Area Under The Curves

1. Calculate the Area $A_1$ between $P_1$ and $P_2$

The two parabolas are given by:

• $P_1: y=4 x^2$

• $P_2: y=x^2+27$

Find the intersection points: Setting the equations equal to each other:

$ 4 x^2=x^2+27 $

$\Rightarrow $$ 3 x^2=27 $

$\Rightarrow $ $ x^2=9 \Rightarrow x= \pm 3 $

$ \begin{aligned} &\text { Find the intersection points : Setting the equations equal to each other : }\\ &\begin{aligned} & 4 x^2=x^2+27 \\ & 3 x^2=27 \quad, \quad x^2=9 \Rightarrow x= \pm 3 \end{aligned} \end{aligned} $

Calculate the area: On the interval $[-3,3]$, the parabola $P_2$ lies above $P_1$. The area $A_1$ is:

$ \begin{gathered} A_1=\int_{-3}^3\left(x^2+27-4 x^2\right) d x=\int_{-3}^3\left(27-3 x^2\right) d x \\ A_1=\left[27 x-x^3\right]_{-3}^3 \\ A_1=\left(27(3)-3^3\right)-\left(27(-3)-(-3)^3\right) \\ A_1=(81-27)-(-81+27)=54-(-54)=108 \end{gathered} $

Calculate the Area $A_2$ between $P_1$ and the line $y=\alpha x$

Find the intersection points:

$ 4 x^2=\alpha x \Longrightarrow x(4 x-\alpha)=0 $

The points are $x=0$ and $x=\frac{\alpha}{4}$.

Calculate the area: On the interval $\left[0, \frac{\alpha}{4}\right]$, the line $y=\alpha x$ lies above $y=4 x^2$. The area $A_2$ is:

$ \begin{aligned} A_2 & =\int_0^{\alpha / 4}\left(\alpha x-4 x^2\right) d x \\ A_2 & =\left[\frac{\alpha x^2}{2}-\frac{4 x^3}{3}\right]_0^{\alpha / 4} \\ A_2 & =\frac{\alpha}{2}\left(\frac{\alpha}{4}\right)^2-\frac{4}{3}\left(\frac{\alpha}{4}\right)^3 \\ A_2 & =\frac{\alpha^3}{32}-\frac{4 \alpha^3}{192}=\frac{\alpha^3}{32}-\frac{\alpha^3}{48}=\frac{3 \alpha^3-2 \alpha^3}{96}=\frac{\alpha^3}{96} \end{aligned} $

The problem states that $A_1$ is six times $A_2$ :

$ \begin{gathered} A_1=6 \times A_2 \\ 108=6 \times\left(\frac{\alpha^3}{96}\right) \\ 108=\frac{\alpha^3}{16} \\ \alpha^3=108 \times 16=1728 \\ \alpha=\sqrt[3]{1728}=12\end{gathered} $

The region is bounded by three curves in the first quadrant ( $x \geq 0$ ) :

• 1. $y=1$ : A horizontal line.

• 2. $y=x^2$ : A parabola opening upwards.

• 3. $y=\frac{8}{x}$ : A rectangular hyperbola.

Find the Intersection Points

We need to know where these curves meet to set our integration limits:

• Parabola and Line $\left(y=x^2\right.$ and $\left.y=1\right)$ : $x^2=1 \Longrightarrow x=1$ (since $x \geq 0$ ). Point: $(1,1)$.

• Parabola and Hyperbola $\left(y=x^2\right.$ and $\left.y=\frac{8}{x}\right): x^2=\frac{8}{x} \Longrightarrow x^3=8 \Longrightarrow x=2$. Point: $(2,4)$.

• Hyperbola and Line $\left(y=\frac{8}{x}\right.$ and $\left.y=1\right): \frac{8}{x}=1 \Longrightarrow x=8$. Point: $(8,1)$.

Understanding the Inequalities

• $x \geq 0$ : The region is located in the first and fourth quadrants (but limited to the first quadrant by $1 \leq y$ ).

• $1 \leq y \leq x^2$ : This gives us a lower bound of the horizontal line $y=1$ and an upper bound of the parabola $y=x^2$.

• $x y \leq 8$ : This can be rewritten as $y \leq \frac{8}{x}$. This represents the area below the rectangular hyperbola.

Setting up the Integrals

The region starts at $x=1$ and ends at $x=8$. Because the "upper" boundary changes at $x=2$, we must split the area into two parts:

• Part 1 (from $x=1$ to $x=2$ ): Bounded above by the parabola $y=x^2$ and below by $y=1$.

• Part 2 (from $x=2$ to $x=8$ ): Bounded above by the hyperbola $y=\frac{8}{x}$ and below by $y=1$.

Area $=\int\limits_1^2\left(x^2-1\right) d x+\int\limits_2^8\left(\frac{8}{x}-1\right) d x$

$=\left[\frac{x^3}{3}-x\right]_1^2+[8 \ln x-x]_2^8$

$=\left[\frac{8}{3}-2-\left\{\frac{1}{3}-1\right\}\right]+[(8 \ln 8-8)-\{8 \ln 2-2\}]$

$=\frac{7}{3}-1+8 \ln 2^3-8 \ln 2-6$

$=\frac{7}{3}-7+24 \ln 2-8 \ln 2 \quad\left\{\ln a^m=m \ln a\right\}$

$=-\frac{14}{3}+16 \ln 2$

$=\frac{2}{3}[24 \ln 2-7]$

Given Curves $y^2=x, y=|\alpha x-5|-|1-\alpha x|+\alpha x^2 x=0$ and $x=1$

The area of region bounded by these curve is $f(\alpha)$.

To find $f(0)+f(1)$.

for $\alpha=0$

$ \begin{aligned} & y=|0 \times x-5|-|1-0 \times x|+0 \times x^2 \\ & y=5-1=4 \end{aligned} $

so the region bounded by $x=0, x=1, y=\sqrt{x}$

$ \begin{aligned} &\text { and } y=4 \text { is shown in graph. }\\ &\begin{gathered} f(0)=\int_0^1(4-\sqrt{x}) d x \\ f(0)=\left[4 x-\frac{2}{3} x^{3 / 2}\right]_0^1=4-\frac{2}{3}=\frac{10}{3} \end{gathered} \end{aligned} $

If, When $\alpha=1$, the second equation becomes:

$ y=|x-5|-|1-x|+x^2 $

For the interval $0 \leq x \leq 1$ :

$ \begin{aligned} & |x-5|=-(x-5)=5-x \\ & |1-x|=1-x \\ & y=(5-x)-(1-x)+x^2=4+x^2 \end{aligned} $

The region is bounded by $x=0, x=1, y=\sqrt{x}$, and $y=4+x^2$ as shown in graph is.

$ \begin{gathered} f(1)=\int_0^1\left(4+x^2-\sqrt{x}\right) d x \\ f(1)=\left[4 x+\frac{x^3}{3}-\frac{2}{3} x^{3 / 2}\right]_0^1=4+\frac{1}{3}-\frac{2}{3}=4-\frac{1}{3}=\frac{11}{3} \end{gathered} $

So, $(f(0)+f(1))=\frac{10}{3}+\frac{11}{3}=\frac{21}{3}=7$

Area $A_1$ enclosed by curves

$y=x^2+2, x+y=8, y$-axis in first quadrant as shown in graph.

