Area bounded by region of single curve Questions in English

(D) To find the area bounded by the curve $y = \ln(x)$ and the lines $y = 0$,$y = \ln(3)$,and $x = 0$,it is easier to integrate with respect to $y$.
Given $y = \ln(x)$,we have $x = e^y$.
The region is bounded by $y = 0$ (the $x$-axis) and $y = \ln(3)$,and the curve $x = e^y$ from $x = 0$ to $x = 3$.
The area $A$ is given by the integral of $x$ with respect to $y$ from $y = 0$ to $y = \ln(3)$:
$A = \int_{0}^{\ln(3)} e^y \, dy$
Evaluating the integral:
$A = [e^y]_{0}^{\ln(3)}$
$A = e^{\ln(3)} - e^0$
$A = 3 - 1$
$A = 2$
Thus,the area is $2$.

(D) The given parabola is $y = 9x^2$,which implies $x^2 = y/9$. Since we are in the first quadrant,$x = \sqrt{y}/3$.
The area $A$ bounded by the curve $x = f(y)$,the $y$-axis $(x = 0)$,and the lines $y = 1$ and $y = 4$ is given by the integral:
$A = \int_{1}^{4} x \, dy = \int_{1}^{4} \frac{\sqrt{y}}{3} \, dy$
$A = \frac{1}{3} \int_{1}^{4} y^{1/2} \, dy$
Using the power rule $\int y^n \, dy = \frac{y^{n+1}}{n+1}$,we get:
$A = \frac{1}{3} \left[ \frac{y^{3/2}}{3/2} \right]_{1}^{4} = \frac{1}{3} \times \frac{2}{3} \left[ y^{3/2} \right]_{1}^{4}$
$A = \frac{2}{9} \left( 4^{3/2} - 1^{3/2} \right)$
$A = \frac{2}{9} (8 - 1) = \frac{2}{9} \times 7 = \frac{14}{9} \text{ sq. units.}$

(B) The function is $y = |\cos x - \sin x|$.
In the interval $[0, \pi/4]$,$\cos x \geq \sin x$,so $y = \cos x - \sin x$.
In the interval $[\pi/4, \pi/2]$,$\sin x \geq \cos x$,so $y = \sin x - \cos x$.
The required area is given by:
$A = \int_{0}^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx$
$A = [\sin x + \cos x]_{0}^{\pi/4} + [-\cos x - \sin x]_{\pi/4}^{\pi/2}$
$A = ((\sin(\pi/4) + \cos(\pi/4)) - (\sin(0) + \cos(0))) + ((-\cos(\pi/2) - \sin(\pi/2)) - (-\cos(\pi/4) - \sin(\pi/4)))$
$A = ((\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1)) + ((0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}))$
$A = (\sqrt{2} - 1) + (-1 + \sqrt{2})$
$A = 2\sqrt{2} - 2$ square units.

(D) The given region is a circle $x^2 + y^2 = 4$ with radius $r = 2$ and center at $(0, 0)$. The line $y = x + 2$ passes through $(-2, 0)$ and $(0, 2)$.
Let $I$ be the smaller portion and $II$ be the greater portion of the circle.
The area of the smaller portion $I$ is given by the integral:
Area of $I = \int_{-2}^{0} [\sqrt{4 - x^2} - (x + 2)] dx$
$= [\frac{x}{2} \sqrt{4 - x^2} + \frac{4}{2} \sin^{-1}(\frac{x}{2})]_{-2}^{0} - [\frac{x^2}{2} + 2x]_{-2}^{0}$
$= [0 + 2 \sin^{-1}(0)] - [(-1) \sqrt{0} + 2 \sin^{-1}(-1)] - [0 - (\frac{4}{2} - 4)]$
$= [0 - 2(-\frac{\pi}{2})] - [0 - (-2)]$
$= \pi - 2$
Now,the area of the greater portion $II$ is:
Area of $II = \text{Area of circle} - \text{Area of } I$
$= 4\pi - (\pi - 2) = 3\pi + 2$
Therefore,the required ratio is:
$\frac{\text{Area of } I}{\text{Area of } II} = \frac{\pi - 2}{3\pi + 2}$

(B) The region is bounded by the curve $y = x^3$,the line $y = 8$,and the $y$-axis $(x = 0)$.
To find the area with respect to the $y$-axis,we express $x$ in terms of $y$: $x = y^{1/3}$.
The limits for $y$ are from $0$ to $8$.
Required Area $= \int\limits_{0}^{8} x \, dy = \int\limits_{0}^{8} y^{1/3} \, dy$
$= \left[ \frac{y^{1/3 + 1}}{1/3 + 1} \right]_{0}^{8} = \left[ \frac{y^{4/3}}{4/3} \right]_{0}^{8} = \left[ \frac{3}{4} y^{4/3} \right]_{0}^{8}$
$= \frac{3}{4} (8^{4/3} - 0^{4/3}) = \frac{3}{4} ((2^3)^{4/3}) = \frac{3}{4} (2^4) = \frac{3}{4} \times 16 = 12 \text{ sq. units.}$

(C) The given region is bounded by $y = x|x| + 1$ and the $x$-axis for $-1 \le x \le 1$.
We can define $y$ as:
$y = \begin{cases} -x^2 + 1, & \text{if } -1 \le x < 0 \\ x^2 + 1, & \text{if } 0 \le x \le 1 \end{cases}$
The area is given by the integral:
$\text{Area} = \int_{-1}^{0} (-x^2 + 1) dx + \int_{0}^{1} (x^2 + 1) dx$
Evaluating the first integral:
$\int_{-1}^{0} (-x^2 + 1) dx = \left[ -\frac{x^3}{3} + x \right]_{-1}^{0} = 0 - (\frac{1}{3} - 1) = \frac{2}{3}$
Evaluating the second integral:
$\int_{0}^{1} (x^2 + 1) dx = \left[ \frac{x^3}{3} + x \right]_{0}^{1} = (\frac{1}{3} + 1) - 0 = \frac{4}{3}$
Total Area $= \frac{2}{3} + \frac{4}{3} = \frac{6}{3} = 2$ square units.

(C) The region $S(\alpha)$ is bounded by the parabola $y^2 = x$ and the vertical line $x = \alpha$.
The area $A(\alpha)$ is given by:
$A(\alpha) = \int_{0}^{\alpha} 2\sqrt{x} \, dx = 2 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{\alpha} = \frac{4}{3} \alpha^{3/2}$.
Given the ratio $A(\lambda) : A(4) = 2 : 5$,we have:
$\frac{\frac{4}{3} \lambda^{3/2}}{\frac{4}{3} 4^{3/2}} = \frac{2}{5}$
$\frac{\lambda^{3/2}}{8} = \frac{2}{5}$
$\lambda^{3/2} = \frac{16}{5}$
$\lambda = \left( \frac{16}{5} \right)^{2/3} = \left( \frac{16^2}{5^2} \right)^{1/3} = \left( \frac{256}{25} \right)^{1/3} = 4 \left( \frac{4}{25} \right)^{1/3}$.
Thus,$\lambda = 4\left(\frac{4}{25}\right)^{1/3}$.

