2$ sq. units.
Therefore,the correct option is $A$.

(B) The area $A$ is given by the integral of $|y| dx$ from $x = 0$ to $x = \frac{3\pi}{2}$.
$A = \int_{0}^{\frac{3\pi}{2}} |\cos x| dx$
We know that $\cos x > 0$ for $x \in [0, \frac{\pi}{2}]$ and $\cos x < 0$ for $x \in (\frac{\pi}{2}, \frac{3\pi}{2}]$.
So,$A = \int_{0}^{\frac{\pi}{2}} \cos x dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (-\cos x) dx$
$A = [\sin x]_{0}^{\frac{\pi}{2}} - [\sin x]_{\frac{\pi}{2}}^{\frac{3\pi}{2}}$
$A = (\sin \frac{\pi}{2} - \sin 0) - (\sin \frac{3\pi}{2} - \sin \frac{\pi}{2})$
$A = (1 - 0) - (-1 - 1)$
$A = 1 - (-2) = 1 + 2 = 3$ square units.
Thus,the correct option is $B$.

(A) The area bounded by the curve $y = \cos x$ from $x = 0$ to $x = \pi$ is given by the integral of the absolute value of the function:
$Area = \int_{0}^{\pi} |\cos x| dx$
Since $\cos x \geq 0$ for $x \in [0, \pi/2]$ and $\cos x \leq 0$ for $x \in [\pi/2, \pi]$,we split the integral:
$Area = \int_{0}^{\pi/2} \cos x dx + \int_{\pi/2}^{\pi} (-\cos x) dx$
$Area = [\sin x]_{0}^{\pi/2} - [\sin x]_{\pi/2}^{\pi}$
$Area = (\sin(\pi/2) - \sin 0) - (\sin \pi - \sin(\pi/2))$
$Area = (1 - 0) - (0 - 1) = 1 + 1 = 2$ sq. units.

(B) Given the equation of the line $y=mx$ and the boundaries $x=1$ and $x=2$.
The required area is given by the definite integral:
$\text{Area} = \int_{1}^{2} mx \, dx = 6$
$\Rightarrow m \left[ \frac{x^2}{2} \right]_{1}^{2} = 6$
$\Rightarrow \frac{m}{2} (2^2 - 1^2) = 6$
$\Rightarrow \frac{m}{2} (4 - 1) = 6$
$\Rightarrow \frac{3m}{2} = 6$
$\Rightarrow 3m = 12$
$\Rightarrow m = 4$

(C) Given equations are:
$y = x^{3} \implies x = y^{1/3}$
$y = 8$
$x = 0$
The area bounded by the curve $x = y^{1/3}$,the $y$-axis $(x=0)$,and the line $y=8$ is given by the integral with respect to $y$ from $y=0$ to $y=8$:
$\text{Area} = \int_{0}^{8} x \, dy$
$\text{Area} = \int_{0}^{8} y^{1/3} \, dy$
Evaluating the integral:
$\text{Area} = \left[ \frac{y^{(1/3) + 1}}{(1/3) + 1} \right]_{0}^{8} = \left[ \frac{3}{4} y^{4/3} \right]_{0}^{8}$
$\text{Area} = \frac{3}{4} (8^{4/3} - 0^{4/3})$
$\text{Area} = \frac{3}{4} ((2^{3})^{4/3}) = \frac{3}{4} (2^{4})$
$\text{Area} = \frac{3}{4} \times 16 = 3 \times 4 = 12 \text{ sq. units}$.

(C) The line $y=2x+1$ intersects the $X$-axis at $x=-\frac{1}{2}$.
For $x \in [-1, -\frac{1}{2}]$,$y \le 0$,and for $x \in [-\frac{1}{2}, 1]$,$y \ge 0$.
The required area is given by:
$\text{Area} = \int_{-1}^{1} |y| dx = \int_{-1}^{-1/2} -(2x+1) dx + \int_{-1/2}^{1} (2x+1) dx$
$= -[x^2+x]_{-1}^{-1/2} + [x^2+x]_{-1/2}^{1}$
$= -[(\frac{1}{4} - \frac{1}{2}) - (1 - 1)] + [(1+1) - (\frac{1}{4} - \frac{1}{2})]$
$= -[-\frac{1}{4}] + [2 - (-\frac{1}{4})]$
$= \frac{1}{4} + 2 + \frac{1}{4} = \frac{1}{2} + 2 = \frac{5}{2} \text{ sq units}$.

(C) The area $A$ is given by the definite integral of the function $y = \sin \left(\frac{x}{3}\right)$ from $x = 0$ to $x = 3\pi$.
$A = \int_0^{3 \pi} \sin \left(\frac{x}{3}\right) dx$
We know that the integral of $\sin(kx)$ is $-\frac{1}{k} \cos(kx)$. Here $k = \frac{1}{3}$,so the integral is $-3 \cos \left(\frac{x}{3}\right)$.
$A = \left[ -3 \cos \left(\frac{x}{3}\right) \right]_0^{3 \pi}$
$A = -3 \left[ \cos \left(\frac{3 \pi}{3}\right) - \cos \left(\frac{0}{3}\right) \right]$
$A = -3 [ \cos(\pi) - \cos(0) ]$
Since $\cos(\pi) = -1$ and $\cos(0) = 1$:
$A = -3 [ -1 - 1 ] = -3 [ -2 ] = 6 \text{ sq. units}$.

