Area bounded by region of single curve Questions in English

(B) The equation of the line is $2y + x = 8$,which can be written as $y = \frac{8 - x}{2}$.
The required area is given by the definite integral $\int_{2}^{4} y \, dx = \int_{2}^{4} \frac{8 - x}{2} \, dx$.
$= \frac{1}{2} \int_{2}^{4} (8 - x) \, dx$
$= \frac{1}{2} \left[ 8x - \frac{x^2}{2} \right]_{2}^{4}$
$= \frac{1}{2} \left( (8(4) - \frac{4^2}{2}) - (8(2) - \frac{2^2}{2}) \right)$
$= \frac{1}{2} \left( (32 - 8) - (16 - 2) \right)$
$= \frac{1}{2} (24 - 14) = \frac{10}{2} = 5 \text{ sq. units}$.

(A) The area $A$ 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,$f(x) = -x^2$,$a = 1$,and $b = 4$.
Since $y = -x^2$ lies below the $x$-axis for $x \in [1, 4]$,the area is given by:
$A = \int_{1}^{4} | -x^2 | \, dx = \int_{1}^{4} x^2 \, dx$
$A = \left[ \frac{x^3}{3} \right]_{1}^{4}$
$A = \frac{4^3}{3} - \frac{1^3}{3}$
$A = \frac{64}{3} - \frac{1}{3} = \frac{63}{3} = 21 \text{ sq. units}$.

(C) The required area is the sum of the absolute values of the integrals over the intervals where the function is negative and positive.
$A = \int_{-1}^{0} |x| dx + \int_{0}^{2} |x| dx$
Since $x < 0$ for $x \in [-1, 0]$ and $x > 0$ for $x \in [0, 2]$,we have:
$A = \int_{-1}^{0} (-x) dx + \int_{0}^{2} (x) dx$
$A = \left[ -\frac{x^2}{2} \right]_{-1}^{0} + \left[ \frac{x^2}{2} \right]_{0}^{2}$
$A = (0 - (-\frac{(-1)^2}{2})) + (\frac{2^2}{2} - 0)$
$A = \frac{1}{2} + 2 = \frac{5}{2} \text{ sq. units}$

(D) The required area is given by the definite integral of the function $y=x^{3}$ from $x=1$ to $x=4$.
$\text{Area} = \int_{1}^{4} x^{3} dx$
Using the power rule for integration,$\int x^{n} dx = \frac{x^{n+1}}{n+1}$,we get:
$\text{Area} = \left[\frac{x^{4}}{4}\right]_{1}^{4}$
Now,apply the limits:
$\text{Area} = \frac{1}{4} [4^{4} - 1^{4}]$
$\text{Area} = \frac{1}{4} [256 - 1]$
$\text{Area} = \frac{255}{4} \text{ sq. units}$

(A) The parabola is $x^2 = 4y$,which implies $x = \pm 2\sqrt{y}$.
Since the area is bounded by the $Y$-axis and the parabola in the first quadrant,we consider $x = 2\sqrt{y}$.
The area $A$ bounded by the curve $x = f(y)$,the $Y$-axis,and the lines $y = 2$ and $y = 4$ is given by:
$A = \int_{2}^{4} x \, dy = \int_{2}^{4} 2\sqrt{y} \, dy$
$A = 2 \int_{2}^{4} y^{1/2} \, dy$
$A = 2 \left[ \frac{y^{3/2}}{3/2} \right]_{2}^{4} = 2 \cdot \frac{2}{3} \left[ y^{3/2} \right]_{2}^{4}$
$A = \frac{4}{3} [4^{3/2} - 2^{3/2}]$
$A = \frac{4}{3} [8 - 2\sqrt{2}]$
Thus,the area is $\frac{4}{3}(8 - 2\sqrt{2})$ sq. units.

(B) We have the curve $y=4x-x^{2}$.
To find the intersection points with the $x$-axis,we set $y=0$:
$x(4-x)=0 \Rightarrow x=0$ or $x=4$.
The required area is the integral of the function from $x=0$ to $x=4$:
$A = \int_{0}^{4} (4x-x^{2}) dx$
$A = \left[ \frac{4x^{2}}{2} - \frac{x^{3}}{3} \right]_{0}^{4}$
$A = \left[ 2x^{2} - \frac{x^{3}}{3} \right]_{0}^{4}$
$A = \left( 2(4)^{2} - \frac{(4)^{3}}{3} \right) - (0 - 0)$
$A = \left( 2(16) - \frac{64}{3} \right) = 32 - \frac{64}{3}$
$A = \frac{96-64}{3} = \frac{32}{3}$ sq. units.

(B) The equation of the parabola is $x^{2}=16y$,which gives $x=4\sqrt{y}$ for the first quadrant.
The area $A$ is bounded by the curve $x=4\sqrt{y}$,the $Y$-axis,and the lines $y=1$ and $y=4$.
$A = \int_{1}^{4} x \, dy = \int_{1}^{4} 4\sqrt{y} \, dy$
$A = 4 \int_{1}^{4} y^{1/2} \, dy$
$A = 4 \left[ \frac{y^{3/2}}{3/2} \right]_{1}^{4}$
$A = 4 \times \frac{2}{3} \left[ y^{3/2} \right]_{1}^{4}$
$A = \frac{8}{3} [4^{3/2} - 1^{3/2}]$
$A = \frac{8}{3} [8 - 1]$
$A = \frac{8}{3} \times 7 = \frac{56}{3} \text{ sq. units}$