(C) The region is bounded by $y = 2^x$ and $y = x + 1$ (since $x \ge 0$ in the first quadrant,$|x + 1| = x + 1$) from $x = 0$ to $x = 1$.
The required area is given by the integral:
$A = \int_{0}^{1} ((x + 1) - 2^x) dx$
Evaluating the integral:
$A = \left[ \frac{x^2}{2} + x - \frac{2^x}{\ln 2} \right]_{0}^{1}$
Substituting the limits:
$A = \left( \frac{1^2}{2} + 1 - \frac{2^1}{\ln 2} \right) - \left( \frac{0^2}{2} + 0 - \frac{2^0}{\ln 2} \right)$
$A = \left( \frac{1}{2} + 1 - \frac{2}{\ln 2} \right) - \left( 0 + 0 - \frac{1}{\ln 2} \right)$
$A = \frac{3}{2} - \frac{2}{\ln 2} + \frac{1}{\ln 2}$
$A = \frac{3}{2} - \frac{1}{\ln 2}$

(A) The given equation of the circle is $x^{2}+y^{2}=a^{2}$.
Since the circle is symmetrical about both the $x$-axis and the $y$-axis,the total area is $4$ times the area of the region $AOBA$ in the first quadrant.
Area $= 4 \int_{0}^{a} y \, dx$
Since $x^{2}+y^{2}=a^{2}$,we have $y = \sqrt{a^{2}-x^{2}}$ in the first quadrant.
Area $= 4 \int_{0}^{a} \sqrt{a^{2}-x^{2}} \, dx$
Using the standard integral formula $\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \left(\frac{x}{a}\right)$:
Area $= 4 \left[ \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \left(\frac{x}{a}\right) \right]_{0}^{a}$
$= 4 \left[ \left( \frac{a}{2} \sqrt{a^{2}-a^{2}} + \frac{a^{2}}{2} \sin^{-1}(1) \right) - (0 + 0) \right]$
$= 4 \left[ 0 + \frac{a^{2}}{2} \left(\frac{\pi}{2}\right) \right]$
$= 4 \left( \frac{\pi a^{2}}{4} \right) = \pi a^{2}$
Thus,the area enclosed by the circle is $\pi a^{2}$ square units.

(A) From the figure,the area of the region $ABA'B'A$ bounded by the ellipse is given by:
Area $= 4 \times (\text{Area of the region } AOBA \text{ in the first quadrant bounded by the curve, } x\text{-axis and the ordinates } x = 0, x = a)$
(as the ellipse is symmetrical about both $x$-axis and $y$-axis)
Area $= 4 \int_{0}^{a} y \, dx$ (taking vertical strips)
Now,$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ gives $y = \pm \frac{b}{a} \sqrt{a^{2}-x^{2}}$. Since the region $AOBA$ lies in the first quadrant,$y$ is taken as positive.
So,the required area is:
Area $= 4 \int_{0}^{a} \frac{b}{a} \sqrt{a^{2}-x^{2}} \, dx$
$= \frac{4b}{a} \left[ \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \left( \frac{x}{a} \right) \right]_{0}^{a}$
$= \frac{4b}{a} \left[ \left( \frac{a}{2} \times 0 + \frac{a^{2}}{2} \sin^{-1}(1) \right) - (0 + 0) \right]$
$= \frac{4b}{a} \left( \frac{a^{2}}{2} \times \frac{\pi}{2} \right) = \pi ab$.

(A) The given curve is $y=x^{2}$,which is a parabola symmetric about the $y$-axis with its vertex at the origin $(0,0)$.
The region is bounded by the curve $y=x^{2}$ and the line $y=4$. The intersection points of the curve and the line are found by substituting $y=4$ into $y=x^{2}$,which gives $x^{2}=4$,so $x=\pm 2$.
Since the parabola is symmetric about the $y$-axis,the total area $A$ is twice the area of the region in the first quadrant bounded by the curve,the $y$-axis,and the line $y=4$.
Using horizontal strips of width $dy$ and length $x = \sqrt{y}$,the area is given by:
$A = 2 \int_{0}^{4} x \, dy = 2 \int_{0}^{4} \sqrt{y} \, dy$
Evaluating the integral:
$A = 2 \left[ \frac{y^{3/2}}{3/2} \right]_{0}^{4} = 2 \times \frac{2}{3} \left[ y^{3/2} \right]_{0}^{4}$
$A = \frac{4}{3} \left( 4^{3/2} - 0^{3/2} \right) = \frac{4}{3} \times 8 = \frac{32}{3}$
Thus,the area of the region is $\frac{32}{3}$ square units.

(A) The given equations are:
$y=x$ ........... $(1)$
$x^{2}+y^{2}=32$ ........... $(2)$
Solving $(1)$ and $(2)$,we substitute $y=x$ in $(2)$ to get $x^{2}+x^{2}=32$,which gives $2x^{2}=32$,so $x^{2}=16$. Since we are in the first quadrant,$x=4$. Thus,the line and the circle meet at $B(4, 4)$.
Draw perpendicular $BM$ to the $x-$ axis,where $M$ is $(4, 0)$.
The required area is the area of the region $OBMO$ $+$ area of the region $BMAB$.
Area of region $OBMO = \int_{0}^{4} y \, dx = \int_{0}^{4} x \, dx = \left[\frac{x^{2}}{2}\right]_{0}^{4} = \frac{16}{2} = 8$.
Area of region $BMAB = \int_{4}^{4\sqrt{2}} y \, dx = \int_{4}^{4\sqrt{2}} \sqrt{(4\sqrt{2})^{2}-x^{2}} \, dx$.
Using the formula $\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2}\sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2}\sin^{-1}\left(\frac{x}{a}\right)$:
$= \left[\frac{x}{2}\sqrt{32-x^{2}} + \frac{32}{2}\sin^{-1}\left(\frac{x}{4\sqrt{2}}\right)\right]_{4}^{4\sqrt{2}}$
$= \left(0 + 16 \sin^{-1}(1)\right) - \left(\frac{4}{2}\sqrt{32-16} + 16 \sin^{-1}\left(\frac{4}{4\sqrt{2}}\right)\right)$
$= 16 \times \frac{\pi}{2} - \left(2 \times 4 + 16 \times \frac{\pi}{4}\right) = 8\pi - (8 + 4\pi) = 4\pi - 8$.
Total area $= 8 + (4\pi - 8) = 4\pi$.

(B) The area of the region bounded by the ellipse and the ordinates $x=0$ and $x=ae$ is symmetric about the $x$-axis.
The area is given by $A = 2 \int_{0}^{ae} y \, dx$.
From the equation of the ellipse,$y = \frac{b}{a} \sqrt{a^{2}-x^{2}}$.
Substituting this into the integral,we get $A = 2 \frac{b}{a} \int_{0}^{ae} \sqrt{a^{2}-x^{2}} \, dx$.
Using the standard integral $\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1}(\frac{x}{a})$,we have:
$A = \frac{2b}{a} \left[ \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1}(\frac{x}{a}) \right]_{0}^{ae}$
$A = \frac{2b}{a} \left[ \frac{ae}{2} \sqrt{a^{2}-a^{2}e^{2}} + \frac{a^{2}}{2} \sin^{-1}(\frac{ae}{a}) - (0 + 0) \right]$
$A = \frac{2b}{a} \left[ \frac{ae}{2} \cdot a\sqrt{1-e^{2}} + \frac{a^{2}}{2} \sin^{-1}(e) \right]$
$A = \frac{2b}{a} \cdot \frac{a^{2}}{2} \left[ e\sqrt{1-e^{2}} + \sin^{-1}(e) \right]$
$A = ab[e\sqrt{1-e^{2}} + \sin^{-1}(e)]$.

(A) The area of the region bounded by the curve $y^{2}=x$,the lines $x=1$ and $x=4$,and the $x$-axis in the first quadrant is given by the integral of $y$ with respect to $x$ from $x=1$ to $x=4$.
Since $y^{2}=x$ and we are in the first quadrant,$y = \sqrt{x}$.
Area $= \int_{1}^{4} y \, dx = \int_{1}^{4} \sqrt{x} \, dx$
$= \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right]_1^4$
$= \frac{2}{3} \left[ x^{\frac{3}{2}} \right]_1^4$
$= \frac{2}{3} \left( 4^{\frac{3}{2}} - 1^{\frac{3}{2}} \right)$
$= \frac{2}{3} (8 - 1)$
$= \frac{2}{3} \times 7 = \frac{14}{3}$ square units.