(B) The curve is $y=x^2$ and the line is $y=16$. The intersection points are found by setting $x^2=16$,which gives $x = \pm 4$.
Since the region is symmetric about the $y$-axis,the area $A$ is given by:
$A = 2 \int_0^{16} x \, dy = 2 \int_0^{16} \sqrt{y} \, dy$
$A = 2 \left[ \frac{y^{3/2}}{3/2} \right]_0^{16}$
$A = 2 \times \frac{2}{3} \left[ y^{3/2} \right]_0^{16}$
$A = \frac{4}{3} \left( 16^{3/2} - 0^{3/2} \right)$
$A = \frac{4}{3} \times (4^2)^{3/2} = \frac{4}{3} \times 4^3$
$A = \frac{4}{3} \times 64 = \frac{256}{3} \text{ sq. units}$

(B) Given equations are $y=3x$ and $y=x^2$.
To find the points of intersection,we set $3x = x^2$,which gives $x^2 - 3x = 0$,so $x(x-3) = 0$.
Thus,the points of intersection are $x=0$ and $x=3$.
The area of the region bounded by these curves is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=3$.
$\text{Area} = \int_{0}^{3} (3x - x^2) dx$
$= \left[ \frac{3x^2}{2} - \frac{x^3}{3} \right]_{0}^{3}$
$= \left( \frac{3(3)^2}{2} - \frac{(3)^3}{3} \right) - (0 - 0)$
$= \left( \frac{27}{2} - \frac{27}{3} \right)$
$= \frac{27}{2} - 9$
$= \frac{27 - 18}{2} = \frac{9}{2} \text{ square units}$.

(D) The area $A$ of the region bounded by the curve $y=f(x)$,the $x$-axis,and the lines $x=a$ and $x=b$ is given by $A = \int_{a}^{b} f(x) dx$.
Here,$f(x) = x+1$,$a=3$,and $b=5$.
$\therefore$ Required area,$A = \int_{3}^{5} (x+1) dx$
$= \left[ \frac{x^2}{2} + x \right]_{3}^{5}$
$= \left( \frac{5^2}{2} + 5 \right) - \left( \frac{3^2}{2} + 3 \right)$
$= \left( \frac{25}{2} + 5 \right) - \left( \frac{9}{2} + 3 \right)$
$= \left( \frac{25+10}{2} \right) - \left( \frac{9+6}{2} \right)$
$= \frac{35}{2} - \frac{15}{2}$
$= \frac{20}{2} = 10 \text{ sq units}$.

(B) The required area is given by the integral of the function $y = \tan x$ from $x = 0$ to $x = \frac{\pi}{3}$.
$\text{Required area} = \int_0^{\pi / 3} \tan x \, dx$
We know that the integral of $\tan x$ is $\log |\sec x|$.
$\text{Required area} = [\log |\sec x|]_0^{\pi / 3}$
Now,substitute the limits:
$= \log |\sec \frac{\pi}{3}| - \log |\sec 0|$
Since $\sec \frac{\pi}{3} = 2$ and $\sec 0 = 1$:
$= \log |2| - \log |1|$
$= \log 2 - 0 = \log 2 \text{ sq units}$.

(A) The equation $y=-\sqrt{16-x^{2}}$ represents the lower semi-circle of the circle $x^{2}+y^{2}=16$,which has a radius $r=4$ and center at the origin $(0,0)$.
Since the area is bounded by this curve and the $X$-axis,we are looking for the area of the semi-circle below the $X$-axis.
The area of a full circle is $\pi r^{2} = \pi(4)^{2} = 16\pi$.
Therefore,the area of the semi-circle is $\frac{1}{2} \times 16\pi = 8\pi$ sq units.
Alternatively,using integration:
$\text{Area} = \left| \int_{-4}^{4} (-\sqrt{16-x^{2}}) dx \right|$
$= \left| \left[ \frac{x}{2} \sqrt{16-x^{2}} + \frac{16}{2} \sin^{-1} \frac{x}{4} \right]_{-4}^{4} \right|$
$= \left| [0 + 8 \sin^{-1}(1)] - [0 + 8 \sin^{-1}(-1)] \right|$
$= \left| 8(\frac{\pi}{2}) - 8(-\frac{\pi}{2}) \right| = |4\pi + 4\pi| = 8\pi$ sq units.

(C) The area bounded by the curve $y = \cos x$ from $x = 0$ to $x = \pi$ is given by the integral of the absolute value of the function:
$\text{Area} = \int_{0}^{\pi} |\cos x| \, dx$
Since $\cos x \ge 0$ for $x \in [0, \pi/2]$ and $\cos x \le 0$ for $x \in [\pi/2, \pi]$,we split the integral:
$\text{Area} = \int_{0}^{\pi/2} \cos x \, dx + \int_{\pi/2}^{\pi} (-\cos x) \, dx$
Evaluating the integrals:
$= [\sin x]_{0}^{\pi/2} + [-\sin x]_{\pi/2}^{\pi}$
$= (\sin(\pi/2) - \sin(0)) + (-(\sin(\pi) - \sin(\pi/2)))$
$= (1 - 0) + (-(0 - 1))$
$= 1 + 1 = 2 \text{ sq. units.}$

(C) The required area is given by the integral $\int_{0}^{3\pi} y \, dx = \int_{0}^{3\pi} \sin \left(\frac{x}{3}\right) \, dx$.
Let $t = \frac{x}{3}$,then $dx = 3 \, dt$.
When $x = 0$,$t = 0$.
When $x = 3\pi$,$t = \pi$.
Substituting these into the integral,we get:
Area $= \int_{0}^{\pi} \sin(t) \cdot 3 \, dt = 3 \int_{0}^{\pi} \sin(t) \, dt$.
$= 3 [-\cos(t)]_{0}^{\pi}$.
$= -3 [\cos(\pi) - \cos(0)]$.
$= -3 [-1 - 1] = -3(-2) = 6$.
Thus,the area is $6$ square units.

(B) The given curve is $y=2x-x^{2}$.
To find the points where the curve intersects the $x$-axis,we set $y=0$:
$2x-x^{2}=0 \implies x(2-x)=0$,which gives $x=0$ and $x=2$.
Thus,the curve intersects the $x$-axis at $(0,0)$ and $(2,0)$.
The required area is given by the integral of $y$ with respect to $x$ from $x=0$ to $x=2$:
$\text{Area} = \int_{0}^{2} (2x-x^{2}) dx$
$= \left[ x^{2} - \frac{x^{3}}{3} \right]_{0}^{2}$
$= \left( 2^{2} - \frac{2^{3}}{3} \right) - (0 - 0)$
$= 4 - \frac{8}{3}$
$= \frac{12-8}{3} = \frac{4}{3} \text{ sq unit}$.