(B) The area $A$ is given by the integral of the absolute value of the function:
$A = \int_{-\pi}^{\frac{3\pi}{2}} |\sin x| dx$
We split the integral based on the sign of $\sin x$:
$A = \int_{-\pi}^{0} |\sin x| dx + \int_{0}^{\pi} |\sin x| dx + \int_{\pi}^{\frac{3\pi}{2}} |\sin x| dx$
Since $\sin x \le 0$ for $x \in [-\pi, 0]$ and $x \in [\pi, \frac{3\pi}{2}]$,and $\sin x \ge 0$ for $x \in [0, \pi]$:
$A = \int_{-\pi}^{0} (-\sin x) dx + \int_{0}^{\pi} (\sin x) dx + \int_{\pi}^{\frac{3\pi}{2}} (-\sin x) dx$
$A = [\cos x]_{-\pi}^{0} + [-\cos x]_{0}^{\pi} + [\cos x]_{\pi}^{\frac{3\pi}{2}}$
$A = (\cos 0 - \cos(-\pi)) + (-\cos \pi + \cos 0) + (\cos(\frac{3\pi}{2}) - \cos \pi)$
$A = (1 - (-1)) + (-(-1) + 1) + (0 - (-1))$
$A = (1 + 1) + (1 + 1) + (0 + 1) = 2 + 2 + 1 = 5 \text{ (unit)}^2$

(C) The required area $A$ is given by the definite integral of the function $y = \sin^{2} x$ from $x = 0$ to $x = \frac{\pi}{2}$.
$A = \int_{0}^{\frac{\pi}{2}} \sin^{2} x \, dx$
Using the trigonometric identity $\sin^{2} x = \frac{1 - \cos 2x}{2}$,we get:
$A = \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos 2x}{2} \, dx$
$A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (1 - \cos 2x) \, dx$
$A = \frac{1}{2} \left[ x - \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{2}}$
$A = \frac{1}{2} \left[ \left( \frac{\pi}{2} - \frac{\sin \pi}{2} \right) - (0 - 0) \right]$
$A = \frac{1}{2} \left[ \frac{\pi}{2} - 0 \right] = \frac{\pi}{4}$ sq. units.

(C) The area $A$ is given by the integral of $|y|$ from $x=0$ to $x=\pi$.
Since $y = \cos x$ is positive in $[0, \pi/2]$ and negative in $[\pi/2, \pi]$,the area is:
$A = \int_0^{\pi/2} \cos x \, dx + \left| \int_{\pi/2}^{\pi} \cos x \, dx \right|$
$A = [\sin x]_0^{\pi/2} + |[\sin x]_{\pi/2}^{\pi}|$
$A = (\sin(\pi/2) - \sin(0)) + |\sin(\pi) - \sin(\pi/2)|$
$A = (1 - 0) + |0 - 1|$
$A = 1 + |-1| = 1 + 1 = 2 \text{ sq. units.}$

(D) The region is bounded by $y=3x+1$ (upper line),$y=2x+1$ (lower line),and $x=4$. The lines intersect at $x=0$,where $y=1$. Thus,the vertices of the triangle are $(0, 1)$,$(4, 9)$,and $(4, 13)$.
The area $A$ can be calculated using integration:
$A = \int_{0}^{4} [(3x+1) - (2x+1)] \, dx$
$A = \int_{0}^{4} x \, dx$
$A = \left[ \frac{x^2}{2} \right]_{0}^{4}$
$A = \frac{16}{2} - 0 = 8 \text{ sq. units}$.
Alternatively,using the formula for the area of a triangle with base on the line $x=4$:
Base length $= 13 - 9 = 4$.
Height $= 4 - 0 = 4$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8 \text{ sq. units}$.

(B) To find the area bounded by the curve $y=2x-x^2$ and the $x$-axis,we first find the intersection points with the $x$-axis by setting $y=0$.
$2x-x^2=0 \Rightarrow x(2-x)=0$,which gives $x=0$ and $x=2$.
The area $A$ is given by the integral:
$A = \int_0^2 (2x-x^2) dx$
$A = [x^2 - \frac{x^3}{3}]_0^2$
$A = (2^2 - \frac{2^3}{3}) - (0^2 - \frac{0^3}{3})$
$A = 4 - \frac{8}{3} = \frac{12-8}{3} = \frac{4}{3} \text{ sq units}$.

(A) The required area is given by the integral of $|y|$ with respect to $x$ from $x=-1$ to $x=2$.
Since $y=x$,we have $|y|=|x|$.
Required Area $= \int_{-1}^{2} |x| dx$
$= \int_{-1}^{0} |x| dx + \int_{0}^{2} |x| dx$
$= \int_{-1}^{0} (-x) dx + \int_{0}^{2} x dx$
$= \left[ -\frac{x^2}{2} \right]_{-1}^{0} + \left[ \frac{x^2}{2} \right]_{0}^{2}$
$= (0 - (-1/2)) + (4/2 - 0)$
$= 1/2 + 2 = 5/2$ sq unit.

(D) The equation of the ellipse is $\frac{x^2}{5^2} + \frac{y^2}{3^2} = 1$.
Here,$a=5$ and $b=3$.
The area of the positive quadrant $OAB$ is $\frac{1}{4} \times \pi ab = \frac{1}{4} \times \pi \times 5 \times 3 = \frac{15\pi}{4}$.
The area of the triangle $OAB$ with vertices $(0,0), (5,0), (0,3)$ is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 3 = \frac{15}{2}$.
The area between the arc $AB$ and the chord $AB$ is the area of the quadrant minus the area of the triangle.
Area $= \frac{15\pi}{4} - \frac{15}{2} = \frac{15}{4}(\pi - 2)$ sq. units.