(A) The area of the region bounded by the curve $y^{2}=9x$,the lines $x=2$ and $x=4$,and the $x$-axis in the first quadrant is given by the integral:
Area $= \int_{2}^{4} y \, dx$
Since $y^{2}=9x$ and we are in the first quadrant,$y = \sqrt{9x} = 3\sqrt{x}$.
Area $= \int_{2}^{4} 3\sqrt{x} \, dx$
$= 3 \int_{2}^{4} x^{1/2} \, dx$
$= 3 \left[ \frac{x^{3/2}}{3/2} \right]_{2}^{4}$
$= 3 \times \frac{2}{3} \left[ x^{3/2} \right]_{2}^{4}$
$= 2 \left[ 4^{3/2} - 2^{3/2} \right]$
$= 2 \left[ 8 - 2\sqrt{2} \right]$
$= 16 - 4\sqrt{2} \text{ square units}$.

(A) The area of the region bounded by the curve $x^{2}=4y$,the lines $y=2$ and $y=4$,and the $y$-axis in the first quadrant is given by the integral of $x$ with respect to $y$ from $y=2$ to $y=4$.
Given $x^{2}=4y$,we have $x = \sqrt{4y} = 2\sqrt{y}$ (since it is in the first quadrant,$x > 0$).
Area $= \int_{2}^{4} x \, dy = \int_{2}^{4} 2\sqrt{y} \, dy$
$= 2 \int_{2}^{4} y^{1/2} \, dy$
$= 2 \left[ \frac{y^{3/2}}{3/2} \right]_{2}^{4} = 2 \times \frac{2}{3} \left[ y^{3/2} \right]_{2}^{4}$
$= \frac{4}{3} \left[ 4^{3/2} - 2^{3/2} \right]$
$= \frac{4}{3} \left[ 8 - 2\sqrt{2} \right]$
$= \left( \frac{32 - 8\sqrt{2}}{3} \right) \text{ square units}$

(A) The given equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$.
Here,$a^2 = 16 \implies a = 4$ and $b^2 = 9 \implies b = 3$.
The ellipse is symmetrical about both the $x-$axis and $y-$axis.
Therefore,the total area bounded by the ellipse is $4 \times$ (Area of the region in the first quadrant).
Area in the first quadrant $= \int_{0}^{4} y \, dx$.
From the equation,$y = 3 \sqrt{1 - \frac{x^2}{16}} = \frac{3}{4} \sqrt{16 - x^2}$.
Area $= \int_{0}^{4} \frac{3}{4} \sqrt{16 - x^2} \, dx = \frac{3}{4} \left[ \frac{x}{2} \sqrt{16 - x^2} + \frac{16}{2} \sin^{-1} \left( \frac{x}{4} \right) \right]_{0}^{4}$.
$= \frac{3}{4} \left[ (0 + 8 \sin^{-1}(1)) - (0 + 8 \sin^{-1}(0)) \right] = \frac{3}{4} \left[ 8 \times \frac{\pi}{2} \right] = \frac{3}{4} [4 \pi] = 3 \pi$.
Total area $= 4 \times 3 \pi = 12 \pi$ square units.

(A) The given equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$.
This can be written as $\frac{y^{2}}{9}=1-\frac{x^{2}}{4} \Rightarrow y^{2}=9\left(1-\frac{x^{2}}{4}\right) \Rightarrow y=3 \sqrt{1-\frac{x^{2}}{4}}$ (taking the positive value for the first quadrant).
Since the ellipse is symmetrical about both the $x$-axis and $y$-axis,the total area is $4$ times the area in the first quadrant.
Area $= 4 \int_{0}^{2} y \, dx = 4 \int_{0}^{2} 3 \sqrt{1-\frac{x^{2}}{4}} \, dx = 4 \times \frac{3}{2} \int_{0}^{2} \sqrt{4-x^{2}} \, dx = 6 \int_{0}^{2} \sqrt{2^{2}-x^{2}} \, dx$.
Using the formula $\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \left(\frac{x}{a}\right)$:
Area $= 6 \left[ \frac{x}{2} \sqrt{4-x^{2}} + \frac{4}{2} \sin^{-1} \left(\frac{x}{2}\right) \right]_{0}^{2}$
$= 6 \left[ (0 + 2 \sin^{-1}(1)) - (0 + 2 \sin^{-1}(0)) \right] = 6 \left[ 2 \times \frac{\pi}{2} \right] = 6 \pi$ square units.

(A) The area of the region bounded by the circle $x^{2}+y^{2}=4$,the line $x=\sqrt{3} y$,and the $x-$ axis in the first quadrant is the area of the region $OAB$ as shown in the figure.
The point of intersection of the line $x=\sqrt{3} y$ and the circle $x^{2}+y^{2}=4$ in the first quadrant is found by substituting $x=\sqrt{3} y$ into the circle equation:
$(\sqrt{3} y)^{2} + y^{2} = 4 \implies 3y^{2} + y^{2} = 4 \implies 4y^{2} = 4 \implies y = 1$.
Thus,$x = \sqrt{3}(1) = \sqrt{3}$. The intersection point is $A(\sqrt{3}, 1)$.
Area $OAB = \text{Area}(\triangle OCA) + \text{Area}(\text{region } ACB)$.
Area of $\triangle OCA = \frac{1}{2} \times OC \times AC = \frac{1}{2} \times \sqrt{3} \times 1 = \frac{\sqrt{3}}{2}$.
Area of region $ACB = \int_{\sqrt{3}}^{2} y \, dx = \int_{\sqrt{3}}^{2} \sqrt{4-x^{2}} \, dx$.
Using the formula $\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1}(\frac{x}{a})$:
$= \left[ \frac{x}{2} \sqrt{4-x^{2}} + \frac{4}{2} \sin^{-1}(\frac{x}{2}) \right]_{\sqrt{3}}^{2}$
$= \left( \frac{2}{2} \sqrt{4-4} + 2 \sin^{-1}(1) \right) - \left( \frac{\sqrt{3}}{2} \sqrt{4-3} + 2 \sin^{-1}(\frac{\sqrt{3}}{2}) \right)$
$= (0 + 2 \times \frac{\pi}{2}) - (\frac{\sqrt{3}}{2} \times 1 + 2 \times \frac{\pi}{3})$
$= \pi - \frac{\sqrt{3}}{2} - \frac{2\pi}{3} = \frac{\pi}{3} - \frac{\sqrt{3}}{2}$.
Total Area $OAB = \frac{\sqrt{3}}{2} + (\frac{\pi}{3} - \frac{\sqrt{3}}{2}) = \frac{\pi}{3}$ square units.

(A) The area of the smaller part of the circle $x^{2}+y^{2}=a^{2}$ cut off by the line $x=\frac{a}{\sqrt{2}}$ is the area of the region bounded by the circle and the line in the first and fourth quadrants.
Since the circle is symmetrical about the $x$-axis,the total area is twice the area of the region in the first quadrant.
Area $= 2 \int_{\frac{a}{\sqrt{2}}}^{a} y \, dx = 2 \int_{\frac{a}{\sqrt{2}}}^{a} \sqrt{a^{2}-x^{2}} \, dx$
Using the formula $\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \left(\frac{x}{a}\right)$:
Area $= 2 \left[ \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \left(\frac{x}{a}\right) \right]_{\frac{a}{\sqrt{2}}}^{a}$
$= 2 \left[ \left( 0 + \frac{a^{2}}{2} \sin^{-1}(1) \right) - \left( \frac{a}{2\sqrt{2}} \sqrt{a^{2}-\frac{a^{2}}{2}} + \frac{a^{2}}{2} \sin^{-1} \left(\frac{1}{\sqrt{2}}\right) \right) \right]$
$= 2 \left[ \frac{a^{2}}{2} \left(\frac{\pi}{2}\right) - \left( \frac{a}{2\sqrt{2}} \cdot \frac{a}{\sqrt{2}} + \frac{a^{2}}{2} \left(\frac{\pi}{4}\right) \right) \right]$
$= 2 \left[ \frac{a^{2}\pi}{4} - \frac{a^{2}}{4} - \frac{a^{2}\pi}{8} \right] = 2 \left[ \frac{2a^{2}\pi - a^{2}\pi}{8} - \frac{a^{2}}{4} \right] = 2 \left[ \frac{a^{2}\pi}{8} - \frac{a^{2}}{4} \right] = \frac{a^{2}\pi}{4} - \frac{a^{2}}{2} = \frac{a^{2}}{2} \left( \frac{\pi}{2} - 1 \right)$.