(C) The given curve is $x = 4 - y^2$. The curve intersects the $Y$-axis where $x = 0$,which gives $4 - y^2 = 0$,so $y = \pm 2$. The points of intersection are $(0, 2)$ and $(0, -2)$.
Since the curve is symmetric about the $X$-axis,the total area $A$ is twice the area in the first quadrant.
$A = 2 \int_{0}^{2} x \, dy$
$A = 2 \int_{0}^{2} (4 - y^2) \, dy$
$A = 2 \left[ 4y - \frac{y^3}{3} \right]_{0}^{2}$
$A = 2 \left[ (4(2) - \frac{2^3}{3}) - (0) \right]$
$A = 2 \left[ 8 - \frac{8}{3} \right]$
$A = 2 \left[ \frac{24 - 8}{3} \right] = 2 \left( \frac{16}{3} \right) = \frac{32}{3} \text{ sq units}$.

(C) The region is bounded by the $y$-axis $(x=0)$,$y = \cos x$,and $y = \sin x$. These curves intersect when $\cos x = \sin x$,which occurs at $x = \frac{\pi}{4}$ in the interval $[0, \frac{\pi}{2}]$.
In the interval $[0, \frac{\pi}{4}]$,$\cos x \geq \sin x$.
Therefore,the required area is given by:
$\text{Area} = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) dx$
$= [\sin x - (-\cos x)]_{0}^{\frac{\pi}{4}}$
$= [\sin x + \cos x]_{0}^{\frac{\pi}{4}}$
$= (\sin \frac{\pi}{4} + \cos \frac{\pi}{4}) - (\sin 0 + \cos 0)$
$= (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1)$
$= \frac{2}{\sqrt{2}} - 1$
$= \sqrt{2} - 1 \text{ sq. units}$

(B) The area bounded by the curve $y=x$,the $x$-axis,and the lines $x=-1$ and $x=2$ is given by the integral of the absolute value of $y$ with respect to $x$:
$Area = \int_{-1}^{2} |y| \, dx = \int_{-1}^{2} |x| \, dx$
We split the integral at $x=0$ because the function changes sign:
$Area = \int_{-1}^{0} |x| \, dx + \int_{0}^{2} |x| \, dx$
Since $|x| = -x$ for $x < 0$ and $|x| = x$ for $x \ge 0$:
$Area = \int_{-1}^{0} (-x) \, dx + \int_{0}^{2} (x) \, dx$
$Area = \left[ -\frac{x^2}{2} \right]_{-1}^{0} + \left[ \frac{x^2}{2} \right]_{0}^{2}$
$Area = (0 - (-\frac{(-1)^2}{2})) + (\frac{2^2}{2} - 0)$
$Area = \frac{1}{2} + \frac{4}{2} = \frac{1}{2} + 2 = \frac{5}{2} \text{ sq. units.}$

(C) To find the area bounded between the parabola $y^{2}=4x$ and the line $y=2x-4$,we first find their points of intersection by substituting $x = \frac{y+4}{2}$ into the parabola equation:
$y^{2} = 4\left(\frac{y+4}{2}\right)$
$y^{2} = 2(y+4)$
$y^{2} - 2y - 8 = 0$
$(y-4)(y+2) = 0$
Thus,the intersection points occur at $y = 4$ and $y = -2$.
The required area is given by the integral of the difference between the line and the parabola with respect to $y$:
$\text{Area} = \int_{-2}^{4} \left( \frac{y+4}{2} - \frac{y^{2}}{4} \right) dy$
$= \left[ \frac{y^{2}}{4} + 2y - \frac{y^{3}}{12} \right]_{-2}^{4}$
$= \left( \frac{16}{4} + 8 - \frac{64}{12} \right) - \left( \frac{4}{4} - 4 - \frac{-8}{12} \right)$
$= \left( 4 + 8 - \frac{16}{3} \right) - \left( 1 - 4 + \frac{2}{3} \right)$
$= \left( 12 - \frac{16}{3} \right) - \left( -3 + \frac{2}{3} \right)$
$= \frac{20}{3} - \left( -\frac{7}{3} \right) = \frac{27}{3} = 9 \text{ sq unit}$.

(D) Given $z=x+iy$,where $x, y \in R$.
Set $A=\{z:|z| \leq 2\}$ represents the interior and boundary of a circle with center $(0,0)$ and radius $2$,i.e.,$x^2+y^2 \leq 4$.
Set $B=\{z:(z+2y)+\bar{z} \geq 4\}$. Substituting $z=x+iy$ and $\bar{z}=x-iy$:
$(x+iy+2y)+(x-iy) \geq 4$
$2x+2y \geq 4 \Rightarrow x+y \geq 2$.
This represents the region on or above the line $x+y=2$.
The intersection $A \cap B$ is the region bounded by the circle $x^2+y^2=4$ and the line $x+y=2$ in the first quadrant.
The points of intersection are found by solving $x^2+(2-x)^2=4$:
$x^2+4-4x+x^2=4$ $\Rightarrow 2x^2-4x=0$ $\Rightarrow 2x(x-2)=0$.
So,$x=0$ (giving $y=2$) and $x=2$ (giving $y=0$).
The area is given by $\int_0^2 (y_{\text{circle}} - y_{\text{line}}) dx = \int_0^2 (\sqrt{4-x^2} - (2-x)) dx$.
$= \left[ \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) - 2x + \frac{x^2}{2} \right]_0^2$
$= (0 + 2\sin^{-1}(1) - 4 + 2) - (0 + 0 - 0 + 0) = 2(\frac{\pi}{2}) - 2 = \pi-2$.

(D) The equation of the parabola is $y^2=16ax$ and the line is $y=4mx$.
Substituting $x = \frac{y^2}{16a}$ into the line equation $y = 4mx$,we get $y = 4m(\frac{y^2}{16a}) = \frac{my^2}{4a}$.
This gives $y^2 = \frac{4ay}{m}$,so $y(y - \frac{4a}{m}) = 0$. Thus,the intersection points are $y=0$ and $y=\frac{4a}{m}$.
The area bounded by the curves is given by:
$\int_0^{\frac{4a}{m}} (\frac{y}{4m} - \frac{y^2}{16a}) dy = \frac{a^2}{12}$
Evaluating the integral:
$[\frac{y^2}{8m} - \frac{y^3}{48a}]_0^{\frac{4a}{m}} = \frac{a^2}{12}$
$\frac{(4a/m)^2}{8m} - \frac{(4a/m)^3}{48a} = \frac{a^2}{12}$
$\frac{16a^2}{8m^3} - \frac{64a^3}{48am^3} = \frac{a^2}{12}$
$\frac{2a^2}{m^3} - \frac{4a^2}{3m^3} = \frac{a^2}{12}$
$\frac{6a^2 - 4a^2}{3m^3} = \frac{a^2}{12}$
$\frac{2a^2}{3m^3} = \frac{a^2}{12}$
$\frac{2}{3m^3} = \frac{1}{12}$
$m^3 = \frac{2 \times 12}{3} = 8$
$m = 2$
Thus,option $(D)$ is correct.