(C) Given equations are $x^2 = 8y$ $(1)$ and $x - 8y + 2 = 0$ $(2)$.
From $(2)$,$8y = x + 2$. Substituting this into $(1)$,we get $x^2 = x + 2$,which implies $x^2 - x - 2 = 0$.
Factoring the quadratic equation: $(x - 2)(x + 1) = 0$,so the intersection points are $x = -1$ and $x = 2$.
The area $A$ is given by the integral of the upper curve minus the lower curve from $x = -1$ to $x = 2$.
$A = \int_{-1}^{2} (\frac{x+2}{8} - \frac{x^2}{8}) dx = \frac{1}{8} \int_{-1}^{2} (x + 2 - x^2) dx$.
$A = \frac{1}{8} [\frac{x^2}{2} + 2x - \frac{x^3}{3}]_{-1}^{2}$.
Evaluating at the limits: $A = \frac{1}{8} [(\frac{4}{2} + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3})]$.
$A = \frac{1}{8} [(2 + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3})] = \frac{1}{8} [\frac{10}{3} - (-\frac{7}{6})] = \frac{1}{8} [\frac{20+7}{6}] = \frac{1}{8} \times \frac{27}{6} = \frac{9}{16}$ sq. units.

(A) The two curves $y=ax^2$ and $x=ay^2$ intersect at $O(0,0)$ and $P\left(\frac{1}{a}, \frac{1}{a}\right)$.
To find the area bounded by these curves,we integrate the difference between the upper curve $y=\sqrt{\frac{x}{a}}$ and the lower curve $y=ax^2$ from $x=0$ to $x=\frac{1}{a}$.
Given area $= \int_0^{\frac{1}{a}} \left(\sqrt{\frac{x}{a}} - ax^2\right) dx = 1$
$\Rightarrow \left[\frac{1}{\sqrt{a}} \cdot \frac{x^{3/2}}{3/2} - \frac{ax^3}{3}\right]_0^{\frac{1}{a}} = 1$
$\Rightarrow \left[\frac{2}{3\sqrt{a}} x^{3/2} - \frac{ax^3}{3}\right]_0^{\frac{1}{a}} = 1$
Substituting the limits:
$\Rightarrow \left(\frac{2}{3\sqrt{a}} \cdot \left(\frac{1}{a}\right)^{3/2} - \frac{a}{3} \cdot \left(\frac{1}{a}\right)^3\right) = 1$
$\Rightarrow \frac{2}{3\sqrt{a} \cdot a\sqrt{a}} - \frac{a}{3a^3} = 1$
$\Rightarrow \frac{2}{3a^2} - \frac{1}{3a^2} = 1$
$\Rightarrow \frac{1}{3a^2} = 1$
$\Rightarrow a^2 = \frac{1}{3}$
Since $a > 0$,we have $a = \frac{1}{\sqrt{3}}$.

(B) The curves $y=3x+1$ and $y=4x+1$ intersect when $3x+1 = 4x+1$,which implies $x=0$.
Thus,the region is bounded by $x=0$ and $x=2$.
The required area is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=2$.
$\text{Required area} = \int_0^2 [(4x+1) - (3x+1)] \, dx$
$= \int_0^2 x \, dx$
$= \left[ \frac{x^2}{2} \right]_0^2$
$= \frac{2^2}{2} - \frac{0^2}{2} = \frac{4}{2} = 2 \text{ sq. units}$.

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

(C) The equation of the circle is $x^2+y^2=a^2$. The line is $x=\frac{a}{\sqrt{2}}$.
Substituting $x=\frac{a}{\sqrt{2}}$ in the circle equation,we get $\frac{a^2}{2}+y^2=a^2$,which implies $y^2=\frac{a^2}{2}$,so $y=\pm\frac{a}{\sqrt{2}}$.
The required area is the area of the region bounded by the circle and the line $x=\frac{a}{\sqrt{2}}$ on the right side.
Area $= 2 \int_{\frac{a}{\sqrt{2}}}^a \sqrt{a^2-x^2} dx$
Using the formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$:
Area $= 2 \left[ \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a}) \right]_{\frac{a}{\sqrt{2}}}^a$
$= 2 \left[ (0 + \frac{a^2}{2}\sin^{-1}(1)) - (\frac{a}{2\sqrt{2}}\sqrt{a^2-\frac{a^2}{2}} + \frac{a^2}{2}\sin^{-1}(\frac{1}{\sqrt{2}})) \right]$
$= 2 \left[ \frac{a^2}{2}(\frac{\pi}{2}) - (\frac{a}{2\sqrt{2}}\cdot\frac{a}{\sqrt{2}} + \frac{a^2}{2}(\frac{\pi}{4})) \right]$
$= 2 \left[ \frac{a^2\pi}{4} - \frac{a^2}{4} - \frac{a^2\pi}{8} \right]$
$= 2 \left[ \frac{a^2\pi}{8} - \frac{a^2}{4} \right] = \frac{a^2\pi}{4} - \frac{a^2}{2} = \frac{a^2}{2}(\frac{\pi}{2}-1)$.