(A) The area bounded by the parabola $x=y^{2}$ and the line $x=4$ is symmetric about the $x$-axis.
The line $x=a$ divides this area into two equal parts.
Total area from $x=0$ to $x=4$ is $2 \int_{0}^{4} \sqrt{x} \, dx = 2 \left[ \frac{2}{3} x^{\frac{3}{2}} \right]_{0}^{4} = \frac{4}{3} (8) = \frac{32}{3}$.
The area from $x=0$ to $x=a$ is half of the total area:
$2 \int_{0}^{a} \sqrt{x} \, dx = \frac{1}{2} \times \frac{32}{3} = \frac{16}{3}$.
$2 \left[ \frac{2}{3} x^{\frac{3}{2}} \right]_{0}^{a} = \frac{16}{3}$.
$\frac{4}{3} a^{\frac{3}{2}} = \frac{16}{3}$.
$a^{\frac{3}{2}} = 4$.
$a = 4^{\frac{2}{3}}$.

(A) The area bounded by the parabola $y=x^{2}$ and the line $y=|x|$ is symmetric about the $y$-axis.
The points of intersection of $y=x^{2}$ and $y=|x|$ are found by setting $x^{2}=|x|$.
For $x \ge 0$,$x^{2}=x \implies x(x-1)=0$,so $x=0$ or $x=1$.
The intersection points are $(0,0)$ and $(1,1)$ in the first quadrant,and $(-1,1)$ in the second quadrant.
The required area is $2 \times \int_{0}^{1} (|x| - x^{2}) dx$.
Since $x \ge 0$ in the first quadrant,$|x|=x$.
Area $= 2 \int_{0}^{1} (x - x^{2}) dx$
$= 2 \left[ \frac{x^{2}}{2} - \frac{x^{3}}{3} \right]_{0}^{1}$
$= 2 \left( \frac{1}{2} - \frac{1}{3} \right)$
$= 2 \left( \frac{3-2}{6} \right) = 2 \left( \frac{1}{6} \right) = \frac{1}{3} \text{ sq. units.}$

(A) The region bounded by the parabola $y^{2}=4x$ and the line $x=3$ is the area $OAC$ (as shown in the figure).
The area is symmetrical about the $x$-axis.
Therefore,the required area $= 2 \times (\text{Area of } OAB)$.
Area $= 2 \int_{0}^{3} y \, dx$
Since $y^{2}=4x$,we have $y = 2\sqrt{x}$ (taking the positive value for the upper half).
Area $= 2 \int_{0}^{3} 2\sqrt{x} \, dx = 4 \int_{0}^{3} x^{1/2} \, dx$
$= 4 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{3} = 4 \times \frac{2}{3} \left[ x^{3/2} \right]_{0}^{3}$
$= \frac{8}{3} \left[ 3^{3/2} - 0^{3/2} \right] = \frac{8}{3} \times 3\sqrt{3} = 8\sqrt{3}$ square units.

(A) The given circle is $x^{2}+y^{2}=4$,which is a circle centered at the origin $(0,0)$ with radius $r=2$.
In the first quadrant,$y = \sqrt{4-x^{2}}$.
The area bounded by the circle,the $y$-axis $(x=0)$,and the line $x=2$ is the area of the quarter circle in the first quadrant.
Area $= \int_{0}^{2} y \, dx = \int_{0}^{2} \sqrt{4-x^{2}} \, dx$.
Using the formula $\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \left(\frac{x}{a}\right)$:
Area $= \left[ \frac{x}{2} \sqrt{4-x^{2}} + \frac{4}{2} \sin^{-1} \left(\frac{x}{2}\right) \right]_{0}^{2}$.
Substituting the limits:
$= \left( \frac{2}{2} \sqrt{4-4} + 2 \sin^{-1} \left(\frac{2}{2}\right) \right) - \left( 0 + 2 \sin^{-1}(0) \right)$.
$= (0 + 2 \sin^{-1}(1)) - (0) = 2 \times \frac{\pi}{2} = \pi$ square units.
Thus,the correct option is $A$.

(B) The area bounded by the curve $y^{2}=4x$,the $y$-axis,and the line $y=3$ is calculated by integrating $x$ with respect to $y$ from $y=0$ to $y=3$.
Given the curve $y^{2}=4x$,we can express $x$ as $x = \frac{y^{2}}{4}$.
The area $A$ is given by the integral:
$A = \int_{0}^{3} x \, dy$
Substituting $x = \frac{y^{2}}{4}$:
$A = \int_{0}^{3} \frac{y^{2}}{4} \, dy$
$A = \frac{1}{4} \left[ \frac{y^{3}}{3} \right]_{0}^{3}$
$A = \frac{1}{4} \left( \frac{3^{3}}{3} - \frac{0^{3}}{3} \right)$
$A = \frac{1}{4} \left( \frac{27}{3} \right)$
$A = \frac{1}{4} (9) = \frac{9}{4} \text{ sq. units}$.
Thus,the correct answer is $B$.

(A) The given equation of the circle $x^{2}+y^{2}=8x$ can be expressed as $(x-4)^{2}+y^{2}=16$. Thus,the centre of the circle is $(4, 0)$ and the radius is $4$.
Its intersection with the parabola $y^{2}=4x$ gives:
$x^{2}+4x=8x$
$x^{2}-4x=0$
$x(x-4)=0$
$x=0, x=4$
Thus,the points of intersection of these two curves are $O(0, 0)$ and $P(4, 4)$ above the $x-$ axis.
The required area of the region $OPQCO$ included between these two curves above the $x-$ axis is:
Area $= \int_{0}^{4} \sqrt{4x} \, dx + \int_{4}^{8} \sqrt{16-(x-4)^{2}} \, dx$
$= 2 \int_{0}^{4} x^{1/2} \, dx + \int_{0}^{4} \sqrt{4^{2}-t^{2}} \, dt$ (where $t = x-4$)
$= 2 \times \frac{2}{3} [x^{3/2}]_{0}^{4} + [\frac{t}{2} \sqrt{16-t^{2}} + \frac{16}{2} \sin^{-1}(\frac{t}{4})]_{0}^{4}$
$= \frac{4}{3} (8) + [0 + 8 \sin^{-1}(1) - 0]$
$= \frac{32}{3} + 8(\frac{\pi}{2}) = \frac{32}{3} + 4\pi = \frac{4}{3}(8+3\pi)$

(A) The given equation of the ellipse is $9x^{2} + y^{2} = 36$,which can be written as $\frac{x^{2}}{4} + \frac{y^{2}}{36} = 1$ or $\frac{x^{2}}{2^{2}} + \frac{y^{2}}{6^{2}} = 1$.
From the equation,$y = \sqrt{36 - 9x^{2}} = 3\sqrt{4 - x^{2}}$.
The equation of the chord $AB$ passing through $A(2, 0)$ and $B(0, 6)$ is given by $\frac{x}{2} + \frac{y}{6} = 1$,which simplifies to $y = 6 - 3x$.
The area of the shaded region is the area under the ellipse minus the area under the line $AB$ from $x = 0$ to $x = 2$.
Area $= \int_{0}^{2} (y_{\text{ellipse}} - y_{\text{line}}) dx = \int_{0}^{2} (3\sqrt{4 - x^{2}} - (6 - 3x)) dx$.
$= 3 \int_{0}^{2} \sqrt{2^{2} - x^{2}} dx - \int_{0}^{2} (6 - 3x) dx$.
Using the formula $\int \sqrt{a^{2} - x^{2}} dx = \frac{x}{2}\sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2}\sin^{-1}(\frac{x}{a})$:
$= 3 \left[ \frac{x}{2}\sqrt{4 - x^{2}} + \frac{4}{2}\sin^{-1}(\frac{x}{2}) \right]_{0}^{2} - \left[ 6x - \frac{3x^{2}}{2} \right]_{0}^{2}$.
$= 3 \left[ (0 + 2\sin^{-1}(1)) - (0 + 0) \right] - \left[ (12 - 6) - (0 - 0) \right]$.
$= 3 \left[ 2 \times \frac{\pi}{2} \right] - 6 = 3\pi - 6$ square units.