(B) The area $A$ is given by the integral $\int_{0}^{2} |y| dx = \int_{0}^{2} |x^2-5x+4| dx$.
First,find the roots of $x^2-5x+4=0$,which are $(x-1)(x-4)=0$,so $x=1$ and $x=4$.
In the interval $[0, 1]$,$x^2-5x+4 \geq 0$.
In the interval $[1, 2]$,$x^2-5x+4 \leq 0$.
Thus,$A = \int_{0}^{1} (x^2-5x+4) dx + \int_{1}^{2} -(x^2-5x+4) dx$.
Evaluating the first integral: $[\frac{x^3}{3} - \frac{5x^2}{2} + 4x]_{0}^{1} = \frac{1}{3} - \frac{5}{2} + 4 = \frac{2-15+24}{6} = \frac{11}{6}$.
Evaluating the second integral: $-[\frac{x^3}{3} - \frac{5x^2}{2} + 4x]_{1}^{2} = -[(\frac{8}{3} - 10 + 8) - (\frac{1}{3} - \frac{5}{2} + 4)] = -[(\frac{2}{3}) - (\frac{11}{6})] = -[\frac{4-11}{6}] = -[-\frac{7}{6}] = \frac{7}{6}$.
Total Area $A = \frac{11}{6} + \frac{7}{6} = \frac{18}{6} = 3$ sq. units.

(B) The curve is given by $y=x^3-19x+30$. Factoring the polynomial,we get $y=(x+5)(x-2)(x-3)$.
The roots are $x=-5, 2, 3$. The curve is above the $x$-axis on the interval $[-5, 2]$ and below the $x$-axis on the interval $[2, 3]$.
The total area $A$ is given by:
$A = \int_{-5}^{2} (x^3-19x+30) dx + \left| \int_{2}^{3} (x^3-19x+30) dx \right|$
$A = \int_{-5}^{2} (x^3-19x+30) dx - \int_{2}^{3} (x^3-19x+30) dx$
Evaluating the integral $\int (x^3-19x+30) dx = \frac{x^4}{4} - \frac{19x^2}{2} + 30x + C$.
For the first part: $\left[ \frac{x^4}{4} - \frac{19x^2}{2} + 30x \right]_{-5}^{2} = (4 - 38 + 60) - (\frac{625}{4} - \frac{475}{2} - 150) = 26 - (\frac{625-950-600}{4}) = 26 - (-\frac{925}{4}) = 26 + 231.25 = 257.25 = \frac{1029}{4}$.
For the second part: $\left[ \frac{x^4}{4} - \frac{19x^2}{2} + 30x \right]_{2}^{3} = (\frac{81}{4} - \frac{171}{2} + 90) - (4 - 38 + 60) = (\frac{81-342+360}{4}) - 26 = \frac{99}{4} - 26 = \frac{99-104}{4} = -\frac{5}{4}$.
Total Area $A = \frac{1029}{4} - (-\frac{5}{4}) = \frac{1034}{4} = \frac{517}{2} \text{ sq. units}$.

(B) The given curve is $y = x|x|$. We can define this as a piecewise function:
$y = \begin{cases} x^2 & \text{if } x \geq 0 \\ -x^2 & \text{if } x < 0 \end{cases}$
We need to find the area enclosed by this curve between $x = -1$ and $x = 1$.
The total area $A$ is given by the integral of the absolute value of $y$ from $-1$ to $1$:
$A = \int_{-1}^{1} |y| dx = \int_{-1}^{0} |-x^2| dx + \int_{0}^{1} |x^2| dx$
$A = \int_{-1}^{0} x^2 dx + \int_{0}^{1} x^2 dx$
Calculating the integrals:
$A = \left[ \frac{x^3}{3} \right]_{-1}^{0} + \left[ \frac{x^3}{3} \right]_{0}^{1}$
$A = (0 - (-\frac{1}{3})) + (\frac{1}{3} - 0)$
$A = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$
Thus,the area is $\frac{2}{3}$ sq units.

(B) Given,the curve $y=ax^2+bx$ passes through the point $(1,2)$.
$\therefore 2 = a(1)^2 + b(1) \Rightarrow a+b=2$ ... $(i)$
Given that the curve lies above the $X$-axis for $0 \leq x \leq 8$,the area enclosed by the curve,the $X$-axis,and the line $x=6$ is given by:
$\int_0^6 (ax^2+bx) dx = 108$
$\Rightarrow \left[ \frac{ax^3}{3} + \frac{bx^2}{2} \right]_0^6 = 108$
$\Rightarrow \frac{a(216)}{3} + \frac{b(36)}{2} = 108$
$\Rightarrow 72a + 18b = 108$
Dividing by $18$,we get $4a + b = 6$ ... $(ii)$
Subtracting $(i)$ from $(ii)$:
$(4a+b) - (a+b) = 6 - 2$
$3a = 4 \Rightarrow a = \frac{4}{3}$
Substituting $a = \frac{4}{3}$ in $(i)$:
$\frac{4}{3} + b = 2 \Rightarrow b = 2 - \frac{4}{3} = \frac{2}{3}$
Now,calculate $2b-a$:
$2b-a = 2\left(\frac{2}{3}\right) - \frac{4}{3} = \frac{4}{3} - \frac{4}{3} = 0$