(A) The given curve is $y = a\sqrt{x} + bx$. Since it passes through $(1, 2)$,we have $2 = a(1) + b(1)$,which gives $a + b = 2$ ...$(i)$.
The area bounded by the curve,$x = 4$,and the $X$-axis is given by $\int_0^4 (a\sqrt{x} + bx) dx = 8$.
Evaluating the integral:
$\int_0^4 (ax^{1/2} + bx) dx = \left[ a \cdot \frac{x^{3/2}}{3/2} + b \cdot \frac{x^2}{2} \right]_0^4 = 8$.
Substituting the limits:
$\left( \frac{2a}{3} \cdot 4^{3/2} + \frac{b}{2} \cdot 4^2 \right) = 8$.
$\frac{2a}{3} \cdot 8 + \frac{b}{2} \cdot 16 = 8$.
$\frac{16a}{3} + 8b = 8$.
Dividing by $8$,we get $\frac{2a}{3} + b = 1$,which simplifies to $2a + 3b = 3$ ...(ii).
Solving equations $(i)$ and (ii):
From $(i)$,$b = 2 - a$. Substituting into (ii):
$2a + 3(2 - a) = 3$.
$2a + 6 - 3a = 3$.
$-a = -3 \Rightarrow a = 3$.
Substituting $a = 3$ into $(i)$:
$3 + b = 2 \Rightarrow b = -1$.
Thus,$a = 3$ and $b = -1$.

(C) To find the area bounded by the curve $y^2=9x$ and the line $y=3x$,we first find the points of intersection.
Substituting $y=3x$ into $y^2=9x$,we get $(3x)^2=9x$,which implies $9x^2=9x$,so $x^2-x=0$,giving $x(x-1)=0$. Thus,$x=0$ and $x=1$.
For $x=0$,$y=0$. For $x=1$,$y=3$.
The area is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=1$.
$\text{Area} = \int_0^1 (\sqrt{9x} - 3x) dx$
$= \int_0^1 (3\sqrt{x} - 3x) dx$
$= 3 \left[ \frac{2}{3}x^{3/2} - \frac{x^2}{2} \right]_0^1$
$= 3 \left( \frac{2}{3}(1)^{3/2} - \frac{(1)^2}{2} \right) - 0$
$= 3 \left( \frac{2}{3} - \frac{1}{2} \right)$
$= 3 \left( \frac{4-3}{6} \right) = 3 \times \frac{1}{6} = \frac{1}{2} \text{ sq.units}$.

(A) The area $A$ of the region bounded by the $y$-axis $(x=0)$,$y=\cos x$,and $y=\sin x$ for $0 \leq x \leq \frac{\pi}{4}$ is given by the integral of the upper curve minus the lower curve.
In the interval $[0, \frac{\pi}{4}]$,$\cos x \geq \sin x$.
Therefore,the area $A$ is:
$A = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx$
$A = [\sin x - (-\cos x)]_{0}^{\frac{\pi}{4}}$
$A = [\sin x + \cos x]_{0}^{\frac{\pi}{4}}$
$A = (\sin \frac{\pi}{4} + \cos \frac{\pi}{4}) - (\sin 0 + \cos 0)$
$A = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1)$
$A = \frac{2}{\sqrt{2}} - 1$
$A = \sqrt{2} - 1$ sq. units.

(D) To find the area bounded by the parabola $y=x^2$ and the line $y=x$,we first determine the points of intersection by setting the equations equal to each other:
$x^2 = x$
$x^2 - x = 0$
$x(x - 1) = 0$
This gives $x = 0$ and $x = 1$. The points of intersection are $O(0, 0)$ and $P(1, 1)$.
In the interval $[0, 1]$,the line $y = x$ lies above the parabola $y = x^2$.
The required area $A$ is given by the integral:
$A = \int_0^1 (x - x^2) dx$
$A = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1$
$A = \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - (0 - 0)$
$A = \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6} \text{ sq. units}$

(A) To find the area of the region bounded by the curve $y^2=4x$ and the line $y=x$,we first determine the points of intersection by substituting $y=x$ into the equation of the parabola:
$x^2 = 4x$
$x^2 - 4x = 0$
$x(x-4) = 0$
Thus,the points of intersection are $x=0$ and $x=4$.
For $x=0$,$y=0$,so the origin $O(0,0)$ is one point.
For $x=4$,$y=4$,so the point $P(4,4)$ is the other point.
The required area $A$ is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=4$:
$A = \int_0^4 (\sqrt{4x} - x) dx$
$A = 2 \int_0^4 x^{1/2} dx - \int_0^4 x dx$
$A = 2 \left[ \frac{x^{3/2}}{3/2} \right]_0^4 - \left[ \frac{x^2}{2} \right]_0^4$
$A = 2 \cdot \frac{2}{3} [x^{3/2}]_0^4 - \left[ \frac{16}{2} - 0 \right]$
$A = \frac{4}{3} (4^{3/2}) - 8$
$A = \frac{4}{3} (8) - 8$
$A = \frac{32}{3} - 8 = \frac{32-24}{3} = \frac{8}{3} \text{ sq. units.}$

(B) The region is bounded by the parabola $y^2=x$,the line $y=4$,and the $Y$-axis $(x=0)$.
To find the area,we integrate with respect to $y$ from $y=0$ to $y=4$.
The equation of the parabola is $x=y^2$.
The area $A$ is given by:
$A = \int_{0}^{4} x \, dy$
$A = \int_{0}^{4} y^2 \, dy$
$A = \left[ \frac{y^3}{3} \right]_{0}^{4}$
$A = \frac{4^3}{3} - \frac{0^3}{3}$
$A = \frac{64}{3} \text{ sq. units}$