(A) Let $A(1,0)$,$B(2,2)$,and $C(3,1)$ be the vertices of a triangle $ABC$.
The equations of the sides are:
Side $AB$: $y - 0 = \frac{2-0}{2-1}(x - 1) \implies y = 2(x-1)$
Side $BC$: $y - 2 = \frac{1-2}{3-2}(x - 2) \implies y - 2 = -1(x-2) \implies y = 4-x$
Side $AC$: $y - 0 = \frac{1-0}{3-1}(x - 1) \implies y = \frac{1}{2}(x-1)$
The area of $\Delta ABC$ is given by:
Area $= \int_{1}^{2} (y_{AB} - y_{AC}) dx + \int_{2}^{3} (y_{BC} - y_{AC}) dx$
$= \int_{1}^{2} (2x - 2 - \frac{x-1}{2}) dx + \int_{2}^{3} (4 - x - \frac{x-1}{2}) dx$
$= \int_{1}^{2} (\frac{3x-3}{2}) dx + \int_{2}^{3} (\frac{9-3x}{2}) dx$
$= \frac{3}{2} [\frac{x^2}{2} - x]_{1}^{2} + \frac{3}{2} [3x - \frac{x^2}{2}]_{2}^{3}$
$= \frac{3}{2} [(2 - 2) - (\frac{1}{2} - 1)] + \frac{3}{2} [(9 - 4.5) - (6 - 2)]$
$= \frac{3}{2} [0.5] + \frac{3}{2} [4.5 - 4] = \frac{3}{2} (0.5) + \frac{3}{2} (0.5) = 0.75 + 0.75 = 1.5$ square units.

(A) The required area is represented by the shaded region $OBCDO$.
Solving the equations $4x^{2}+4y^{2}=9$ and $x^{2}=4y$,we substitute $x^{2}=4y$ into the circle equation:
$4(4y)+4y^{2}=9 \implies 4y^{2}+16y-9=0$.
Using the quadratic formula,$y = \frac{-16 \pm \sqrt{256 - 4(4)(-9)}}{8} = \frac{-16 \pm \sqrt{400}}{8} = \frac{-16 \pm 20}{8}$.
Since $y \ge 0$ for the parabola,$y = \frac{4}{8} = \frac{1}{2}$.
Then $x^{2} = 4(\frac{1}{2}) = 2$,so $x = \pm \sqrt{2}$. The intersection points are $B(\sqrt{2}, \frac{1}{2})$ and $D(-\sqrt{2}, \frac{1}{2})$.
The area is symmetric about the $y$-axis,so Area $= 2 \times \int_{0}^{\sqrt{2}} (y_{circle} - y_{parabola}) dx$.
$y_{circle} = \sqrt{\frac{9-4x^{2}}{4}} = \frac{1}{2}\sqrt{9-4x^{2}}$ and $y_{parabola} = \frac{x^{2}}{4}$.
Area $= 2 \int_{0}^{\sqrt{2}} (\frac{1}{2}\sqrt{9-4x^{2}} - \frac{x^{2}}{4}) dx = \int_{0}^{\sqrt{2}} \sqrt{9-4x^{2}} dx - \frac{1}{2} \int_{0}^{\sqrt{2}} x^{2} dx$.
Using $\int \sqrt{a^{2}-u^{2}} du = \frac{u}{2}\sqrt{a^{2}-u^{2}} + \frac{a^{2}}{2}\sin^{-1}(\frac{u}{a})$,let $u=2x, du=2dx$:
$= \frac{1}{2} [\frac{2x}{2}\sqrt{9-4x^{2}} + \frac{9}{2}\sin^{-1}(\frac{2x}{3})]_{0}^{\sqrt{2}} - \frac{1}{2} [\frac{x^{3}}{3}]_{0}^{\sqrt{2}}$
$= [\frac{x}{2}\sqrt{9-4x^{2}} + \frac{9}{4}\sin^{-1}(\frac{2x}{3})]_{0}^{\sqrt{2}} - \frac{1}{6}(\sqrt{2})^{3}$
$= (\frac{\sqrt{2}}{2}\sqrt{9-8} + \frac{9}{4}\sin^{-1}(\frac{2\sqrt{2}}{3})) - \frac{2\sqrt{2}}{6}$
$= \frac{\sqrt{2}}{2} + \frac{9}{4}\sin^{-1}(\frac{2\sqrt{2}}{3}) - \frac{\sqrt{2}}{3} = \frac{\sqrt{2}}{6} + \frac{9}{4}\sin^{-1}(\frac{2\sqrt{2}}{3})$ sq. units.

(A) The area bounded by the curves $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$ is represented by the shaded region.
On solving the equations $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$,we get:
$x^2 - 2x + 1 + y^2 = 1$
Since $x^2 + y^2 = 1$,we have $1 - 2x + 1 = 1$,which gives $2x = 1$,so $x = \frac{1}{2}$.
Substituting $x = \frac{1}{2}$ in $x^2 + y^2 = 1$,we get $y^2 = 1 - \frac{1}{4} = \frac{3}{4}$,so $y = \pm \frac{\sqrt{3}}{2}$.
The intersection points are $A\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and $B\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$.
The required area is symmetric about the $x$-axis.
Area $= 2 \times \int_{0}^{1/2} \sqrt{1 - (x-1)^2} dx + 2 \times \int_{1/2}^{1} \sqrt{1 - x^2} dx$.
Using the formula $\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$:
Area $= 2 \left[ \frac{x-1}{2}\sqrt{1-(x-1)^2} + \frac{1}{2}\sin^{-1}(x-1) \right]_0^{1/2} + 2 \left[ \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1}(x) \right]_{1/2}^1$
$= 2 \left[ (-\frac{1}{4}\frac{\sqrt{3}}{2} + \frac{1}{2}\sin^{-1}(-1/2)) - (0 + \frac{1}{2}\sin^{-1}(-1)) \right] + 2 \left[ (0 + \frac{1}{2}\sin^{-1}(1)) - (\frac{1}{4}\frac{\sqrt{3}}{2} + \frac{1}{2}\sin^{-1}(1/2)) \right]$
$= 2 \left[ -\frac{\sqrt{3}}{8} - \frac{\pi}{12} + \frac{\pi}{4} \right] + 2 \left[ \frac{\pi}{4} - \frac{\sqrt{3}}{8} - \frac{\pi}{12} \right]$
$= 2 \left[ \frac{\pi}{6} - \frac{\sqrt{3}}{8} \right] + 2 \left[ \frac{\pi}{6} - \frac{\sqrt{3}}{8} \right] = 4 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{8} \right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}$ sq. units.