(A) To find the area bounded by the curve $y=1-x-6x^2$ and the $X$-axis,we first find the intersection points with the $X$-axis by setting $y=0$.
$1-x-6x^2=0$
$6x^2+x-1=0$
$6x^2+3x-2x-1=0$
$3x(2x+1)-1(2x+1)=0$
$(3x-1)(2x+1)=0$
So,$x = \frac{1}{3}$ and $x = -\frac{1}{2}$.
The required area is given by the integral:
$\text{Area} = \int_{-1/2}^{1/3} (1-x-6x^2) dx$
$= [x - \frac{x^2}{2} - 2x^3]_{-1/2}^{1/3}$
$= (\frac{1}{3} - \frac{(1/3)^2}{2} - 2(1/3)^3) - (-\frac{1}{2} - \frac{(-1/2)^2}{2} - 2(-1/2)^3)$
$= (\frac{1}{3} - \frac{1}{18} - \frac{2}{27}) - (-\frac{1}{2} - \frac{1}{8} + \frac{1}{4})$
$= (\frac{18-3-4}{54}) - (\frac{-4-1+2}{8})$
$= \frac{11}{54} - (-\frac{3}{8}) = \frac{11}{54} + \frac{3}{8}$
$= \frac{44+81}{216} = \frac{125}{216} \text{ sq. units}$.

(D) Given the curve is $x = \log |y|$.
This can be rewritten as $|y| = e^x$,which implies $y = \pm e^x$.
The curve is symmetric about the $x$-axis.
The area bounded by the curve,the lines $x = -1$ and $x = 0$ is given by the integral of $y$ with respect to $x$ from $-1$ to $0$.
Since the curve is symmetric,the total area is twice the area above the $x$-axis:
$\text{Area} = 2 \int_{-1}^{0} |y| \, dx = 2 \int_{-1}^{0} e^x \, dx$
Evaluating the integral:
$\text{Area} = 2 \left[ e^x \right]_{-1}^{0}$
$= 2 (e^0 - e^{-1})$
$= 2 (1 - e^{-1})$
Thus,the required area is $2(1 - e^{-1})$ square units.

(A) The equation of the curve is $y = x^2 + 2x + 1 = (x + 1)^2$.
First,we find the equation of the tangent at $(1, 4)$.
Differentiating $y$ with respect to $x$,we get $\frac{dy}{dx} = 2x + 2$.
At the point $(1, 4)$,the slope of the tangent is $m = 2(1) + 2 = 4$.
The equation of the tangent line is $y - 4 = 4(x - 1)$,which simplifies to $y = 4x$.
The area bounded by the curve,the tangent,and the $Y$-axis is the area under the curve from $x = 0$ to $x = 1$ minus the area of the triangle formed by the tangent line,the $X$-axis,and the vertical line $x = 1$.
Area $A = \int_0^1 (x^2 + 2x + 1) dx - \int_0^1 (4x) dx$.
Calculating the first integral: $\int_0^1 (x^2 + 2x + 1) dx = \left[ \frac{x^3}{3} + x^2 + x \right]_0^1 = \frac{1}{3} + 1 + 1 = \frac{7}{3}$.
Calculating the second integral (area of the triangle): $\int_0^1 4x dx = \left[ 2x^2 \right]_0^1 = 2(1)^2 - 0 = 2$.
Therefore,the required area $A = \frac{7}{3} - 2 = \frac{7 - 6}{3} = \frac{1}{3} \ sq. \ units$.

(C) The region is bounded by the parabola $x^2=y$,the line $y=x+2$,and the $X$-axis.
The intersection points of the parabola $y=x^2$ and the line $y=x+2$ are found by setting $x^2=x+2$,which gives $x^2-x-2=0$,so $(x-2)(x+1)=0$. Thus,$x=-1$ and $x=2$.
The line $y=x+2$ intersects the $X$-axis at $x=-2$ (where $y=0$).
The required area is the region bounded by the line from $x=-2$ to $x=-1$ and the parabola from $x=-1$ to $x=0$ (since the region is bounded by the $X$-axis,we consider the area under the curve down to $y=0$).
Area $= \int_{-2}^{-1} (x+2) dx + \int_{-1}^{0} x^2 dx$
$= \left[ \frac{x^2}{2} + 2x \right]_{-2}^{-1} + \left[ \frac{x^3}{3} \right]_{-1}^{0}$
$= \left( (\frac{1}{2} - 2) - (\frac{4}{2} - 4) \right) + \left( 0 - (-\frac{1}{3}) \right)$
$= (-\frac{3}{2} - (-2)) + \frac{1}{3}$
$= \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \text{ sq. units}$.

(D) The given curve is $ay^2 = x^2(a - x)$,where $a > 0$.
Since the curve is symmetric about the $x$-axis,the total area is twice the area above the $x$-axis.
From the equation,$y^2 = \frac{x^2(a - x)}{a}$,so $y = \pm x \sqrt{\frac{a - x}{a}}$.
The loop exists for $x \in [0, a]$.
The required area $A$ is given by:
$A = 2 \int_0^a y \, dx = 2 \int_0^a x \sqrt{\frac{a - x}{a}} \, dx$
$A = \frac{2}{\sqrt{a}} \int_0^a x \sqrt{a - x} \, dx$
Let $a - x = t^2$,then $dx = -2t \, dt$. When $x = 0, t = \sqrt{a}$ and when $x = a, t = 0$.
$A = \frac{2}{\sqrt{a}} \int_{\sqrt{a}}^0 (a - t^2) t (-2t \, dt) = \frac{4}{\sqrt{a}} \int_0^{\sqrt{a}} (at^2 - t^4) \, dt$
$A = \frac{4}{\sqrt{a}} \left[ \frac{at^3}{3} - \frac{t^5}{5} \right]_0^{\sqrt{a}}$
$A = \frac{4}{\sqrt{a}} \left( \frac{a(\sqrt{a})^3}{3} - \frac{(\sqrt{a})^5}{5} \right) = \frac{4}{\sqrt{a}} \left( \frac{a^2 \sqrt{a}}{3} - \frac{a^2 \sqrt{a}}{5} \right)$
$A = 4 \left( \frac{a^2}{3} - \frac{a^2}{5} \right) = 4 \left( \frac{5a^2 - 3a^2}{15} \right) = 4 \left( \frac{2a^2}{15} \right) = \frac{8}{15} a^2$.
Thus,the correct option is $(d)$.