Solve $y=x^2+2$ and $x+y=8$ to find point of intersection.

Substituting $y=8-x$ in $y=x^2+2$

$ \begin{aligned} & 8-x=x^2+2 \\ & x^2+x-6=0 \\ & x^2+3 x-2 x-6=0 \\ & x(x+3)-2(x+3)=0 \\ & (x+3)(x-2)=0 \\ & x=-3,2 \end{aligned} $

$ \text { Since area in first quadrant, point of intersection at } x=2, y=8-2=6 \Rightarrow(2,6) \text {. } $

$ \begin{aligned} & A_1=\int_0^2\left[(8-x)-\left(x^2+2\right)\right] d x \\ & =\int_0^2\left(-x^2-x+6\right) d x \\ & =\left[-\frac{x^3}{3}-\frac{x^2}{2}+6 x\right]_0^2 \\ & =\left[-\frac{8}{3}-\frac{4}{2}+12-\{0-0-0\}\right] \\ & =\frac{-16-12+72}{6}=\frac{72-28}{6}=\frac{44}{6} \\ & A_1=\frac{22}{3} \end{aligned} $

Now, $A_2$ is area enclosed by curves

$ y=x^2+2, y^2=x, x=2 $

Solve $y=x^2+2$ and $y=\sqrt{x}$

$ y=y^4+2 $

$ y^4-y+2=0 $

$ D=(-1)^2-4 \times 1 \times 2=1-8=-7 $

discriminant (D) is negative. This means curves do not intersect.

$ \begin{aligned} &\text { So } A_2 \text { is bounded by curves y-axis, } x=2, y=x^2+2 \text { and } y=\sqrt{x} \text { in first quadrant is: }\\ &\begin{aligned} & A_2=\int_0^2\left(x^2+2-\sqrt{x}\right) d x \\ & =\left[\frac{x^3}{3}+2 x-\frac{2 x^{3 / 2}}{3}\right]_0^2 \\ & =\left[\frac{8}{3}+2 \times 2-\frac{2 \times 2 \sqrt{2}}{3}-\{0+0-0\}\right] \\ & =\left[\frac{8}{3}+4-\frac{4 \sqrt{2}}{3}\right]=\left[\frac{20}{3}-\frac{4 \sqrt{2}}{3}\right] \\ & A_1-A_2=\frac{22}{3}-\left(\frac{20}{3}-\frac{4 \sqrt{2}}{3}\right) \\ & =\frac{22}{3}-\frac{20}{3}+\frac{4 \sqrt{2}}{3}=\frac{2}{3}+\frac{4 \sqrt{2}}{3} \\ & =\frac{2}{3}(2 \sqrt{2}+1)\end{aligned} \end{aligned} $

$ \begin{aligned} & x^2+y^2=4 \Longrightarrow C_1(0,0), r=2 \\ & x^2+(y-2)^2=4 \Longrightarrow C_2(0,2), r=2 \end{aligned} $

Intersection points:

$ \begin{aligned} & x^2+y^2=x^2+(y-2)^2 \\ & y^2=y^2-4 y+4 \Longrightarrow 4 y=4 \Longrightarrow y=1 \\ & x^2+1=4 \Longrightarrow x^2=3 \Longrightarrow x= \pm \sqrt{3} \end{aligned} $

Points are $(\sqrt{3}, 1)$ and $(-\sqrt{3}, 1)$

For the region between $y=0$ and $y=2$ :

$ \begin{aligned} & y_{\text {upper }}=\sqrt{4-x^2}\left(\text { from } x^2+y^2=4\right) \\ & y_{\text {lower }}=2-\sqrt{4-x^2}\left(\text { from } x^2+(y-2)^2=4\right) \end{aligned} $

$ \begin{aligned} & \text { Area }=\int_{-\sqrt{3}}^{\sqrt{3}}\left(y_{\text {upper }}-y_{\text {lower }}\right) d x \\ & \text { Area }=\int_{-\sqrt{3}}^{\sqrt{3}}\left(\sqrt{4-x^2}-\left(2-\sqrt{4-x^2}\right)\right) d x \\ & \text { Area }=\int_{-\sqrt{3}}^{\sqrt{3}}\left(2 \sqrt{4-x^2}-2\right) d x \end{aligned} $

Property: $\int_{-a}^a f(x) d x=2 \int_0^a f(x) d x$ (for even functions)

$ \begin{aligned} & \text { Area }=2 \int_0^{\sqrt{3}}\left(2 \sqrt{4-x^2}-2\right) d x \\ & \text { Area }=4 \int_0^{\sqrt{3}}\left(\sqrt{4-x^2}-1\right) d x \\ & \int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x}{a}\right) \\ & \text { Area }=4\left[\frac{x}{2} \sqrt{4-x^2}+2 \sin ^{-1}\left(\frac{x}{2}\right)-x\right]_0^{\sqrt{3}} \\ & \text { Area }=4\left[\frac{\sqrt{3}}{2}(1)+2\left(\frac{\pi}{3}\right)-\sqrt{3}\right] \\ & \text { Area }=4\left[\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right] \\ & \text { Area }=\frac{8 \pi}{3}-2 \sqrt{3}=\frac{2}{3}(4 \pi-3 \sqrt{3}) \end{aligned} $

$ A=2 \int_0^1 2 \sqrt{x} d x+2 \int_1^{\sqrt{2}} \sqrt{8-4 x^2} d x=\frac{8}{3}\left(X^{\frac{3}{2}}\right)_0^1+4 \int_1^{\sqrt{2}} \sqrt{2-x^2} d x $

$ \begin{aligned} &\frac{8}{3}+4 \times \frac{1}{2}\left[x \sqrt{2-x^2}+2 \sin ^{-1}\left(\frac{x}{\sqrt{2}}\right)\right]_1^{\sqrt{2}}\\ &\frac{8}{3}+2\left[2 \times \frac{\pi}{2}-1-2 \times \frac{\pi}{4}\right]=\frac{8}{3}+2 \pi-2-\pi=\pi+\frac{2}{3} \text { Sq. unit } \end{aligned} $

$ \begin{aligned} & m=\int_{-2}^{-1}(3-x)-\left(1+x^2\right) d x \\ & m=2+\frac{3}{2}-\frac{7}{3}=\frac{7}{6} \quad M=\int_{-1}^1(3-x)-\left(1+x^2\right) d x=4-\frac{1}{2}(0)-\frac{1}{3}(1+1)=4-\frac{2}{3}=\frac{10}{3} \\ & m: M=\frac{7}{6}: \frac{10}{3}=\frac{7}{20} \\ & m+n=27 \end{aligned} $

$ \int_0^2\left(4-x^2\right) d x-\frac{1}{2} \times 1 \times \frac{1}{2} $

$ \begin{aligned} & E: \frac{x^2}{4}+\frac{y^2}{1}=1 \\ & \text { Area inside } E=2 \pi \\ & \begin{aligned} \text { Area } P Q R S= & (\sqrt{2})^2 \\ & =2 \\ \text { Required area } & =2 \pi-2 \\ & =2(\pi-1) \end{aligned} \end{aligned} $

We first focus on the orbit periodicity of $f^n(\boldsymbol{x})$ :