(A) The area bounded by the curves $y=x^{2}+2, \,y=x, \,x=0,$ and $x=3$ is given by the integral of the difference between the upper curve and the lower curve within the given limits.
Area $= \int_{0}^{3} ((x^{2}+2) - x) \, dx$
$= \int_{0}^{3} (x^{2} - x + 2) \, dx$
$= \left[ \frac{x^{3}}{3} - \frac{x^{2}}{2} + 2x \right]_{0}^{3}$
$= \left( \frac{3^{3}}{3} - \frac{3^{2}}{2} + 2(3) \right) - (0)$
$= \left( \frac{27}{3} - \frac{9}{2} + 6 \right)$
$= 9 - 4.5 + 6$
$= 10.5 = \frac{21}{2} \text{ sq. units}$

(A) $BL$ and $CM$ are drawn perpendicular to the $x$-axis.
It can be observed in the figure that,
Area $(\Delta ABC) = \text{Area}(ALBA) + \text{Area}(BLMCB) - \text{Area}(AMCA)$ ...... $(1)$
The equation of the line segment $AB$ passing through $(-1, 0)$ and $(1, 3)$ is:
$y - 0 = \frac{3 - 0}{1 - (-1)}(x - (-1)) \implies y = \frac{3}{2}(x + 1)$
$\text{Area}(ALBA) = \int_{-1}^{1} \frac{3}{2}(x + 1) dx = \frac{3}{2} \left[ \frac{x^2}{2} + x \right]_{-1}^{1} = \frac{3}{2} \left[ (\frac{1}{2} + 1) - (\frac{1}{2} - 1) \right] = \frac{3}{2} [2] = 3$ sq. units.
The equation of the line segment $BC$ passing through $(1, 3)$ and $(3, 2)$ is:
$y - 3 = \frac{2 - 3}{3 - 1}(x - 1) \implies y - 3 = -\frac{1}{2}(x - 1) \implies y = -\frac{1}{2}x + \frac{7}{2}$
$\text{Area}(BLMCB) = \int_{1}^{3} (-\frac{1}{2}x + \frac{7}{2}) dx = \left[ -\frac{x^2}{4} + \frac{7x}{2} \right]_{1}^{3} = (-\frac{9}{4} + \frac{21}{2}) - (-\frac{1}{4} + \frac{7}{2}) = (\frac{33}{4}) - (\frac{13}{4}) = \frac{20}{4} = 5$ sq. units.
The equation of the line segment $AC$ passing through $(-1, 0)$ and $(3, 2)$ is:
$y - 0 = \frac{2 - 0}{3 - (-1)}(x - (-1)) \implies y = \frac{2}{4}(x + 1) \implies y = \frac{1}{2}(x + 1)$
$\text{Area}(AMCA) = \int_{-1}^{3} \frac{1}{2}(x + 1) dx = \frac{1}{2} \left[ \frac{x^2}{2} + x \right]_{-1}^{3} = \frac{1}{2} \left[ (\frac{9}{2} + 3) - (\frac{1}{2} - 1) \right] = \frac{1}{2} [\frac{15}{2} + \frac{1}{2}] = \frac{1}{2} [8] = 4$ sq. units.
Therefore,from equation $(1)$,
$\text{Area}(\Delta ABC) = 3 + 5 - 4 = 4$ sq. units.

(B) The equations of the sides of the triangle are $y=2x+1$,$y=3x+1$,and $x=4$.
On solving these equations,we obtain the vertices of the triangle as $A(0, 1)$,$B(4, 13)$,and $C(4, 9)$.
It can be observed that the area of the triangular region is the area between the two lines $y=3x+1$ and $y=2x+1$ from $x=0$ to $x=4$.
$\text{Area} = \int_{0}^{4} [(3x+1) - (2x+1)] \, dx$
$= \int_{0}^{4} (3x + 1 - 2x - 1) \, dx$
$= \int_{0}^{4} x \, dx$
$= \left[ \frac{x^2}{2} \right]_{0}^{4}$
$= \frac{16}{2} - 0$
$= 8 \text{ sq. units}$

(A) The circle is $x^{2}+y^{2}=2^{2}$,which has a radius of $2$ and center at $(0,0)$. The line $x+y=2$ passes through $(2,0)$ and $(0,2)$.
The smaller area enclosed by the circle and the line is the area of the circular segment in the first quadrant.
Area $= \int_{0}^{2} (y_{\text{circle}} - y_{\text{line}}) dx$
$= \int_{0}^{2} (\sqrt{4-x^{2}} - (2-x)) dx$
$= \int_{0}^{2} \sqrt{4-x^{2}} dx - \int_{0}^{2} (2-x) dx$
Using the formula $\int \sqrt{a^{2}-x^{2}} dx = \frac{x}{2}\sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2}\sin^{-1}(\frac{x}{a})$:
$= [\frac{x}{2}\sqrt{4-x^{2}} + \frac{4}{2}\sin^{-1}(\frac{x}{2})]_{0}^{2} - [2x - \frac{x^{2}}{2}]_{0}^{2}$
$= [0 + 2\sin^{-1}(1)] - [0 + 2\sin^{-1}(0)] - [(4 - 2) - (0 - 0)]$
$= 2(\frac{\pi}{2}) - 2 = \pi - 2$ sq. units.
Thus,the correct answer is $A$.

(B) To find the area between the curves $y^{2}=4x$ and $y=2x$,we first find their points of intersection.
Substituting $y=2x$ into $y^{2}=4x$,we get $(2x)^{2}=4x$,which simplifies to $4x^{2}-4x=0$,or $4x(x-1)=0$.
Thus,the points of intersection are $x=0$ and $x=1$.
For $x=0$,$y=0$,and for $x=1$,$y=2$. So the intersection points are $(0,0)$ and $(1,2)$.
The area $A$ is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=1$.
$A = \int_{0}^{1} (\text{upper curve} - \text{lower curve}) dx$
$A = \int_{0}^{1} (\sqrt{4x} - 2x) dx$
$A = \int_{0}^{1} (2\sqrt{x} - 2x) dx$
$A = 2 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1} - 2 \left[ \frac{x^{2}}{2} \right]_{0}^{1}$
$A = 2 \left( \frac{2}{3} \right) - 1 = \frac{4}{3} - 1 = \frac{1}{3}$ square units.
Thus,the correct answer is $B$.

(A) The vertex of the parabola $y^{2}=4ax$ is at the origin $(0,0)$.
The equation of the latus rectum $LSL^{\prime}$ is $x=a$.
The parabola is symmetrical about the $x$-axis.
The required area of the region $OLL^{\prime}O$ is given by:
Area $= 2 \times (\text{Area of the region } OLSO)$
Area $= 2 \int_{0}^{a} y \, dx = 2 \int_{0}^{a} \sqrt{4ax} \, dx$
Area $= 2 \times 2 \sqrt{a} \int_{0}^{a} \sqrt{x} \, dx$
Area $= 4 \sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{a}$
Area $= 4 \sqrt{a} \times \frac{2}{3} \left[ x^{3/2} \right]_{0}^{a}$
Area $= \frac{8}{3} \sqrt{a} \times a^{3/2} = \frac{8}{3} a^{2}$ square units.

(A) As shown in the figure,the line $y=3x+2$ meets the $x$-axis at $x=-2/3$. The graph lies below the $x$-axis for $x \in (-1, -2/3)$ and above the $x$-axis for $x \in (-2/3, 1)$.
The required area is given by:
$\text{Area} = \left| \int_{-1}^{-2/3} (3x+2) dx \right| + \int_{-2/3}^{1} (3x+2) dx$
First,evaluate the integral $\int (3x+2) dx = \frac{3x^2}{2} + 2x$.
For the first part:
$\left| \left[ \frac{3x^2}{2} + 2x \right]_{-1}^{-2/3} \right| = \left| \left( \frac{3(-2/3)^2}{2} + 2(-2/3) \right) - \left( \frac{3(-1)^2}{2} + 2(-1) \right) \right|$
$= \left| \left( \frac{3(4/9)}{2} - 4/3 \right) - \left( 3/2 - 2 \right) \right| = \left| (2/3 - 4/3) - (-1/2) \right| = \left| -2/3 + 1/2 \right| = \left| -1/6 \right| = 1/6$.
For the second part:
$\left[ \frac{3x^2}{2} + 2x \right]_{-2/3}^{1} = \left( \frac{3(1)^2}{2} + 2(1) \right) - \left( \frac{3(-2/3)^2}{2} + 2(-2/3) \right)$
$= (3/2 + 2) - (-1/6) = 7/2 + 1/6 = 21/6 + 1/6 = 22/6 = 11/3$.
Total Area $= 1/6 + 11/3 = 1/6 + 22/6 = 23/6$. Wait,let's re-calculate: $1/6 + 22/6 = 23/6$. Let's re-check the integral: $\int_{-2/3}^{1} (3x+2) dx = [3/2 + 2] - [2/3 - 4/3] = 7/2 - (-2/3) = 7/2 + 2/3 = 21/6 + 4/6 = 25/6$.
Total Area $= 1/6 + 25/6 = 26/6 = 13/3$.