(A) The given parabola is $y^2 = 8x$. Comparing this with $y^2 = 4ax$,we get $4a = 8$,so $a = 2$.
The latus rectum of the parabola is the line $x = a = 2$.
The area bounded by the curve and the latus rectum is symmetric about the $x$-axis.
The required area is $A = 2 \int_0^2 y \, dx = 2 \int_0^2 \sqrt{8x} \, dx$.
$A = 2 \times 2\sqrt{2} \int_0^2 x^{1/2} \, dx = 4\sqrt{2} \left[ \frac{x^{3/2}}{3/2} \right]_0^2$.
$A = 4\sqrt{2} \times \frac{2}{3} \times (2)^{3/2} = \frac{8\sqrt{2}}{3} \times 2\sqrt{2} = \frac{16 \times 2}{3} = \frac{32}{3}$ sq units.

(C) The area is given by the integral $A = \int_0^{\frac{\pi}{2}} |\sin x - \cos x| \, dx$.
Since $\cos x \geq \sin x$ for $0 \leq x \leq \frac{\pi}{4}$ and $\sin x \geq \cos x$ for $\frac{\pi}{4} \leq x \leq \frac{\pi}{2}$,we split the integral:
$A = \int_0^{\frac{\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin x - \cos x) \, dx$.
Evaluating the first part: $[\sin x + \cos x]_0^{\frac{\pi}{4}} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \sqrt{2} - 1$.
Evaluating the second part: $[-\cos x - \sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = (-0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = -1 + \sqrt{2} = \sqrt{2} - 1$.
Adding both parts: $A = (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2(\sqrt{2} - 1)$ square units.

(A) The area $A$ of the region enclosed by the curve $x^2+y^2=16$ and the lines $x=2$ and $x=3$ is given by the integral:
$A = \int_2^3 \sqrt{16-x^2} dx$
Using the standard integral formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$:
$A = \left[\frac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}\left(\frac{x}{4}\right)\right]_2^3$
Evaluating the definite integral:
$A = \left(\frac{3}{2}\sqrt{16-9} + 8\sin^{-1}\left(\frac{3}{4}\right)\right) - \left(\frac{2}{2}\sqrt{16-4} + 8\sin^{-1}\left(\frac{2}{4}\right)\right)$
$A = \left(\frac{3}{2}\sqrt{7} + 8\sin^{-1}\left(\frac{3}{4}\right)\right) - \left(\sqrt{12} + 8\sin^{-1}\left(\frac{1}{2}\right)\right)$
Since $\sqrt{12} = 2\sqrt{3}$ and $\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$:
$A = \frac{3}{2}\sqrt{7} + 8\sin^{-1}\left(\frac{3}{4}\right) - 2\sqrt{3} - 8\left(\frac{\pi}{6}\right)$
$A = \frac{3}{2}\sqrt{7} - 2\sqrt{3} - \frac{4\pi}{3} + 8\sin^{-1}\left(\frac{3}{4}\right)$
Comparing this with the given expression $\left(3 \sqrt{7}-4 \sqrt{3}-\frac{8 \pi}{3}+k\right)$:
Note that the given expression is $2 \times \left(\frac{3}{2}\sqrt{7} - 2\sqrt{3} - \frac{4\pi}{3} + \frac{k}{2}\right)$.
Actually,evaluating the integral result directly: $A = \frac{3}{2}\sqrt{7} - 2\sqrt{3} - \frac{4\pi}{3} + 8\sin^{-1}\left(\frac{3}{4}\right)$.
To match the form $\left(3 \sqrt{7}-4 \sqrt{3}-\frac{8 \pi}{3}+k\right)$,we multiply the expression by $2$ and divide by $2$:
$A = \frac{1}{2} \left(3\sqrt{7} - 4\sqrt{3} - \frac{8\pi}{3} + 16\sin^{-1}\left(\frac{3}{4}\right)\right)$
Thus,$k = 16\sin^{-1}\left(\frac{3}{4}\right)$.

(A) The area $A$ of the region bounded by the curve $y=f(x)$,the $x$-axis,and the lines $x=a$ and $x=b$ is given by $A = \int_{a}^{b} |f(x)| \, dx$.
Here,$f(x) = x^3$,$a = -2$,and $b = 4$.
The function $x^3$ is negative for $x < 0$ and positive for $x > 0$.
Thus,the integral is split at $x = 0$:
$A = \int_{-2}^{0} |x^3| \, dx + \int_{0}^{4} |x^3| \, dx = \int_{-2}^{0} (-x^3) \, dx + \int_{0}^{4} x^3 \, dx$.
Evaluating the first integral: $\int_{-2}^{0} (-x^3) \, dx = [-\frac{x^4}{4}]_{-2}^{0} = 0 - (-\frac{(-2)^4}{4}) = \frac{16}{4} = 4$.
Evaluating the second integral: $\int_{0}^{4} x^3 \, dx = [\frac{x^4}{4}]_{0}^{4} = \frac{4^4}{4} - 0 = 4^3 = 64$.
Total area $A = 4 + 64 = 68$ square units.

(C) The curve $y = 8x^3 - 1$ intersects the $x$-axis $(y=0)$ at $8x^3 - 1 = 0$,which gives $x^3 = \frac{1}{8}$,so $x = \frac{1}{2}$.
For $x \in [-1, \frac{1}{2}]$,$y \le 0$,so the area is $\int_{-1}^{\frac{1}{2}} -(8x^3 - 1) dx = \int_{-1}^{\frac{1}{2}} (1 - 8x^3) dx$.
For $x \in [\frac{1}{2}, 1]$,$y \ge 0$,so the area is $\int_{\frac{1}{2}}^{1} (8x^3 - 1) dx$.
Total Area = $\int_{-1}^{\frac{1}{2}} (1 - 8x^3) dx + \int_{\frac{1}{2}}^{1} (8x^3 - 1) dx$.
$= [x - 2x^4]_{-1}^{\frac{1}{2}} + [2x^4 - x]_{\frac{1}{2}}^{1}$.
$= ((\frac{1}{2} - 2(\frac{1}{16})) - (-1 - 2(1))) + ((2(1) - 1) - (2(\frac{1}{16}) - \frac{1}{2}))$.
$= ((\frac{1}{2} - \frac{1}{8}) - (-3)) + (1 - (\frac{1}{8} - \frac{1}{2}))$.
$= (\frac{3}{8} + 3) + (1 - (-\frac{3}{8}))$.
$= \frac{27}{8} + \frac{11}{8} = \frac{38}{8} = \frac{19}{4}$ sq. units.