$ f^1(x)=f(x)=\frac{x-1}{x+1} ; x \in(1, \infty) \ldots . . (i) $

$f^2(x)=f\left\{f^1(x)\right\}=f\{f(x)\}=f\left(\frac{x-1}{x+1}\right)=\frac{\left(\frac{x-1}{x+1}\right)-1}{\left(\frac{x-1}{x+1}\right)+1}=-\frac{2}{2 x}=-\frac{1}{x} ; x \in(1, \infty) \ldots . .(i i)$

$\boldsymbol{f}^3(\boldsymbol{x})=f\left\{\boldsymbol{f}^2(\boldsymbol{x})\right\}=f\left(-\frac{1}{\boldsymbol{x}}\right)=\frac{\left(-\frac{1}{\boldsymbol{x}}\right)-1}{\left(-\frac{1}{\boldsymbol{x}}\right)+1}=\frac{1+\boldsymbol{x}}{1-\boldsymbol{x}} ; \boldsymbol{x} \in(1, \infty) \ldots . .(\boldsymbol{i \boldsymbol { i i }})$

$f^4(x)=f\left\{f^3(x)\right\}=f\left(\frac{1+x}{1-x}\right)=\frac{\left(\frac{1+x}{1-x}\right)-1}{\left(\frac{1+x}{1-x}\right)+1}=\frac{2 x}{2}=x ; x \in(1, \infty) \ldots . .(i v)$

$ f^5(x)=f\left\{f^4(x)\right\}=f(x)=\frac{x-1}{x+1} ; x \in(1, \infty) \ldots . . (v)$

From the nature of table which completes the 25 iterations in 6 cycles with orbit periodicity 4 :

$ f^{25}(x)=f(x)=f^1(x) \Rightarrow f^{26}(x)=f^2(x)=-\frac{1}{x} \ldots \ldots(v i) $

Equation (vi) exactly identifies the function $\boldsymbol{g}(\boldsymbol{x})$ defined under SOW :

$ \boldsymbol{g}(\boldsymbol{x})+\boldsymbol{f}^{26}(\boldsymbol{x})=0 \Rightarrow \boldsymbol{g}(\boldsymbol{x})=-\boldsymbol{f}^{26}(\boldsymbol{x})=\frac{1}{\boldsymbol{x}} \ldots \ldots(v i i) $

Equation (vii) has identified the function $\boldsymbol{g}(\boldsymbol{x})$ and we can mathematically plot the bounded area between the curves :

$ \begin{array}{|l|l|l|l|} \hline g(x) & \text { Slant line } & \text { Horizontal line } & \text { Vertical line } \\ \hline \frac{1}{x} & y=x-\left(\frac{3}{2}\right) & X \text {-axis } & x=4 \\ \hline \end{array} $

$ \begin{array}{|l|l|l|l|l|} \hline \text { Pair } & \boldsymbol{g}(\boldsymbol{x}) & \text { Slant line } & \text { Feasible } & \text { Intersection Points } \\ \hline 1 & \frac{1}{\boldsymbol{x}} & \boldsymbol{y}=\boldsymbol{x}-\left(\frac{3}{2}\right) & \text { Yes } & \boldsymbol{A}, \boldsymbol{K} \\ \hline \end{array} $

$\Rightarrow \frac{1}{x}=x-\left(\frac{3}{2}\right) \Rightarrow 2 x^2-3 x-2=0 \Rightarrow x=\left(2,-\frac{1}{2}\right) \Rightarrow y=\left(\frac{1}{2},-2\right)$

$ \begin{array}{|l|l|l|l|l|} \hline \text { Intersections } & \boldsymbol{A} & \boldsymbol{K} & \text { Discarded point } & \text { Reason } \\ \hline \text { Coordinates } & \left(2, \frac{1}{2}\right) & \left(-\frac{1}{2},-2\right) & \boldsymbol{K} & \text { Falls outside the scope of bounded area } \\ \hline \end{array} $

$ \begin{array}{|l|l|l|l|l|} \hline \text { Pair } & \boldsymbol{g}(\boldsymbol{x}) & \text { Vertical line } & \text { Feasible } & \text { Intersection Points } \\ \hline 2 & \frac{1}{\boldsymbol{x}} & \boldsymbol{x}=4 & \text { Yes } & \boldsymbol{B} \\ \hline \end{array} $

$\Rightarrow x=4 \Rightarrow y=0.25$

$ \begin{array}{|l|l|} \hline \text { Intersection } & \boldsymbol{B} \\ \hline \text { Coordinate } & (4,0.25) \\ \hline \end{array} $

$ \begin{array}{|l|l|l|l|l|} \hline \text { Pair } & \text { Horizontal line } & \text { Vertical line } & \text { Feasible } & \text { Intersection point } \\ \hline 3 & \boldsymbol{X} \text {-axis } & \boldsymbol{x}=4 & \text { Yes } & \boldsymbol{C} \\ \hline \end{array} $

$\Rightarrow x=4, y=0$

$ \begin{array}{|l|l|} \hline \text { Intersection } & \boldsymbol{C} \\ \hline \text { Coordinate } & (4,0) \\ \hline \end{array} $

$ \begin{array}{|l|l|l|l|l|} \hline \text { Pair } & \text { Slant line } & \text { Horizontal line } & \text { Feasible } & \text { Intersection point } \\ \hline 4 & \boldsymbol{y}=x-\left(\frac{3}{2}\right) & X \text {-axis } & \text { Yes } & D \\ \hline \end{array} $

$\Rightarrow \boldsymbol{y}=0 \Rightarrow \boldsymbol{x}=\frac{3}{2}$

$ \begin{array}{|l|l|} \hline \text { Intersection } & \boldsymbol{D} \\ \hline \text { Coordinate } & \left(\frac{3}{2}, 0\right) \\ \hline \end{array} $

We can visualize that both $\boldsymbol{A}$ and $\boldsymbol{D}$ lie on the slant line but $\boldsymbol{A}$ is ahead of $\boldsymbol{D}$ along the $\boldsymbol{X}$-axis. From $\boldsymbol{A}$ we can drop a ⟂ on the $\boldsymbol{X}$-axis at $\boldsymbol{M}$ to form a right angled $\boldsymbol{\Delta} \boldsymbol{A} \boldsymbol{M} \boldsymbol{D}$. From $\boldsymbol{A}$ and $\boldsymbol{D}$ the remaining area upto the vertical $\boldsymbol{x}=4$ lies under the rectangular hyperbola $g(x)=\frac{1}{x}$ :

$\begin{aligned} {\text { Area }}=\left(\frac{1}{2}\right) \cdot(\boldsymbol{D} \boldsymbol{M}) \cdot(\boldsymbol{A} \boldsymbol{M})+\int_2^4 \boldsymbol{g}(\boldsymbol{x}) \boldsymbol{d} \boldsymbol{x} & =\left(\frac{1}{2}\right) \cdot(2-1 \cdot 5) \cdot\left(\frac{1}{2}\right)+\int_2^4\left(\frac{1}{x}\right) d \boldsymbol{x} \\ & =\frac{1}{8}+[\ln |\boldsymbol{x}|]_2^4 \\ & =\frac{1}{8}+\ln 2\end{aligned}$

For the region

$ \{(x,y): x^2-8x \le y \le -x\}, $

the lower curve is

$ y=x^2-8x $

and the upper curve is

$ y=-x. $

To find the area between the curves, we first find their points of intersection.