(A) The area bounded by the curve $y=\cos x$ between $x=0$ and $x=2 \pi$ is given by the integral of the absolute value of the function.
Required Area $= \int_{0}^{2 \pi} |\cos x| \, dx$
From the graph,the curve is above the $x$-axis from $x=0$ to $x=\frac{\pi}{2}$,below the $x$-axis from $x=\frac{\pi}{2}$ to $x=\frac{3 \pi}{2}$,and above the $x$-axis from $x=\frac{3 \pi}{2}$ to $x=2 \pi$.
Thus,the area is:
$= \int_{0}^{\frac{\pi}{2}} \cos x \, dx + \left| \int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} \cos x \, dx \right| + \int_{\frac{3 \pi}{2}}^{2 \pi} \cos x \, dx$
$= [\sin x]_{0}^{\frac{\pi}{2}} + \left| [\sin x]_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} \right| + [\sin x]_{\frac{3 \pi}{2}}^{2 \pi}$
$= (\sin \frac{\pi}{2} - \sin 0) + |(\sin \frac{3 \pi}{2} - \sin \frac{\pi}{2})| + (\sin 2 \pi - \sin \frac{3 \pi}{2})$
$= (1 - 0) + |(-1 - 1)| + (0 - (-1))$
$= 1 + |-2| + 1 = 1 + 2 + 1 = 4$ square units.

(N/A) The points of intersection of the parabolas $y^{2}=4x$ and $x^{2}=4y$ are found by substituting $y = \frac{x^{2}}{4}$ into $y^{2}=4x$,which gives $(\frac{x^{2}}{4})^{2} = 4x$,so $x^{4} = 64x$. This implies $x(x^{3}-64) = 0$,so $x=0$ or $x=4$. Thus,the intersection points are $(0,0)$ and $(4,4)$.
$1$. The area of the region bounded by the curves $y^{2}=4x$ and $x^{2}=4y$ is:
$\int_{0}^{4} (2\sqrt{x} - \frac{x^{2}}{4}) dx = [2 \times \frac{2}{3} x^{\frac{3}{2}} - \frac{x^{3}}{12}]_{0}^{4} = \frac{32}{3} - \frac{16}{3} = \frac{16}{3}$ square units.
$2$. The area of the region bounded by the curve $x^{2}=4y$,the $x$-axis,and the lines $x=0$ and $x=4$ is:
$\int_{0}^{4} \frac{x^{2}}{4} dx = \frac{1}{12} [x^{3}]_{0}^{4} = \frac{64}{12} = \frac{16}{3}$ square units.
$3$. The area of the region bounded by the curve $y^{2}=4x$,the $y$-axis,and the lines $y=0$ and $y=4$ is:
$\int_{0}^{4} \frac{y^{2}}{4} dy = \frac{1}{12} [y^{3}]_{0}^{4} = \frac{64}{12} = \frac{16}{3}$ square units.
Since the total area of the square is $4 \times 4 = 16$ square units and each of the three regions has an area of $\frac{16}{3}$ square units,the curves divide the square into three equal parts.

(A) The region is defined by the intersection of $y \leq x^2 + 1$,$y \leq x + 1$,and $0 \leq x \leq 2$.
First,find the intersection points of the curves $y = x^2 + 1$ and $y = x + 1$:
$x^2 + 1 = x + 1 \implies x^2 - x = 0 \implies x(x - 1) = 0$.
So,the curves intersect at $x = 0$ and $x = 1$.
For $0 \leq x \leq 1$,the region is bounded by $y \leq x + 1$ (since $x+1 \geq x^2+1$ in this interval).
For $1 \leq x \leq 2$,the region is bounded by $y \leq x^2 + 1$ (since $x^2+1 \geq x+1$ in this interval).
Wait,looking at the graph provided,the region is bounded by the lower of the two curves at any point $x$.
For $0 \leq x \leq 1$,the lower curve is $y = x^2 + 1$.
For $1 \leq x \leq 2$,the lower curve is $y = x + 1$.
Area $= \int_{0}^{1} (x^2 + 1) dx + \int_{1}^{2} (x + 1) dx$.
$= [\frac{x^3}{3} + x]_{0}^{1} + [\frac{x^2}{2} + x]_{1}^{2}$.
$= (\frac{1}{3} + 1) - 0 + ((\frac{4}{2} + 2) - (\frac{1}{2} + 1))$.
$= \frac{4}{3} + (4 - \frac{3}{2}) = \frac{4}{3} + \frac{5}{2} = \frac{8 + 15}{6} = \frac{23}{6}$.

(A) The required area is represented by the shaded region bounded by the curve $y=x^{2}$,the lines $x=1$,$x=2$,and the $x$-axis.
The area is given by the definite integral:
$\text{Area} = \int_{1}^{2} y \, dx$
Substituting $y = x^{2}$:
$\text{Area} = \int_{1}^{2} x^{2} \, dx$
Evaluating the integral:
$\text{Area} = \left[ \frac{x^{3}}{3} \right]_{1}^{2}$
Applying the limits:
$\text{Area} = \left( \frac{2^{3}}{3} \right) - \left( \frac{1^{3}}{3} \right)$
$\text{Area} = \frac{8}{3} - \frac{1}{3}$
$\text{Area} = \frac{7}{3} \text{ sq. units}$

(A) The required area is given by the definite integral of the function $y=x^{4}$ from $x=1$ to $x=5$.
$\text{Area} = \int_{1}^{5} x^{4} \, dx$
$= \left[ \frac{x^{5}}{5} \right]_{1}^{5}$
$= \frac{(5)^{5}}{5} - \frac{(1)^{5}}{5}$
$= \frac{3125}{5} - \frac{1}{5}$
$= 625 - 0.2$
$= 624.8 \text{ sq. units}$

(A) The area in the first quadrant bounded by $y=4x^2$,$x=0$,$y=1$,and $y=4$ is calculated by integrating with respect to $y$.
Given $y=4x^2$,we have $x^2 = \frac{y}{4}$,which implies $x = \frac{\sqrt{y}}{2}$ (since $x \ge 0$ in the first quadrant).
The area $A$ is given by the integral:
$A = \int_{1}^{4} x \, dy$
$A = \int_{1}^{4} \frac{\sqrt{y}}{2} \, dy$
$A = \frac{1}{2} \int_{1}^{4} y^{1/2} \, dy$
$A = \frac{1}{2} \left[ \frac{y^{3/2}}{3/2} \right]_{1}^{4}$
$A = \frac{1}{2} \cdot \frac{2}{3} \left[ y^{3/2} \right]_{1}^{4}$
$A = \frac{1}{3} [4^{3/2} - 1^{3/2}]$
$A = \frac{1}{3} [8 - 1]$
$A = \frac{7}{3} \text{ sq. units}$.

(A) The given equation is $y=|x+3|$.
The corresponding values of $x$ and $y$ are given in the following table:

$X$	$-6$	$-5$	$-4$	$-3$	$-2$	$-1$	$0$
$Y$	$3$	$2$	$1$	$0$	$1$	$2$	$3$

On plotting these points,we obtain the graph of $y=|x+3|$.
It is known that $(x+3) \leq 0$ for $-6 \leq x \leq -3$ and $(x+3) \geq 0$ for $-3 \leq x \leq 0$.
Therefore,$\int_{-6}^{0}|x+3| d x = \int_{-6}^{-3}-(x+3) d x + \int_{-3}^{0}(x+3) d x$.
$= -\left[\frac{x^2}{2} + 3x\right]_{-6}^{-3} + \left[\frac{x^2}{2} + 3x\right]_{-3}^{0}$.
$= -\left[\left(\frac{9}{2} - 9\right) - \left(\frac{36}{2} - 18\right)\right] + \left[(0) - \left(\frac{9}{2} - 9\right)\right]$.
$= -\left[-\frac{9}{2} - 0\right] + \left[0 - (-\frac{9}{2})\right]$.
$= \frac{9}{2} + \frac{9}{2} = 9$.