(D) The given curve is $y = 2 - x - 3x^2$. The region is bounded by the $X$-axis $(y = 0)$,the $Y$-axis $(x = 0)$,and the line $x = -2$.
First,find the roots of the curve by setting $y = 0$:
$2 - x - 3x^2 = 0 \implies 3x^2 + x - 2 = 0 \implies (3x - 2)(x + 1) = 0$.
The roots are $x = -1$ and $x = 2/3$.
For $x \in [-2, -1]$,$y = 2 - x - 3x^2$ is negative (e.g.,at $x = -1.5$,$y = 2 + 1.5 - 3(2.25) = 3.5 - 6.75 = -3.25 < 0$).
For $x \in [-1, 0]$,$y = 2 - x - 3x^2$ is positive (e.g.,at $x = -0.5$,$y = 2 + 0.5 - 3(0.25) = 2.5 - 0.75 = 1.75 > 0$).
The total area is given by:
Area $= \int_{-2}^{-1} -(2 - x - 3x^2) dx + \int_{-1}^{0} (2 - x - 3x^2) dx$
$= \int_{-2}^{-1} (3x^2 + x - 2) dx + \int_{-1}^{0} (2 - x - 3x^2) dx$
$= [x^3 + \frac{x^2}{2} - 2x]_{-2}^{-1} + [2x - \frac{x^2}{2} - x^3]_{-1}^{0}$
$= [(-1 + 0.5 + 2) - (-8 + 2 + 4)] + [(0) - (-2 - 0.5 + 1)]$
$= [1.5 - (-2)] + [0 - (-1.5)]$
$= 3.5 + 1.5 = 5$.
Thus,the correct option is $D$.

(A) Given parabola is $y=x^2+3$.
First,find the equation of the tangent at $(3,12)$.
The derivative is $\frac{dy}{dx} = 2x$. At $x=3$,the slope $m = 2(3) = 6$.
The equation of the tangent is $y - 12 = 6(x - 3)$,which simplifies to $y = 6x - 6$.
The tangent intersects the $x$-axis at $y=0$,so $6x - 6 = 0 \Rightarrow x = 1$.
The region bounded by the parabola,the tangent,and the coordinate axes in the first quadrant is the area under the parabola from $x=0$ to $x=3$ minus the area of the triangle formed by the tangent line,the $x$-axis,and the vertical line $x=3$.
Alternatively,the area is $\int_0^3 (x^2+3) dx - \int_1^3 (6x-6) dx$.
Calculating the first integral: $\int_0^3 (x^2+3) dx = [\frac{x^3}{3} + 3x]_0^3 = (9 + 9) - 0 = 18$.
Calculating the second integral (area of the triangle): $\int_1^3 (6x-6) dx = [3x^2 - 6x]_1^3 = (27 - 18) - (3 - 6) = 9 - (-3) = 12$.
Required area = $18 - 12 = 6$ sq. units.

(D) The required area $A$ is given by the integral $A = \int_0^{2\pi} |\sin 2x| \, dx$.
Since the function $f(x) = |\sin 2x|$ is periodic with period $\frac{\pi}{2}$,the area over $[0, 2\pi]$ consists of $4$ identical humps,each over an interval of length $\frac{\pi}{2}$.
Thus,$A = 4 \int_0^{\pi/2} \sin 2x \, dx$.
Evaluating the integral: $A = 4 \left[ -\frac{\cos 2x}{2} \right]_0^{\pi/2}$.
$A = 4 \left( -\frac{1}{2} (\cos \pi - \cos 0) \right) = 4 \left( -\frac{1}{2} (-1 - 1) \right) = 4 \left( -\frac{1}{2} (-2) \right) = 4(1) = 4$.
Therefore,the area is $4$ square units.
Hence,option $D$ is correct.

(D) Given the curve $y = x^4 - x^2$.
To find the minimum points,we calculate the derivative:
$\frac{dy}{dx} = 4x^3 - 2x = 2x(2x^2 - 1) = 0$.
This gives $x = 0$ and $x = \pm \frac{1}{\sqrt{2}}$.
Using the second derivative test,$y'' = 12x^2 - 2$.
At $x = \pm \frac{1}{\sqrt{2}}$,$y'' = 12(\frac{1}{2}) - 2 = 4 > 0$,so these are the minimum points.
The area is given by $\int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} |x^4 - x^2| dx$.
Since the function is even,the area is $2 \int_{0}^{\frac{1}{\sqrt{2}}} -(x^4 - x^2) dx$ (as $x^4 - x^2 < 0$ in this interval).
Area $= 2 \int_{0}^{\frac{1}{\sqrt{2}}} (x^2 - x^4) dx = 2 \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_{0}^{\frac{1}{\sqrt{2}}}$.
$= 2 \left( \frac{1}{3(2\sqrt{2})} - \frac{1}{5(4\sqrt{2})} \right) = 2 \left( \frac{1}{6\sqrt{2}} - \frac{1}{20\sqrt{2}} \right)$.
$= 2 \left( \frac{10 - 3}{60\sqrt{2}} \right) = 2 \left( \frac{7}{60\sqrt{2}} \right) = \frac{7}{30\sqrt{2}}$.