(A) To find the area bounded between the curve $x^2=y$ and the line $y=4x$,we first find their points of intersection.
Setting $x^2 = 4x$,we get $x^2 - 4x = 0$,which implies $x(x-4) = 0$. Thus,$x=0$ and $x=4$.
For $x=0$,$y=0$,and for $x=4$,$y=16$. So the points of intersection are $(0,0)$ and $(4,16)$.
The required area $A$ is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=4$:
$A = \int_{0}^{4} (4x - x^2) dx$
$A = \left[ \frac{4x^2}{2} - \frac{x^3}{3} \right]_{0}^{4}$
$A = \left[ 2x^2 - \frac{x^3}{3} \right]_{0}^{4}$
$A = \left( 2(4)^2 - \frac{(4)^3}{3} \right) - (0 - 0)$
$A = \left( 32 - \frac{64}{3} \right)$
$A = \frac{96 - 64}{3} = \frac{32}{3} \text{ sq. units}$

(C) Given the parabola $y^{2}=8x$. Comparing this with $y^{2}=4ax$,we get $4a=8$,so $a=2$.
The latus rectum is the line $x=a$,which is $x=2$.
The points of intersection of the parabola and the latus rectum are $(2, 4)$ and $(2, -4)$.
The area $A$ bounded by the parabola and the latus rectum is symmetric about the $x$-axis.
$A = 2 \int_{0}^{2} y \, dx = 2 \int_{0}^{2} \sqrt{8x} \, dx$
$A = 2 \times 2\sqrt{2} \int_{0}^{2} x^{1/2} \, dx = 4\sqrt{2} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{2}$
$A = 4\sqrt{2} \times \frac{2}{3} \times (2)^{3/2} = \frac{8\sqrt{2}}{3} \times 2\sqrt{2} = \frac{8 \times 2 \times 2}{3} = \frac{32}{3}$ sq. units.

(A) The given equation of the circle is $x^{2}+y^{2}=16$.
Since the circle is symmetric about both the $x$-axis and $y$-axis,the total area bounded by the lines $x=0$ and $x=2$ is twice the area of the region in the first quadrant.
$\text{Area} = 2 \int_{0}^{2} y \, dx = 2 \int_{0}^{2} \sqrt{16-x^{2}} \, dx$.
Using the formula $\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \left(\frac{x}{a}\right)$:
$\text{Area} = 2 \left[ \frac{x}{2} \sqrt{16-x^{2}} + \frac{16}{2} \sin^{-1} \left(\frac{x}{4}\right) \right]_{0}^{2}$
$= 2 \left[ \left( \frac{2}{2} \sqrt{16-4} + 8 \sin^{-1} \left(\frac{2}{4}\right) \right) - (0 + 8 \sin^{-1}(0)) \right]$
$= 2 \left[ \sqrt{12} + 8 \sin^{-1} \left(\frac{1}{2}\right) \right]$
$= 2 \left[ 2\sqrt{3} + 8 \left(\frac{\pi}{6}\right) \right]$
$= 4\sqrt{3} + 8 \left(\frac{\pi}{3}\right) = 4\sqrt{3} + \frac{8\pi}{3} \text{ sq. units.}$

(B) The given parabola is $y^{2}=16x$. Comparing this with $y^{2}=4ax$,we get $4a=16$,which implies $a=4$.
The latus rectum is the line $x=a$,so $x=4$.
The points of intersection of the parabola and the latus rectum are $(4, 8)$ and $(4, -8)$.
In the first quadrant,the area is bounded by the curve $y=\sqrt{16x}=4\sqrt{x}$,the $x$-axis,and the line $x=4$.
The required area is given by the integral:
$\text{Area} = \int_{0}^{4} y \, dx = \int_{0}^{4} 4\sqrt{x} \, dx$
$= 4 \int_{0}^{4} x^{\frac{1}{2}} \, dx$
$= 4 \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right]_{0}^{4}$
$= 4 \times \frac{2}{3} \left[ x^{\frac{3}{2}} \right]_{0}^{4}$
$= \frac{8}{3} \left[ 4^{\frac{3}{2}} - 0^{\frac{3}{2}} \right]$
$= \frac{8}{3} \times 8 = \frac{64}{3} \text{ sq. units}$.

(B) 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 the integral $A = \int_{a}^{b} f(x) dx$.
Here,$f(x) = x^{2}+1$,$a=1$,and $b=2$.
Therefore,the area is:
$A = \int_{1}^{2} (x^{2}+1) dx$
$A = \left[ \frac{x^{3}}{3} + x \right]_{1}^{2}$
$A = \left( \frac{2^{3}}{3} + 2 \right) - \left( \frac{1^{3}}{3} + 1 \right)$
$A = \left( \frac{8}{3} + 2 \right) - \left( \frac{1}{3} + 1 \right)$
$A = \left( \frac{8+6}{3} \right) - \left( \frac{1+3}{3} \right)$
$A = \frac{14}{3} - \frac{4}{3}$
$A = \frac{10}{3} \text{ sq. units}$

(A) The required area is given by the definite integral of the function $y$ with respect to $x$ from $x=1$ to $x=5$. Since the curve is above the $x$-axis in the interval $[1, 5]$,the area is:
$\text{Area} = \int_{1}^{5} (4x^{3}-6x^{2}+4x+1) dx$
$= \left[\frac{4x^{4}}{4} - \frac{6x^{3}}{3} + \frac{4x^{2}}{2} + x\right]_{1}^{5}$
$= \left[x^{4} - 2x^{3} + 2x^{2} + x\right]_{1}^{5}$
$= [5^{4} - 2(5)^{3} + 2(5)^{2} + 5] - [1^{4} - 2(1)^{3} + 2(1)^{2} + 1]$
$= [625 - 250 + 50 + 5] - [1 - 2 + 2 + 1]$
$= 430 - 2 = 428 \text{ sq. units}$
Thus,the correct option is $A$.