(B) The graph of $y=\sin x$ is shown in the figure. The area bounded by the curve between $x=0$ and $x=2 \pi$ is given by the sum of the areas of the two loops.
$\text{Required Area} = \int_{0}^{\pi} \sin x \, dx + \left| \int_{\pi}^{2 \pi} \sin x \, dx \right|$
$= [-\cos x]_{0}^{\pi} + \left| [-\cos x]_{\pi}^{2 \pi} \right|$
$= (-\cos \pi - (-\cos 0)) + |(-\cos 2 \pi - (-\cos \pi))|$
$= (-(-1) + 1) + |(-1 - (1))|$
$= (1 + 1) + |-2|$
$= 2 + 2 = 4 \text{ sq. units}$.

(A) The area enclosed between the parabola $y^{2}=4ax$ and the line $y=mx$ is found by determining the points of intersection.
Substituting $y=mx$ into $y^{2}=4ax$,we get:
$(mx)^{2}=4ax \implies m^{2}x^{2}-4ax=0 \implies x(m^{2}x-4a)=0$.
Thus,the points of intersection are $x=0$ and $x=\frac{4a}{m^{2}}$.
The area $A$ is given by the integral of the difference between the upper curve (parabola) and the lower curve (line) from $x=0$ to $x=\frac{4a}{m^{2}}$:
$A = \int_{0}^{\frac{4a}{m^{2}}} (\sqrt{4ax} - mx) dx$
$A = 2\sqrt{a} \int_{0}^{\frac{4a}{m^{2}}} x^{1/2} dx - m \int_{0}^{\frac{4a}{m^{2}}} x dx$
$A = 2\sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{\frac{4a}{m^{2}}} - m \left[ \frac{x^{2}}{2} \right]_{0}^{\frac{4a}{m^{2}}}$
$A = \frac{4\sqrt{a}}{3} \left( \frac{4a}{m^{2}} \right)^{3/2} - \frac{m}{2} \left( \frac{4a}{m^{2}} \right)^{2}$
$A = \frac{4\sqrt{a}}{3} \cdot \frac{8a^{3/2}}{m^{3}} - \frac{m}{2} \cdot \frac{16a^{2}}{m^{4}}$
$A = \frac{32a^{2}}{3m^{3}} - \frac{8a^{2}}{m^{3}} = \frac{32a^{2} - 24a^{2}}{3m^{3}} = \frac{8a^{2}}{3m^{3}}$ sq. units.

(A) The area of the smaller region bounded by the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$ is the area of the region $ABC$ shown in the figure.
The points of intersection are $A(0, 2)$ and $B(3, 0)$.
Area of the region $ABC = \int_{0}^{3} (y_{\text{ellipse}} - y_{\text{line}}) dx$
From the ellipse equation,$y = 2\sqrt{1-\frac{x^{2}}{9}} = \frac{2}{3}\sqrt{9-x^{2}}$.
From the line equation,$y = 2(1-\frac{x}{3}) = 2 - \frac{2x}{3}$.
Area $= \int_{0}^{3} \left( \frac{2}{3}\sqrt{9-x^{2}} - (2 - \frac{2x}{3}) \right) dx$
$= \frac{2}{3} \int_{0}^{3} \sqrt{3^{2}-x^{2}} dx - \int_{0}^{3} (2 - \frac{2x}{3}) dx$
$= \frac{2}{3} \left[ \frac{x}{2}\sqrt{9-x^{2}} + \frac{9}{2}\sin^{-1}(\frac{x}{3}) \right]_{0}^{3} - \left[ 2x - \frac{x^{2}}{3} \right]_{0}^{3}$
$= \frac{2}{3} \left[ (0 + \frac{9}{2} \cdot \frac{\pi}{2}) - 0 \right] - \left[ (6 - 3) - 0 \right]$
$= \frac{2}{3} \cdot \frac{9\pi}{4} - 3 = \frac{3\pi}{2} - 3 = \frac{3}{2}(\pi-2)$ sq. units.

(A) The area of the smaller region bounded by the ellipse,$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,$ and the line,$\frac{x}{a}+\frac{y}{b}=1,$ is represented by the shaded region $BCAB$ as:
Area $BCAB = \text{Area}(OBCAO) - \text{Area}(OBAO)$
$= \int_{0}^{a} b \sqrt{1-\frac{x^{2}}{a^{2}}} \, dx - \int_{0}^{a} b\left(1-\frac{x}{a}\right) \, dx$
$= \frac{b}{a} \int_{0}^{a} \sqrt{a^{2}-x^{2}} \, dx - \frac{b}{a} \int_{0}^{a}(a-x) \, dx$
$= \frac{b}{a} \left[ \left\{ \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \frac{x}{a} \right\}_{0}^{a} - \left\{ ax - \frac{x^{2}}{2} \right\}_{0}^{a} \right]$
$= \frac{b}{a} \left[ \left\{ \frac{a^{2}}{2} \left(\frac{\pi}{2}\right) \right\} - \left\{ a^{2} - \frac{a^{2}}{2} \right\} \right]$
$= \frac{b}{a} \left[ \frac{a^{2}\pi}{4} - \frac{a^{2}}{2} \right]$
$= \frac{ba^{2}}{2a} \left[ \frac{\pi}{2} - 1 \right]$
$= \frac{ab}{2} \left[ \frac{\pi-2}{2} \right] = \frac{ab}{4}(\pi-2) \text{ sq. units.}$

(A) The area of the region enclosed by the parabola $x^{2}=y,$ the line $y=x+2,$ and the $x-$ axis is the shaded region shown in the figure.
The line $y=x+2$ intersects the $x-$ axis at $x=-2$ (where $y=0$).
The parabola $x^{2}=y$ and the line $y=x+2$ intersect where $x^{2}=x+2.$
$x^{2}-x-2=0 \Rightarrow (x-2)(x+1)=0.$
Since the region is in the second quadrant,we take the intersection point $x=-1.$
The required area is the sum of the area under the line from $x=-2$ to $x=-1$ and the area under the parabola from $x=-1$ to $x=0$ (or simply the area under the line from $x=-2$ to $x=-1$ plus the area under the parabola from $x=-1$ to $x=0$ is not correct; looking at the figure,it is the area under the line from $x=-2$ to $x=-1$ plus the area under the parabola from $x=-1$ to $x=0$ is actually the area bounded by the line and the $x-$ axis from $x=-2$ to $x=-1$ plus the area bounded by the parabola and the $x-$ axis from $x=-1$ to $x=0$).
Area $= \int_{-2}^{-1} (x+2) dx + \int_{-1}^{0} x^{2} dx$
$= \left[ \frac{x^{2}}{2} + 2x \right]_{-2}^{-1} + \left[ \frac{x^{3}}{3} \right]_{-1}^{0}$
$= \left( (\frac{1}{2} - 2) - (\frac{4}{2} - 4) \right) + (0 - (-\frac{1}{3}))$
$= (-\frac{3}{2} - (-2)) + \frac{1}{3}$
$= \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \text{ sq. units.}$

(A) The curve $|x|+|y|=1$ represents a square with vertices at $A(0,1)$,$B(1,0)$,$C(0,-1)$,and $D(-1,0)$.
The curve is symmetric about both the $x$-axis and the $y$-axis.
Therefore,the total area is $4$ times the area of the region in the first quadrant $(OBA)$.
In the first quadrant,$x \ge 0$ and $y \ge 0$,so the equation becomes $x+y=1$,or $y=1-x$.
$\text{Area} = 4 \int_{0}^{1} y \, dx = 4 \int_{0}^{1} (1-x) \, dx$
$= 4 \left[ x - \frac{x^2}{2} \right]_{0}^{1}$
$= 4 \left( (1 - \frac{1}{2}) - (0 - 0) \right)$
$= 4 \left( \frac{1}{2} \right) = 2 \text{ sq. units}$.