Set

$ x^2-8x=-x $

$ x^2-7x=0 $

$ x(x-7)=0 $

So, the curves intersect at

$ x=0 \quad \text{and} \quad x=7. $

Now, on the interval $[0,7]$, the line $y=-x$ lies above the parabola $y=x^2-8x$.

Hence, the required area is

$ \text{Area}=\int_0^7 \left[(-x)-(x^2-8x)\right]dx. $

Simplify the integrand:

$ (-x)-(x^2-8x)=-x-x^2+8x=7x-x^2. $

Therefore,

$ \text{Area}=\int_0^7 (7x-x^2)\,dx. $

Now integrate:

$ \int (7x-x^2)\,dx=\frac{7x^2}{2}-\frac{x^3}{3}. $

Applying the limits from $0$ to $7$:

$ \text{Area}=\left[\frac{7x^2}{2}-\frac{x^3}{3}\right]_0^7 $

$ =\left(\frac{7\cdot 49}{2}-\frac{343}{3}\right)-0 $

$ =\frac{343}{2}-\frac{343}{3} $

Taking LCM $6$:

$ =\frac{1029-686}{6} =\frac{343}{6}. $

Hence, the correct answer is

$ \boxed{\frac{343}{6}} $

So, the correct option is A.

We need to find the area of the region

$ \{(x,y): 0 \le y \le 6-x,\; y^2 \ge 4x-3,\; x \ge 0\}. $

Let us understand each condition carefully.

1. From $0 \le y \le 6-x$

This means:

• $y \ge 0$, so the region lies above the $x$-axis.

• $y \le 6-x$, or

$ x+y \le 6, $

so the region lies below the line

$ y=6-x. $

2. From $y^2 \ge 4x-3$

Rewrite it in terms of $x$:

$ 4x-3 \le y^2 $

$ 4x \le y^2+3 $

$ x \le \frac{y^2+3}{4}. $

So the region lies to the left of the parabola

$ x=\frac{y^2+3}{4}. $

Also, we are given

$ x \ge 0. $

So for a fixed $y$, $x$ lies between $0$ and whichever is smaller:

$ x=\frac{y^2+3}{4} \quad \text{and} \quad x=6-y. $

So the horizontal width is

$ 0 \le x \le \min\left(\frac{y^2+3}{4},\,6-y\right). $

Now we find where these two curves intersect.

3. Point of intersection

Set

$ \frac{y^2+3}{4}=6-y. $

Multiply by $4$:

$ y^2+3=24-4y $

$ y^2+4y-21=0 $

$ (y+7)(y-3)=0. $

So,

$ y=3 \quad \text{or} \quad y=-7. $

Since $y \ge 0$, we take

$ y=3. $

Now compare the two right boundaries:

• For $0 \le y \le 3$, we have

$ \frac{y^2+3}{4} \le 6-y, $

so width is

$ \frac{y^2+3}{4}. $

• For $3 \le y \le 6$, we have

$ 6-y \le \frac{y^2+3}{4}, $

so width is

$ 6-y. $

Hence area is

$ A=\int_0^3 \frac{y^2+3}{4}\,dy+\int_3^6 (6-y)\,dy. $

4. Evaluate the integrals

First integral:

$ \int_0^3 \frac{y^2+3}{4}\,dy = \frac14 \int_0^3 (y^2+3)\,dy $

$ = \frac14 \left[\frac{y^3}{3}+3y\right]_0^3 = \frac14 \left(\frac{27}{3}+9\right) = \frac14 (9+9) = \frac{18}{4} = \frac92. $

Second integral:

$ \int_3^6 (6-y)\,dy = \left[6y-\frac{y^2}{2}\right]_3^6 $

$ = \left(36-18\right)-\left(18-\frac92\right) = 18-\frac{27}{2} = \frac{9}{2}. $

So total area is

$ A=\frac92+\frac92=9. $

Therefore, the correct answer is:

$ \boxed{9} $

So, Option B is correct.

Since $f:\{1,2,3,4\}\to \{1,e,e^2,e^3\}$ is bijective and strictly decreasing, we match the largest element of the domain to the smallest element of the codomain.

So,

$ f(1)=e^3,\quad f(2)=e^2,\quad f(3)=e,\quad f(4)=1 $

Hence,

$ f^{-1}(1)=4,\quad f^{-1}(e)=3,\quad f^{-1}(e^2)=2,\quad f^{-1}(e^3)=1 $

Now $g:\{1,e,e^2,e^3\}\to \left\{1,\frac12,\frac13,\frac14\right\}$ is bijective and strictly increasing.

Since

$ 1

and

$ \frac14<\frac13<\frac12<1 $

for $g$ to be strictly increasing, we must assign:

$ g(1)=\frac14,\quad g(e)=\frac13,\quad g(e^2)=\frac12,\quad g(e^3)=1 $

Therefore,

$ g^{-1}\left(\frac12\right)=e^2 $

Now,

$ f^{-1}\left(g^{-1}\left(\frac12\right)\right)=f^{-1}(e^2)=2 $

So,

$ \phi(x)=\left[f^{-1}\left\{g^{-1}\left(\frac12\right)\right\}\right]^x=2^x $

Thus the region is

$ R=\{(x,y):x^2\le y\le 2^x,\ 0\le x\le 1\} $

Its area is

$ \int_0^1 (2^x-x^2)\,dx $

Now,

$ \int 2^x\,dx=\frac{2^x}{\ln 2} $

So,

$ \int_0^1 2^x\,dx=\left[\frac{2^x}{\ln 2}\right]_0^1=\frac{2-1}{\ln 2}=\frac{1}{\ln 2} $

Also,

$ \int_0^1 x^2\,dx=\left[\frac{x^3}{3}\right]_0^1=\frac13 $

Hence area

$ =\frac{1}{\ln 2}-\frac13 $

Taking LCM,

$ =\frac{3-\ln 2}{3\ln 2} $

Therefore, the required area is

$ \boxed{\frac{3-\log_e 2}{3\log_e 2}} $

So the correct option is $\boxed{\text{A}}$.

$\begin{aligned} & \text { Area }=\int_1^3\left(x^2-1\right) d x+\int_3^{27}\left(\frac{27}{x}-1\right) d x \\ & =\left[\frac{x^3}{3}-x\right]_1^3+[27 \ell n x-x]_3^{27} \\ & =\frac{26}{3}-2+27 \ln 9-24 \\ & =27 \ln 9-\frac{26 \times 2}{3} \\ & =54 \ln 3-\frac{52}{3}\end{aligned}$

Given curves are

$ x+3y^2=0 \quad \text{and} \quad x+4y^2=1 $

First, write them in the form $x=$:

$ x=-3y^2 $

and

$ x=1-4y^2 $

To find the bounded area, we first find their points of intersection.

Set the two values of $x$ equal:

$ -3y^2=1-4y^2 $

$ y^2=1 $

$ y=\pm 1 $

So the curves intersect at $y=-1$ and $y=1$.