(A) Given curves are $y=2x-x^2$ and $y=x$.
To find the points of intersection,set $2x-x^2 = x$.
$x-x^2 = 0 \implies x(1-x) = 0$.
Thus,the points of intersection are $x=0$ and $x=1$.
The required area is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=1$.
$\text{Area} = \int_0^1 (y_{\text{upper}} - y_{\text{lower}}) dx = \int_0^1 ((2x-x^2) - x) dx$.
$\text{Area} = \int_0^1 (x-x^2) dx$.
Evaluating the integral: $\left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1$.
$= (\frac{1}{2} - \frac{1}{3}) - (0 - 0) = \frac{3-2}{6} = \frac{1}{6}$ square units.

(B) The region is bounded by the parabola $x^2=4y$,the lines $y=1$ and $y=4$,and the y-axis in the first quadrant.
From $x^2=4y$,we have $x = \sqrt{4y} = 2\sqrt{y}$ (since $x > 0$ in the first quadrant).
The required area $A$ is given by the integral with respect to $y$:
$A = \int_{1}^{4} x \, dy = \int_{1}^{4} 2\sqrt{y} \, dy$
$A = 2 \int_{1}^{4} y^{1/2} \, dy$
$A = 2 \left[ \frac{y^{3/2}}{3/2} \right]_{1}^{4}$
$A = 2 \times \frac{2}{3} \left[ y^{3/2} \right]_{1}^{4}$
$A = \frac{4}{3} (4^{3/2} - 1^{3/2})$
$A = \frac{4}{3} (8 - 1)$
$A = \frac{4}{3} (7) = \frac{28}{3}$ square units.

(B) The line $y=4$ meets the parabola $y^{2}=x$ at point $A$. Substituting $y=4$ into the equation of the parabola,we get $4^{2}=x$,so $x=16$. Thus,point $A$ is $(16, 4)$.
The required area is bounded by the parabola $x=y^{2}$,the $y$-axis $(x=0)$,and the line $y=4$ from $y=0$ to $y=4$.
Required Area $= \int_{0}^{4} x \, dy = \int_{0}^{4} y^{2} \, dy$
$= \left[ \frac{y^{3}}{3} \right]_{0}^{4}$
$= \frac{4^{3}}{3} - \frac{0^{3}}{3} = \frac{64}{3} \text{ sq. units}$.

(C) The curves are $y=x^{2}$ (or $x=\sqrt{y}$) and $x=y^{2}$.
They intersect at $(0,0)$ and $(1,1)$.
When revolving about the $y$-axis,the volume $V$ is given by the formula $V = \pi \int_{a}^{b} (x_{outer}^{2} - x_{inner}^{2}) dy$.
Here,for $y \in [0,1]$,the outer curve is $x = \sqrt{y}$ and the inner curve is $x = y^{2}$.
Thus,$V = \pi \int_{0}^{1} ((\sqrt{y})^{2} - (y^{2})^{2}) dy$
$V = \pi \int_{0}^{1} (y - y^{4}) dy$
$V = \pi \left[ \frac{y^{2}}{2} - \frac{y^{5}}{5} \right]_{0}^{1}$
$V = \pi \left( \frac{1}{2} - \frac{1}{5} \right) = \pi \left( \frac{5-2}{10} \right) = \frac{3}{10} \pi$.

(C) The curves are $y=(x+1)^2$ and $y=(x-1)^2$. The line is $y=\frac{1}{4}$.
To find the intersection of $y=(x-1)^2$ and $y=\frac{1}{4}$,we set $(x-1)^2 = \frac{1}{4}$,which gives $x-1 = \pm \frac{1}{2}$,so $x = \frac{1}{2}$ or $x = \frac{3}{2}$.
Similarly,for $y=(x+1)^2$ and $y=\frac{1}{4}$,we get $x = -\frac{1}{2}$ or $x = -\frac{3}{2}$.
The region is symmetric about the $y$-axis. The bounded region is between $x = -\frac{1}{2}$ and $x = \frac{1}{2}$.
For $x \in [0, \frac{1}{2}]$,the upper boundary is $y = \min((x+1)^2, (x-1)^2)$ and the lower boundary is $y = \frac{1}{4}$.
Specifically,for $x \in [0, \frac{1}{2}]$,the curve $y=(x-1)^2$ is the upper boundary.
$\text{Required Area} = 2 \int_0^{\frac{1}{2}} \left[ (x-1)^2 - \frac{1}{4} \right] dx$
$= 2 \left[ \frac{(x-1)^3}{3} - \frac{x}{4} \right]_0^{\frac{1}{2}}$
$= 2 \left[ \left( \frac{(-1/2)^3}{3} - \frac{1/2}{4} \right) - \left( \frac{(-1)^3}{3} - 0 \right) \right]$
$= 2 \left[ \left( -\frac{1}{24} - \frac{1}{8} \right) - \left( -\frac{1}{3} \right) \right]$
$= 2 \left[ -\frac{4}{24} + \frac{1}{3} \right] = 2 \left[ -\frac{1}{6} + \frac{2}{6} \right] = 2 \left( \frac{1}{6} \right) = \frac{1}{3} \text{ sq. units.}$