(A) The given curve is $y = x^3 - 3x^2 + 2x$.
To find the points where the curve intersects the $X$-axis,we set $y = 0$:
$x(x^2 - 3x + 2) = 0$
$x(x - 1)(x - 2) = 0$
So,the curve intersects the $X$-axis at $x = 0, 1, 2$.
The area is given by $\int_0^2 |y| dx = \int_0^1 (x^3 - 3x^2 + 2x) dx + \left| \int_1^2 (x^3 - 3x^2 + 2x) dx \right|$.
First part: $\int_0^1 (x^3 - 3x^2 + 2x) dx = \left[ \frac{x^4}{4} - x^3 + x^2 \right]_0^1 = (\frac{1}{4} - 1 + 1) - 0 = \frac{1}{4}$.
Second part: $\int_1^2 (x^3 - 3x^2 + 2x) dx = \left[ \frac{x^4}{4} - x^3 + x^2 \right]_1^2 = (\frac{16}{4} - 8 + 4) - (\frac{1}{4} - 1 + 1) = (4 - 8 + 4) - \frac{1}{4} = 0 - \frac{1}{4} = -\frac{1}{4}$.
The absolute value is $|-\frac{1}{4}| = \frac{1}{4}$.
Total area = $\frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$ square units.

(B) The given curves are $y=|x|$ and $y=1-|x|$.
To find the points of intersection,set $|x| = 1-|x|$,which gives $2|x| = 1$,so $|x| = \frac{1}{2}$.
This implies $x = \frac{1}{2}$ or $x = -\frac{1}{2}$.
When $x = \frac{1}{2}$,$y = \frac{1}{2}$. When $x = -\frac{1}{2}$,$y = \frac{1}{2}$.
The intersection points are $A(\frac{1}{2}, \frac{1}{2})$ and $C(-\frac{1}{2}, \frac{1}{2})$.
The curves also intersect the $y$-axis at $O(0,0)$ and $B(0,1)$.
The region is a square with vertices $O(0,0)$,$A(\frac{1}{2}, \frac{1}{2})$,$B(0,1)$,and $C(-\frac{1}{2}, \frac{1}{2})$.
The side length of the square is the distance $OA = \sqrt{(\frac{1}{2}-0)^2 + (\frac{1}{2}-0)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
The area of the square is $(\text{side})^2 = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2} \text{ sq unit}$.

(D) The given curves are $y=x^2+1$ and $y=2x-2$.
The region is bounded by the vertical lines $x=-1$ and $x=2$.
For the interval $[-1, 2]$,we check if the curves intersect. Setting $x^2+1 = 2x-2$,we get $x^2-2x+3=0$. The discriminant $D = (-2)^2 - 4(1)(3) = 4-12 = -8 < 0$. Thus,there is no intersection point,and $x^2+1 > 2x-2$ for all $x$.
The required area is given by:
$\text{Area} = \int_{-1}^{2} [(x^2+1) - (2x-2)] \, dx$
$\text{Area} = \int_{-1}^{2} (x^2 - 2x + 3) \, dx$
$\text{Area} = \left[ \frac{x^3}{3} - x^2 + 3x \right]_{-1}^{2}$
$\text{Area} = \left( \frac{8}{3} - 4 + 6 \right) - \left( \frac{-1}{3} - 1 - 3 \right)$
$\text{Area} = \left( \frac{8}{3} + 2 \right) - \left( -\frac{1}{3} - 4 \right)$
$\text{Area} = \frac{14}{3} - \left( -\frac{13}{3} \right)$
$\text{Area} = \frac{14}{3} + \frac{13}{3} = \frac{27}{3} = 9$ sq units.

(C) Given $h = 1$ (the difference between consecutive $x$ values).
Let the values of $f(x)$ be $y_0, y_1, y_2, y_3, y_4, y_5$.
Here,$y_0 = 2, y_1 = 3, y_2 = 6, y_3 = 11, y_4 = 18, y_5 = 27$.
By the Trapezoidal rule,the area $A$ is given by:
$A = \frac{h}{2} [ (y_0 + y_5) + 2(y_1 + y_2 + y_3 + y_4) ]$
Substituting the values:
$A = \frac{1}{2} [ (2 + 27) + 2(3 + 6 + 11 + 18) ]$
$A = \frac{1}{2} [ 29 + 2(38) ]$
$A = \frac{1}{2} [ 29 + 76 ]$
$A = \frac{1}{2} [ 105 ] = 52.5$
Thus,the approximate area is $52.5$ square units.

(D) The region is bounded by $y=\sin x$,$y=\cos x$ and the $x$-axis for $0 \leq x \leq \frac{\pi}{2}$.
The area $A_1$ is the area under $y=\sin x$ from $x=0$ to $x=\frac{\pi}{4}$:
$A_1 = \int_0^{\pi/4} \sin x \, dx = [-\cos x]_0^{\pi/4} = -(\cos \frac{\pi}{4} - \cos 0) = -(\frac{1}{\sqrt{2}} - 1) = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2}-1}{\sqrt{2}}$.
The area $A_2$ is the area under $y=\cos x$ from $x=\frac{\pi}{4}$ to $x=\frac{\pi}{2}$:
$A_2 = \int_{\pi/4}^{\pi/2} \cos x \, dx = [\sin x]_{\pi/4}^{\pi/2} = \sin \frac{\pi}{2} - \sin \frac{\pi}{4} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2}-1}{\sqrt{2}}$.
Thus,the ratio $A_1 : A_2 = \frac{\sqrt{2}-1}{\sqrt{2}} : \frac{\sqrt{2}-1}{\sqrt{2}} = 1 : 1$.

(B) The given curve is $2x = y^2 - 1$, which can be rewritten as $x = \frac{y^2 - 1}{2}$.
The curve $x = 0$ is the $y$-axis.
To find the intersection points of the curve $2x = y^2 - 1$ and $x = 0$, substitute $x = 0$ into the equation:
$0 = y^2 - 1 \implies y^2 = 1 \implies y = \pm 1$.
Thus, the region is bounded by $y = -1$ to $y = 1$.
The area $A$ is given by the integral of the distance between the curves with respect to $y$:
$A = \int_{-1}^{1} |x_{\text{right}} - x_{\text{left}}| dy = \int_{-1}^{1} |0 - \frac{y^2 - 1}{2}| dy = \int_{-1}^{1} \frac{1 - y^2}{2} dy$.
Since the function is symmetric about the $x$-axis, we can write:
$A = 2 \int_{0}^{1} \frac{1 - y^2}{2} dy = \int_{0}^{1} (1 - y^2) dy$.
$A = [y - \frac{y^3}{3}]_{0}^{1} = (1 - \frac{1}{3}) - (0 - 0) = \frac{2}{3} \text{ sq unit}$.