Now, for a fixed value of $y$, the right curve is

$ x=1-4y^2 $

and the left curve is

$ x=-3y^2 $

So the horizontal width of the region is

$ (1-4y^2)-(-3y^2)=1-y^2 $

Hence, area of the enclosed region is

$ A=\int_{-1}^{1}(1-y^2)\,dy $

Now evaluate:

$ A=\left[y-\frac{y^3}{3}\right]_{-1}^{1} $

$ A=\left(1-\frac{1}{3}\right)-\left(-1+\frac{1}{3}\right) $

$ A=\frac{2}{3}-\left(-\frac{2}{3}\right) $

$ A=\frac{4}{3} $

Therefore, the required area is

$ \boxed{\frac{4}{3}} $

So, the correct option is:

$ \boxed{\text{Option C}} $

We need the area of the region

$ \{(x,y): y\le \pi-|x|,\; y\le |x\sin x|,\; y\ge 0\}. $

This means the required area is the area under the curve

$ y=\min(\pi-|x|,\; |x\sin x|) $

and above the $x$-axis.

So the area is

$ A=\int \min(\pi-|x|,\; |x\sin x|)\,dx $

over the values of $x$ where the region exists.

Since $y\ge 0$ and $y\le \pi-|x|$, we must have

$ \pi-|x|\ge 0 \implies |x|\le \pi. $

Hence the region lies in $[-\pi,\pi]$.

Also, both functions $\pi-|x|$ and $|x\sin x|$ are even functions, so the region is symmetric about the $y$-axis. Therefore,

$ A=2\int_0^\pi \min(\pi-x,\; x\sin x)\,dx. $

Now for $0\le x\le \pi$, we have $\sin x\ge 0$, so

$ |x\sin x|=x\sin x. $

Thus,

$ A=2\int_0^\pi \min(\pi-x,\; x\sin x)\,dx. $

Now compare $\pi-x$ and $x\sin x$ on $[0,\pi]$.

At $x=0$,

$ \pi-x=\pi,\qquad x\sin x=0. $

At $x=\pi$,

$ \pi-x=0,\qquad x\sin x=0. $

Check where they meet:

$ x\sin x=\pi-x. $

It is easy to see that $x=\dfrac{\pi}{2}$ satisfies this, because

$ \frac{\pi}{2}\sin\frac{\pi}{2}=\frac{\pi}{2}=\pi-\frac{\pi}{2}. $

Also, on $[0,\pi/2]$, we have $x\sin x\le \pi-x$, and on $[\pi/2,\pi]$, we have $\pi-x\le x\sin x$.

So,

$ A=2\left[\int_0^{\pi/2} x\sin x\,dx+\int_{\pi/2}^\pi (\pi-x)\,dx\right]. $

Now evaluate each integral.

First,

$ \int x\sin x\,dx=-x\cos x+\int \cos x\,dx=-x\cos x+\sin x. $

Therefore,

$ \int_0^{\pi/2} x\sin x\,dx =\left[-x\cos x+\sin x\right]_0^{\pi/2} =1. $

Next,

$ \int_{\pi/2}^\pi (\pi-x)\,dx. $

Let $u=\pi-x$. Then as $x$ goes from $\pi/2$ to $\pi$, $u$ goes from $\pi/2$ to $0$. Or directly,

$ \int_{\pi/2}^\pi (\pi-x)\,dx =\left[\pi x-\frac{x^2}{2}\right]_{\pi/2}^{\pi}. $

Compute:

At $x=\pi$,

$ \pi^2-\frac{\pi^2}{2}=\frac{\pi^2}{2}. $

At $x=\frac{\pi}{2}$,

$ \pi\cdot \frac{\pi}{2}-\frac{1}{2}\left(\frac{\pi}{2}\right)^2 =\frac{\pi^2}{2}-\frac{\pi^2}{8} =\frac{3\pi^2}{8}. $

So,

$ \int_{\pi/2}^\pi (\pi-x)\,dx =\frac{\pi^2}{2}-\frac{3\pi^2}{8} =\frac{\pi^2}{8}. $

Hence,

$ A=2\left(1+\frac{\pi^2}{8}\right) =2+\frac{\pi^2}{4}. $

Therefore, the correct answer is

$ \boxed{2+\frac{\pi^2}{4}} $

So, Option B is correct.

$\begin{aligned} & A=\int_{-2}^1\left(x+7-x^2-1\right) d x+\int_1^2\left(11+3 x-x^2-1\right) d x \\ & =\left[\frac{x^2}{2}+6 x-\frac{x^3}{3}\right]_{-2}^1+\left[10 x-\frac{3 x^2}{2}-\frac{x^3}{3}\right]_1^2 \\ & =\frac{50}{3} \Rightarrow 3 A=50 \quad \text { Option (1) } \end{aligned}$

$y=4-\frac{x^2}{4}$ and $y=\frac{x-4}{2}$

$\begin{aligned} & \text { Area }=\int_\limits{-6}^4\left(4-\frac{x^2}{4}-\frac{x}{2}+2\right) d x \\ & =\left[6 x-\frac{x^3}{12}-\frac{x^2}{4}\right]_{-6}^4 \\ & =6(4+6)-\left(\frac{64}{12}+\frac{216}{12}\right)-\left(\frac{16}{4}-\frac{36}{4}\right) \\ & =60-\frac{70}{3}+5 \\ & \alpha=\frac{125}{3} \\ & 6 \alpha=6 \times \frac{125}{3}=250 \end{aligned}$

$\because f(x)=1-2 x+\int_0^x e^{x-t} f(t) d t$

or, $f(x)=1-2 x+e^x \int_0^x e^{-t} f(t) d t$

on differentiating both sides w.r.t. $x$ we get

$\begin{aligned} & f^{\prime}(x)=-2+e^x \int_0^x e^{-t} f(t) d t+e^x \cdot e^{-x} f(x) \\ & f^{\prime}(x)=-2+f(x)+2 x-1+f(x)\{\text { from eq. (1) \}} \end{aligned}$

$\begin{aligned} & \therefore \quad f(x)-2 f(x)=2 x-3 \\ & \text { I.F. }=e^{\int-2 d x}=e^{-2 x} \\ & \therefore \quad e^{-2 x} \cdot f(x)=\int e^{-2 x}(2 x-3) d x \\ & e^{-2 x} \cdot f(x)=(2 x-3) \cdot \frac{e^{-2 x}}{-2}-\int 2 \cdot \frac{e^{-2 x}}{-2} d x \\ & e^{-2 x} \cdot f(x)=\frac{(2 x-3) e^{-2 x}}{-2}+\frac{e^{-2 x}}{-2}+c \\ & f(x)=-x+1+c^{\prime} e^{2 x} \\ & \because \quad f(x)=1 \text { from eq. (1) } \\ & \therefore \quad 1=0+1+c^{\prime} \Rightarrow c^{\prime}=0 \\ & \therefore \quad f(x)=-x+1 \end{aligned}$

$\Rightarrow \quad$ Area $=\frac{1}{2}$

$|x-y| \leq y \leq 4 \sqrt{x}$

$\begin{aligned} & \text { Area }=\int_0^{64}\left(4 \sqrt{x}-\frac{x}{2}\right) d x \\ & \left.=\frac{4 x^{3 / 2}}{\frac{3}{2}}-\frac{x^2}{4}\right]_0^{64} \end{aligned}$

$\begin{aligned} & C_1:|y|=1-x^2 \\ & C_2: x^2+y^2=1 \end{aligned}$

$\therefore$ Required Area

$=\alpha=4\left[\right.$ Area of circle in $1^{\text {st }}$ quad. $\left.-\int_0^1\left(1-\mathrm{x}^2\right) \mathrm{dx}\right]$