(D) The required area is bounded by the curve $y=x^2+2$ (upper curve) and the line $y=x+1$ (lower curve) between the vertical lines $x=0$ and $x=2$.
$\text{Required Area} = \int_0^2 [(x^2+2) - (x+1)] dx$
$= \int_0^2 (x^2 - x + 1) dx$
$= \left[ \frac{x^3}{3} - \frac{x^2}{2} + x \right]_0^2$
$= \left( \frac{2^3}{3} - \frac{2^2}{2} + 2 \right) - (0)$
$= \frac{8}{3} - 2 + 2 = \frac{8}{3} \text{ sq. units}$

(A) The given circle is $x^2+y^2=4$,which has a radius $r=2$. The area is bounded by $x=0$ (the $y$-axis) and $x=2$ in the first quadrant. The area $A$ is given by the integral of $y$ with respect to $x$ from $0$ to $2$. Since $y^2 = 4-x^2$,we have $y = \sqrt{4-x^2}$ in the first quadrant.
$A = \int_0^2 \sqrt{4-x^2} dx$
Using the standard integral formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a}) + C$:
$A = \left[ \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}(\frac{x}{2}) \right]_0^2$
$A = \left( \frac{2}{2}\sqrt{4-4} + 2\sin^{-1}(1) \right) - \left( 0 + 2\sin^{-1}(0) \right)$
$A = (0 + 2 \times \frac{\pi}{2}) - 0 = \pi \text{ square units.}$

(B) Given the curve $y = 4x^2$,we can express $x$ in terms of $y$ as $x = \sqrt{\frac{y}{4}} = \frac{\sqrt{y}}{2}$.
Since the region is in the first quadrant and bounded by $x=0$,$y=2$,and $y=4$,the area $A$ is given by the integral with respect to $y$:
$A = \int_{2}^{4} x \, dy = \int_{2}^{4} \frac{\sqrt{y}}{2} \, dy$.
Evaluating the integral:
$A = \frac{1}{2} \int_{2}^{4} y^{1/2} \, dy = \frac{1}{2} \left[ \frac{y^{3/2}}{3/2} \right]_{2}^{4} = \frac{1}{2} \cdot \frac{2}{3} \left[ y^{3/2} \right]_{2}^{4}$.
$A = \frac{1}{3} [4^{3/2} - 2^{3/2}] = \frac{1}{3} [8 - 2\sqrt{2}]$ sq. units.

(A) The parabola is $y^2=4ax$. The latus-rectum is the line $x=a$.
To find the area,we integrate with respect to $x$ from $0$ to $a$.
Since the parabola is symmetric about the $x$-axis,the total area $A$ is twice the area in the first quadrant.
$A = 2 \int_{0}^{a} y \, dx = 2 \int_{0}^{a} \sqrt{4ax} \, dx$
$A = 2 \int_{0}^{a} 2\sqrt{a} \sqrt{x} \, dx = 4\sqrt{a} \int_{0}^{a} x^{1/2} \, dx$
$A = 4\sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{a} = 4\sqrt{a} \cdot \frac{2}{3} \cdot a^{3/2}$
$A = \frac{8}{3} \sqrt{a} \cdot a \sqrt{a} = \frac{8}{3} a^2 \text{ sq. units}$

(B) The required area $A$ is given by the integral of the function $y = \log x$ from $x=1$ to $x=e$.
$A = \int_{1}^{e} \log x \, dx$
Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = \log x$ and $dv = dx$:
$A = [x \log x]_{1}^{e} - \int_{1}^{e} x \cdot \frac{1}{x} \, dx$
$A = [x \log x]_{1}^{e} - \int_{1}^{e} 1 \, dx$
$A = [x \log x - x]_{1}^{e}$
Substituting the limits:
$A = (e \log e - e) - (1 \log 1 - 1)$
Since $\log e = 1$ and $\log 1 = 0$:
$A = (e(1) - e) - (0 - 1)$
$A = (e - e) - (-1)$
$A = 0 + 1 = 1 \text{ Sq. Units}$

(A) The volume $V$ of a solid generated by rotating a curve $y = f(x)$ about the $x$-axis between $x = a$ and $x = b$ is given by the formula: $V = \int_{a}^{b} \pi y^{2} dx$.
Given the curve $y^{2} = 4x$ and the limits $x = 4$ to $x = 5$.
Substituting the values into the formula:
$V = \int_{4}^{5} \pi (4x) dx$
$V = 4\pi \int_{4}^{5} x dx$
$V = 4\pi \left[ \frac{x^{2}}{2} \right]_{4}^{5}$
$V = 2\pi [x^{2}]_{4}^{5}$
$V = 2\pi (5^{2} - 4^{2})$
$V = 2\pi (25 - 16)$
$V = 2\pi (9) = 18\pi$ cubic units.