$\begin{aligned} & =4\left[\frac{\pi}{4}-\left[\mathrm{x}-\frac{\mathrm{x}^3}{3}\right]_0^1\right] \\ & \alpha=\pi-\frac{8}{3} \\ & \therefore 3 \alpha=3 \pi-8 \\ & \therefore 9 \alpha=9 \pi-24 \\ & \therefore \beta=9, \gamma=-24 \\ & \therefore|\beta-\gamma|=33 \end{aligned}$

$A \Rightarrow$ Rectangle ABDE $-$ Area of region EDC

$\begin{aligned} & A \Rightarrow 4-2 \int_0^1(3-x)-\left(\frac{x^2+3}{2}\right) d x \\ & A \Rightarrow 4-2\left\{3 x-\frac{x^2}{2}-\frac{x^3}{6}-\frac{3}{2} x\right\}_0^1 \\ & A \Rightarrow 4-2\left\{3-\frac{1}{2}-\frac{1}{6}-\frac{3}{2}\right\}=\frac{7}{3} \end{aligned}$

So $6 \mathrm{~A}=14$

$\begin{aligned} & x\left(1+y^2\right)=1 \quad\text{..... (1)}\\ & y^2=2 x\quad\text{..... (2)} \end{aligned}$

$\begin{aligned} &\text { From equation (1) & (2) }\\ &\begin{aligned} x(1+2 x)=1 & \Rightarrow 2 x^2+x-1=0 \\ & \Rightarrow x=\frac{1}{2}, x=-1 \text { (Reject) } \\ & \Rightarrow y^2=2\left(\frac{1}{2}\right) \\ & \Rightarrow y= \pm 1 \end{aligned} \end{aligned}$

$\begin{aligned} \text { Area bounded } & =\int_{-1}^1\left(\frac{1}{1+y^2}-\frac{y^2}{2}\right) d y \\ & =\left.\left(\tan ^{-1} y-\frac{y^3}{6}\right)\right|_{-1} ^1 \\ & =\frac{\pi}{2}-\frac{1}{3} \end{aligned}$

$\begin{aligned} & \text { Area }=2\left[\int_0^2\left(\mathrm{x}^2+1\right) \mathrm{dx}+\int_2^3(2 \mathrm{x}+1) \mathrm{dx}\right] \\ & \Rightarrow \frac{64}{3} \quad \therefore(2) \end{aligned}$

$\begin{aligned} &\text { For Area } \int_{-\ln 2}^0\left[e^x-\left(1-e^x\right)\right] d x\\ &\begin{aligned} & \int_{-\operatorname{nn} 2}^0\left(2 \mathrm{e}^{\mathrm{x}}-1\right) \mathrm{dx}=\left[2 \mathrm{e}^{\mathrm{x}}-\mathrm{x}\right]_{-\ln 2}^0 \\ & =(2-(1+\ell \mathrm{n} 2)) \\ & =1-\ell \mathrm{n} 2 \end{aligned} \end{aligned}$

$x^2+4 x+2 \leq y \leq|x+2|$

The area bounded between

$y=x^2+4 x+2=(x+2)^2-2$

and $y=|x+2|$ is same as area bounded between $y=x^2-2$ and $y=|x|$

For P.O.I $|x|^2-2=|x|$

$\begin{aligned} & \Rightarrow|x|=2 \Rightarrow x= \pm 2 \\ & \therefore \text { Required area }=-\int_{-2}^2\left(x^2-2\right) d x+\int_{-2}^2|x| d x \\ & =-2 \int_0^2\left(x^2-2\right) d x+2 \int_0^2 x \cdot d x \\ & =-2\left[\frac{x^3}{3}-2 x\right]_0^2+2\left[\frac{x^2}{2}\right]_0^2 \\ & =--2\left[\frac{8}{3}-4\right]+2\left[\frac{4}{2}\right] \\ & =-2 \times\left(\frac{-4}{3}\right)+4 \\ & =\frac{20}{3} \end{aligned}$

$\begin{aligned} & \text { required area is } a+\int_\limits0^1\left(a+e^x-e^{-x}\right) d x \\ & a+\left[a+e^x+e^{-x}\right]_0^1 \\ & 2 a+e-1+e^{-1}-1=e+8+\frac{1}{e} \\ & 2 a=10 \Rightarrow a=5 \end{aligned}$

Consider the curves

$ y = x^2 - 4x + 4 = (x-2)^2 $

and

$ y^2 = 16 - 8x. $

Notice that the second equation can be rewritten in terms of $x$:

$ 8x = 16 - y^2 \quad \Longrightarrow \quad x = 2 - \frac{y^2}{8}. $

Step 1. Find the Intersection Points

To find the points where the curves intersect, substitute

$ y = (x-2)^2 $

into

$ y^2 = 16 - 8x. $

Let

$ u = x - 2 \quad \text{so that} \quad y = u^2. $

Then

$ y^2 = u^4 $

and

$ x = u + 2. $

Substitute into the second curve:

$ u^4 = 16 - 8(u+2). $

Simplify the right side:

$ 16 - 8(u+2) = 16 - 8u - 16 = -8u. $

Thus, the equation becomes

$ u^4 + 8u = 0 \quad \Longrightarrow \quad u(u^4/ u? \,\, **{correcting factor}**). $

In fact, factor by taking out a common factor $u$:

$ u(u^3 + 8) = 0. $

Thus, either

$ u = 0 \quad \text{or} \quad u^3 = -8. $

For $u = 0$:

$ x = u + 2 = 2, \quad y = u^2 = 0. $

For $u^3 = -8$:

$ u = -2, \quad \text{so} \quad x = -2 + 2 = 0, \quad y = (-2)^2 = 4. $

The curves intersect at the points $ (2, 0) $ and $ (0, 4). $

Step 2. Express the Curves in Terms of $y$

It is easier to integrate horizontally by expressing $x$ as a function of $y$.

From the second curve:

$ x = 2 - \frac{y^2}{8}. $

From the first curve, solving

$ y = (x-2)^2 $

for $x$ gives

$ x-2 = \pm\sqrt{y}. $

Since at the intersection $ (0, 4) $ the $x$–value is less than 2, we take the negative branch:

$ x = 2 - \sqrt{y}. $

Thus, for a fixed $y$ between 0 and 4, the left boundary is

$ x_{\text{left}} = 2 - \sqrt{y}, $

and the right boundary is

$ x_{\text{right}} = 2 - \frac{y^2}{8}. $

Step 3. Set Up the Integral for the Area

The horizontal distance between the curves at a given $y$ is

$ \Delta x = x_{\text{right}} - x_{\text{left}} = \left(2 - \frac{y^2}{8}\right) - \left(2 - \sqrt{y}\right) = \sqrt{y} - \frac{y^2}{8}. $

Integrate with respect to $y$ from $y = 0$ to $y = 4$:

$ A = \int_{0}^{4} \left(\sqrt{y} - \frac{y^2}{8}\right) \, dy. $

Step 4. Evaluate the Integral

Write the integral as

$ A = \int_{0}^{4} y^{1/2}\, dy - \frac{1}{8} \int_{0}^{4} y^2\, dy. $

Compute each term:

For the first integral:

$ \int y^{1/2}\, dy = \frac{2}{3} y^{3/2}. $

For the second integral:

$ \int y^2\, dy = \frac{y^3}{3}. $

Thus,

$ A = \left[\frac{2}{3} y^{3/2}\right]_0^4 - \frac{1}{8}\left[\frac{y^3}{3}\right]_0^4. $

Substitute $y = 4$ (note that at $y = 0$ both terms vanish):

Compute $4^{3/2}$:

$ 4^{3/2} = \left( \sqrt{4} \right)^3 = 2^3 = 8. $

Compute the first term:

$ \frac{2}{3} \times 8 = \frac{16}{3}. $

Compute the second term:

$ \frac{1}{8} \cdot \frac{64}{3} = \frac{64}{24} = \frac{8}{3}. $

Therefore, the area is

$ A = \frac{16}{3} - \frac{8}{3} = \frac{8}{3}. $

Final Answer

The area of the region enclosed by the curves is

$ \frac{8}{3}. $

$\begin{aligned} & \text { Required area }=2 \int_0^{2 \sqrt{3}}\left(\sqrt{4 \sqrt{3} x-x^2}-\sqrt{2 \sqrt{3} x}\right) d x \\\\ & =2 \int_0^{2 \sqrt{3}}\left(\sqrt{12-(x-2 \sqrt{3})^2}-\sqrt{2 \sqrt{3} x}\right) d x\end{aligned}$

$\begin{array}{r}=2\left[\frac{x-2 \sqrt{3}}{2} \sqrt{12-(x-2 \sqrt{3})^2}+\frac{12}{2} \sin ^{-1}\left(\frac{x-2 \sqrt{3}}{2 \sqrt{3}}\right)\right. \left.-\frac{\sqrt{2 \sqrt{3}} x^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^{2 \sqrt{3}}\end{array}$

$\begin{aligned} & =2\{3 \pi-8\} \\\\ & =6 \pi-16 \text { sq. units. }\end{aligned}$

$\begin{aligned} & \text { Area }=\int_\limits0^{3 \sqrt{2}} x d x+\int_\limits{3 / \sqrt{2}}^{3 \sqrt{2}} \sqrt{\frac{18-x^2}{3}} d x \\ & =\frac{1}{2}\left(x^2\right)_0^{3 \sqrt{2}}+\frac{1}{\sqrt{3}}\left[\frac{x}{2} \sqrt{18-x^2}+9 \sin ^{-1}\left(\frac{x}{3 \sqrt{2}}\right)\right]_{\frac{3}{\sqrt{2}}}^{3 \sqrt{2}} \\ & =\frac{1}{2}\left(\frac{9}{2}\right)+\frac{1}{\sqrt{3}}\left[9 \sin ^{-1}(1)-\frac{3}{2 \sqrt{2}} \cdot \frac{3 \sqrt{3}}{\sqrt{2}}-9 \sin ^{-1}\left(\frac{1}{2}\right)\right] \\ & =\frac{9}{4}+\frac{1}{\sqrt{3}}\left(\frac{9 \pi}{2}-\frac{9 \sqrt{3}}{4}-\frac{9 \pi}{6}\right)=\sqrt{3} \pi \end{aligned}$

The point of intersection of $y^2=4 x$ and $x^2+y^2=5$ are $(1,2)$ and $(1,-2)$.

$\because$ Area of smaller region bounded by $y^2=4 x$ and $x^2+y^2=5$

$=2\{\text {area of } O A C O+\text { area of } C A B C\}$

$\begin{aligned} & =2\left[\int_\limits0^1 2 \sqrt{x} d x+\int_\limits1^{\sqrt{5}} \sqrt{5-x^2} d x\right. \\ & =2\left[\left|\frac{4}{3} x^{\frac{3}{2}}\right|_0^1+\left(\frac{1}{2} x \sqrt{5-x^2}+\frac{5}{2} \sin ^{-1} \frac{x}{\sqrt{5}}\right)\right]_1^{\sqrt{5}} \\ & =2\left[\left(\frac{4}{3}-0\right)+\left(0+\frac{5 \pi}{4}\right)-\left(1+\frac{5}{2} \sin ^{-1} \frac{1}{\sqrt{5}}\right)\right] \\ & =2\left[\frac{1}{3}+\frac{5 \pi}{4}-\frac{5}{2} \sin ^{-1} \frac{1}{\sqrt{5}}\right]=\frac{2}{3}+\frac{5 \pi}{2}-5 \sin ^{-1} \frac{1}{\sqrt{5}} \\ & =\frac{2}{3}+5 \cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)=\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right) \end{aligned}$

We have, $x^2+y^2=8$ and $y^2=2 x$

Area of shaded region

$\begin{aligned} & =\int_\limits0^2 \frac{y^2}{2} d y+\int_\limits2^{2 \sqrt{2}} \sqrt{8-y^2} d y \\ & =\frac{1}{6}\left[y^3\right]_0^2+\left[\frac{y}{2} \sqrt{8-y^2}+4 \sin ^{-1}\left(\frac{y}{2 \sqrt{2}}\right)\right]_2^{2 \sqrt{2}} \\ & =\frac{4}{3}+\pi-2=\pi-\frac{2}{3} \end{aligned}$

$\left\{(x, y): \frac{a}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2,0< a<1\right\}$

$ \Rightarrow \int\limits_1^2 {\left( {{1 \over x} - {a \over {{x^2}}}} \right)dx = \left| {\ln |x| + {a \over x}} \right|_1^2} $

$\begin{aligned} & \left(\ln 2+\frac{a}{2}\right)-(\ln 1+a)=\ln 2-\frac{a}{2} \\ & \Rightarrow \quad \frac{a}{2}=\frac{1}{7} \\ & \Rightarrow \quad a=\frac{2}{7} \\ & \quad 7 a-3=\frac{2}{7} \times 7-3=-1 \end{aligned}$

$\begin{aligned} & y^{\prime}=3-\frac{3}{2} \sqrt{x} \\ & y^{\prime}=3\left(1-\frac{\sqrt{x}}{2}\right) \end{aligned}$

$\begin{aligned} & \text { Area }=\int_\limits0^3\left(3 x-\left(3 x-x^{3 / 2}\right)\right) d x+\int_\limits3^9 \frac{27-3 x}{2}-\left(3 x-x^{3 / 2}\right) d x \\ & =\int_\limits0^3 x^{3 / 2} d x+\frac{1}{2} \int_\limits3^9\left(27-9 x+2 x^{3 / 2}\right) d x \\ & =\left[\frac{2}{5} x^{5 / 2}\right]_0^3+\frac{1}{2}\left[27 x-\frac{9 x^2}{2}+2 \times \frac{2}{5} x^{5 / 2}\right]_3^9 \\ & =\frac{2}{5} \times 9 \sqrt{3}+\frac{1}{2}\left[243-\frac{729}{2}+\frac{4}{5} \times 81 \times 3-81+\frac{81}{2}-\frac{4}{5} \times 9 \sqrt{3}\right] \\ & =\frac{1}{2}\left[\frac{486-729-81}{2}+\frac{972}{5}\right]=\frac{81}{5} \\ & A=\frac{81}{5} \\ & \therefore 10 A=10 \times \frac{81}{5}=162 \end{aligned}$

$y=x|x|$ & $y=x-|x|$

$\begin{aligned} & y=x|x|= \begin{cases}x^2, & x>0 \\ -x^2, & x<0\end{cases} \\ & y=x-|x|= \begin{cases}0, & x>0 \\ 2 x, & x<0\end{cases} \end{aligned}$

Point of intersections are $(0,0)$ & $(-2,4)$.

$\begin{aligned} \therefore \quad & \text { Area }=\int_\limits{-2}^0\left(-x^2-2 x\right) d x \\ & =\left[\frac{-x^3}{3}-\frac{2 x^2}{2}\right]_{-2}^0 \\ & =-\left[\frac{8}{3}-4\right]=\frac{4}{3} \text { sq. unit } \end{aligned}$

$\text { Area }=\int_\limits{-\frac{1}{2}}^1\left(\frac{y+1}{4}-\frac{y^2}{2}\right) d y$

$\begin{aligned} & =\left[\frac{y^2}{8}+\frac{y}{4}-\frac{y^3}{6}\right]_{-\frac{1}{2}}^1 \\ & =\left(\frac{1}{8}+\frac{1}{4}-\frac{1}{6}\right)-\left(\frac{1}{32}-\frac{1}{8}+\frac{1}{48}\right) \\ & =\frac{5}{24}+\frac{7}{96} \\ & =\frac{27}{96} \\ & =\frac{9}{32} \end{aligned}$

Solving curves $y=1+3 x-2 x^2 ~\& ~y=\frac{1}{x}$

$\begin{aligned} & 2 x^3-3 x^2-x+1=0 \\ & \Rightarrow \quad(2 x-1)\left(x^2-x-1\right)=0 \\ & \Rightarrow \quad x=\frac{1}{2}, x=\frac{1 \pm \sqrt{5}}{2} \end{aligned}$

Area $ = \int\limits_{{1 \over 2}}^{{{1 + \sqrt 5 } \over 2}} {\left( {1 + 3x - 2{x^2} - {1 \over x}dx} \right)} $

$ = \left[ {x + {{3{x^2}} \over 2} - {{2x} \over 3} - \ln x} \right]$

$ = {{\sqrt 5 + 1} \over 2} + {3 \over 8}{(\sqrt 5 + 1)^2} - {1 \over {12}}(\sqrt 5 + 1) - \ln \left( {{{\sqrt 5 + 1} \over 2}} \right) - \left( {{1 \over 2} + {3 \over 8} - {1 \over {12}} - \ln {1 \over 2}} \right)$

$ = {1 \over {24}}\left[ {12(\sqrt 5 + 1) + 9{{(\sqrt 5 + 1)}^2} - 2(\sqrt 5 + 1) - 12 - 9 + 2] - \ln \left( {{{\sqrt 5 + 1} \over 2} \times 2} \right)} \right]$

$\begin{aligned} = & \frac{1}{24}[12(\sqrt{5}+1)+9(6+2 \sqrt{5})- \\ & 2(5 \sqrt{5}+1+3 \sqrt{5}(\sqrt{5}+1)-19)]-\ln (\sqrt{5}+1) \\ = & \frac{1}{24}[14 \sqrt{5}+15]-\ln (\sqrt{5}+1) \\ \therefore \quad & \quad l=14, m=15, n=1 \\ & \quad I+m+n=30 \end{aligned}$

To find the enclosed area between the two curves $ x y+4 y=16 $ and $ x+y=6 $, we need to determine the region of intersection and integrate the difference of the functions over the interval where they intersect.

First, let's solve the equations simultaneously to find the points of intersection.

The second curve $ x+y=6 $ can be rewritten as $ y=6-x $.

Substitute $ y=6-x $ into the first equation:

$ x(6-x) + 4(6-x) = 16$

$ 6x - x^2 + 24 - 4x = 16 $

$ -x^2 + 2x + 8 = 0 $

This can be simplified to:

$ x^2 - 2x - 8 = 0 $

Using the quadratic formula, we can find the roots of this equation:

$ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)} $

$ x = \frac{2 \pm \sqrt{4 + 32}}{2} $

$ x = \frac{2 \pm \sqrt{36}}{2} $

$ x = \frac{2 \pm 6}{2} $

So we have two solutions:

$ x_1 = \frac{2 + 6}{2} = 4 $

$ x_2 = \frac{2 - 6}{2} = -2 $

For $ x_1 = 4 $, substitute this back into the line equation $ x+y=6 $:

$ 4+y=6 $

$ y=2 $

For $ x_2 = -2 $, do the same:

$ -2+y=6 $

$ y=8 $

So we have two points of intersection: $ (4,2) $ and $ (-2,8) $.

Now, to find the area between $ x y + 4y = 16 $ and $ x + y = 6 $, we'll integrate the top function minus the bottom function from $ x = -2 $ to $ x = 4 $. First, we need to express $ y $ from both equations in terms of $ x $:

For $ x y + 4 y = 16 $, express $ y $ in terms of $ x $:

$ y(x+4) = 16 $

$ y = \frac{16}{x+4} $

For $ x + y = 6 $, we have:

$ y = 6 - x $

The area $ A $ is therefore given by:

$ A = \int\limits_{-2}^{4}\left( 6 - x - \frac{16}{x+4} \right) dx $

Now, we integrate:

$ A = \int\limits_{-2}^{4} (6 - x) dx - \int_{-2}^{4} \frac{16}{x + 4} dx $

$ A = \left[6x - \frac{x^2}{2}\right]_{-2}^{4} - \left[16 \ln|x + 4|\right]_{-2}^{4} $

$ A = \left( 6 \times 4 - \frac{1}{2} \times 4^2 \right) - \left( 6 \times -2 - \frac{1}{2} \times -2^2 \right) - 16 \left( \ln|4+4| - \ln|-2+4| \right) $

$ A = \left( 24 - 8 \right) - \left( -12 - 2 \right) - 16 \left( \ln 8 - \ln 2 \right) $

$ A = 16 + 14 - 16 \left( \ln (2^3) - \ln 2 \right) $

$ A = 30 - 16 \left( 3 \ln 2 - \ln 2 \right) $

$ A = 30 - 16 (2 \ln 2) $

$ A = 30- 32 \ln 2 $

$\begin{aligned} & \text { Area }=\left\lvert\, \int_\limits1^4\left[\left(4 x-x^2\right)-\frac{(x-4)^2}{3}\right] d x\right. \\ & \text { Area }=\left|\frac{4 x^2}{2}-\frac{x^3}{3}-\frac{(x-4)^3}{9}\right|_1^4 \\ & =\left|\left(\frac{64}{2}-\frac{64}{3}-\frac{4}{2}+\frac{1}{3}-\frac{27}{9}\right)\right| \\ & \Rightarrow(27-21)=6 \end{aligned}$

$\begin{aligned} & y^2 \leq 4 x, x<4 \\ & \frac{x y(x-1)(x-2)}{(x-3)(x-4)}>0 \\ & \text { Case - I}: y>0 \\ & \frac{x(x-1)(x-2)}{(x-3)(x-4)}>0 \\ & x \in(0,1) \cup(2,3) \\ & \text { Case }- \text { II : y<0 } \\ & \frac{x(x-1)(x-2)}{(x-3)(x-4)}<0, x \in(1,2) \cup(3,4) \end{aligned}$

$\begin{aligned} & \text { Area }=2 \int_\limits0^4 \sqrt{x} d x \\ & =2 \cdot \frac{2}{3}\left[x^{3 / 2}\right]_0^4=\frac{32}{3} \end{aligned}$