(A) The area $R_1$ is given by $\int_0^b (1-x)^2 \, dx$ and the area $R_2$ is given by $\int_b^1 (1-x)^2 \, dx$.
Given $R_1 - R_2 = \frac{1}{4}$,we have:
$\int_0^b (1-x)^2 \, dx - \int_b^1 (1-x)^2 \, dx = \frac{1}{4}$
Evaluating the integrals:
$\left[ \frac{-(1-x)^3}{3} \right]_0^b - \left[ \frac{-(1-x)^3}{3} \right]_b^1 = \frac{1}{4}$
$\left( \frac{-(1-b)^3}{3} - \frac{-(1-0)^3}{3} \right) - \left( \frac{-(1-1)^3}{3} - \frac{-(1-b)^3}{3} \right) = \frac{1}{4}$
$\left( \frac{1 - (1-b)^3}{3} \right) - \left( \frac{(1-b)^3}{3} \right) = \frac{1}{4}$
$\frac{1 - 2(1-b)^3}{3} = \frac{1}{4}$
$1 - 2(1-b)^3 = \frac{3}{4}$
$2(1-b)^3 = 1 - \frac{3}{4} = \frac{1}{4}$
$(1-b)^3 = \frac{1}{8}$
$1-b = \frac{1}{2}$
$b = 1 - \frac{1}{2} = \frac{1}{2}$

(D) The area $A$ is given by the integral of the absolute value of the function over the given interval:
$A = \int_{-\pi/2}^{\pi/2} |\cos x| \, dx$.
Since $\cos x \geq 0$ for $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,we have:
$A = \int_{-\pi/2}^{\pi/2} \cos x \, dx$.
Evaluating the integral:
$A = [\sin x]_{-\pi/2}^{\pi/2} = \sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2})$.
$A = 1 - (-1) = 1 + 1 = 2$.
Thus,the area is $2$ square units.

(B) The given equation of the parabola is $x^2 = 4y$,which implies $y = \frac{x^2}{4}$.
To find the area bounded by the curve $y = \frac{x^2}{4}$,the $X$-axis,and the line $x = 3$,we integrate the function with respect to $x$ from $x = 0$ to $x = 3$.
Area $= \int_{0}^{3} y \, dx = \int_{0}^{3} \frac{x^2}{4} \, dx$.
Area $= \frac{1}{4} \left[ \frac{x^3}{3} \right]_{0}^{3}$.
Area $= \frac{1}{4} \left( \frac{3^3}{3} - \frac{0^3}{3} \right) = \frac{1}{4} \left( \frac{27}{3} \right) = \frac{1}{4} \times 9 = \frac{9}{4}$ sq. units.
Therefore,the correct option is $B$.

(A) The given equation of the ellipse is $9x^2 + 16y^2 = 1$.
This can be rewritten in the standard form $\frac{x^2}{(1/3)^2} + \frac{y^2}{(1/4)^2} = 1$.
Here,$a = \frac{1}{3}$ and $b = \frac{1}{4}$.
The total area of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $\pi ab$.
The area in the first quadrant is $\frac{1}{4}$ of the total area.
Area $= \frac{1}{4} \pi ab = \frac{1}{4} \pi \left(\frac{1}{3}\right) \left(\frac{1}{4}\right) = \frac{\pi}{48}$ sq. units.
Thus,the correct option is $A$.

(C) The area $A$ is given by the integral of $|y|$ with respect to $x$ from $x = 2$ to $x = 5$.
$A = \int_{2}^{5} |3 - x| \, dx$.
Since the line $y = 3 - x$ crosses the $X$-axis at $x = 3$,we split the integral:
$A = \int_{2}^{3} (3 - x) \, dx + \int_{3}^{5} -(3 - x) \, dx$.
$A = \left[ 3x - \frac{x^2}{2} \right]_{2}^{3} + \left[ \frac{x^2}{2} - 3x \right]_{3}^{5}$.
$A = \left( (9 - 4.5) - (6 - 2) \right) + \left( (12.5 - 15) - (4.5 - 9) \right)$.
$A = (4.5 - 4) + (-2.5 + 4.5) = 0.5 + 2 = 2.5 = \frac{5}{2}$ square units.
Thus,the correct option is $C$.

(C) The given equation of the ellipse is $9x^2 + 4y^2 = 1$.
This can be rewritten in the standard form $\frac{x^2}{(1/3)^2} + \frac{y^2}{(1/2)^2} = 1$.
Here,$a = \frac{1}{3}$ and $b = \frac{1}{2}$.
The area of the entire ellipse is given by the formula $A = \pi ab$.
$A = \pi \times \frac{1}{3} \times \frac{1}{2} = \frac{\pi}{6}$.
Since the ellipse is symmetric about both axes,the area in the first quadrant is one-fourth of the total area.
Area in the first quadrant $= \frac{1}{4} \times \frac{\pi}{6} = \frac{\pi}{24}$.
Thus,the correct option is $C$.

(C) The given curve is $y^2 = 4x$ and the line is $x = 3$.
Since the curve is symmetric about the $x$-axis,the total area $A$ is given by $2 \times \int_{0}^{3} y \, dx$.
From $y^2 = 4x$,we have $y = \sqrt{4x} = 2\sqrt{x}$.
Thus,$A = 2 \int_{0}^{3} 2\sqrt{x} \, dx = 4 \int_{0}^{3} x^{1/2} \, dx$.
Evaluating the integral: $A = 4 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{3} = 4 \times \frac{2}{3} \times [x^{3/2}]_{0}^{3}$.
$A = \frac{8}{3} \times (3)^{3/2} = \frac{8}{3} \times 3\sqrt{3} = 8\sqrt{3}$ sq. units.
Therefore,the correct option is $C$.