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_{-2}^0-x^3 d x+\int_0^4 x^3 d x=\frac{1}{4} \times 2^4+\frac{4^4}{4} \\ & =4+64=68 \end{aligned} $

Given, curves, $y=x^3, y=x^2$

for point of intersection

$ \begin{array}{ll} & x^3=x^2 \\ \Rightarrow \quad & x^2(x-1)=0 \Rightarrow x=0,1 \\ \therefore \quad & y=0,1 \end{array} $

point of intersection are $(0,0)$ and $(1,1)$

$\therefore$ Required area

$ \begin{aligned} & =\int_0^1\left(x^2-x^3\right) d x+\int_1^2\left(x^3-x^2\right) d x \\ & =\left(\frac{x^3}{3}-\frac{x^4}{4}\right)_0^1+\left(\frac{x^4}{4}-\frac{x^3}{3}\right)_1^2 \\ & =\left(\frac{1}{3}-\frac{1}{4}\right)+\left[\left(\frac{16}{4}-\frac{8}{3}\right)-\left(\frac{1}{4}-\frac{1}{3}\right)\right] \\ & =\frac{1}{12}+\frac{15}{4}-\frac{7}{3} \\ & =\frac{1+45-28}{12}=\frac{18}{12}=\frac{3}{2} \text { sq. units } \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \begin{aligned} & R=\left\{(x, y): \frac{y^2}{2} \leq x \leq y+4\right\} \\ & x \geq \frac{y^2}{2} \Rightarrow y^2 \leq 2 x \\ & \text { and } x \leq y+4 \Rightarrow y-x+4 \geq 0 \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Solving } y^2=2 x \text { and } y-x+4=0\\ &\begin{aligned} & \quad y-\frac{y^2}{2}+4=0 \\ & \Rightarrow \quad y^2-2 y-8=0 \\ & \Rightarrow(y-4)(y+2)=0 \\ & \Rightarrow \quad y=-2,4 \\ & \text { Area required }=\int_{-2}^4\left(x_L-x_p\right) d y \\ & =\int_{-2}^4\left(y+4-\frac{y^2}{2}\right) d y \\ & =\left[\frac{y^2}{2}+4 y-\frac{y^3}{6}\right]_{-2}^4 \\ & =\left(8+16-\frac{32}{3}\right)-\left(2-8+\frac{4}{3}\right) \\ & =24-\frac{32}{3}+6-\frac{4}{3} \\ & =30-\frac{36}{3}=30-12=18 \end{aligned} \end{aligned} $

Given:

The circle is $x^2 + y^2 = 16$.

The parabola is $y^2 = 6x$.

Finding Intersection Points:

To find where these curves meet, put $y^2$ from the parabola into the circle's equation:

$ \begin{array}{ll} \rightarrow & x^2 + y^2 = 16 \\ & x^2 + 6x = 16 \\ & x^2 + 6x - 16 = 0 \\ & (x + 8)(x - 2) = 0 \\ & x = -8,\ 2 \end{array} $

We only use $x = 2$ (since $x = -8$ is outside the region for $y^2 = 6x$ and gives no real $y$ for $x^2 + y^2 = 16$).

For $x = 2$, $y^2 = 6 \cdot 2 = 12$. So $y = \sqrt{12}$ and $y = -\sqrt{12}$.

Finding the Area:

We want the area inside the circle and outside the parabola, between $x = 0$ and $x = 4$.

To get this area, we split it into two parts:

• From $x = 0$ to $x = 2$ (where the parabola is below the circle), the top curve is the parabola: $y = \sqrt{6x}$
• From $x = 2$ to $x = 4$ (where only the circle remains), the top curve is the circle: $y = \sqrt{16 - x^2}$

So, the area is:

Area $= 2 \left[\int_0^2 y_\text{parabola} \,dx + \int_2^4 y_\text{circle} \,dx \right]$

So, $ = 2 \int_0^2 \sqrt{6x}\,dx + 2 \int_2^4 \sqrt{16 - x^2}\,dx $

Solving the Integrals:

First, $\int_0^2 \sqrt{6x}\,dx$:

$\sqrt{6x} = \sqrt{6} \sqrt{x}$, so

$ \int_0^2 \sqrt{6x} dx = \sqrt{6} \int_0^2 x^{1/2} dx = \sqrt{6} \cdot \left[\frac{2}{3}x^{3/2}\right]_0^2 = \sqrt{6} \cdot \frac{2}{3} \left[(2)^{3/2} - 0\right] $

Second, $\int_2^4 \sqrt{16 - x^2}\,dx$:

The formula for this is:

$ \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) $

Here $a = 4$. Plug in $x = 4$ and $x = 2$ to get the answer.

Now, combine both solutions:

$ \begin{aligned} &= 2 \sqrt{6} \cdot \frac{2}{3} \left(2^{3/2}\right) + 2\left[ \frac{x}{2}\sqrt{16 - x^2} + 8 \sin^{-1}\left(\frac{x}{4}\right)\Big|_{x=2}^{x=4} \right] \\ &= \frac{4}{3}\sqrt{6}\cdot 2\sqrt{2} + 2 \left[ \left(0 + 8 \frac{\pi}{2}\right) - \left(2\sqrt{3} + 8 \frac{\pi}{6}\right) \right] \\ &= \frac{16\sqrt{3}}{3} + 2\left(\frac{8\pi}{3} - 2\sqrt{3}\right) \\ &= \frac{16\sqrt{3}}{3} + \frac{16\pi}{3} - 4\sqrt{3} \\ &= \frac{4\sqrt{3}}{3} + \frac{16\pi}{3} \\ &= \frac{4}{3}(4\pi + \sqrt{3}) \end{aligned} $

Given, curves are $y=x^2$ and

$ y=8-x^2 $

Equating these two equations, we get

$ x^2=8-x^2 \Rightarrow 2 x^2=8 $

$ \begin{aligned} & \Rightarrow \quad x^2=4 \Rightarrow x= \pm 2 \\ & \text { Thus, area }=\int_{-2}^2\left[\left(8-x^2\right)-x^2\right] d x \\ & \qquad \begin{array}{l} \quad=\int_{-2}^2\left(8-x^2-x^2\right) d x=\int_{-2}^2\left(8-2 x^2\right) d x \\ \quad=\left[8 x-\frac{2}{3} x^3\right]_{-2}^2 \\ \quad=\left\{8(2)-\frac{2}{3}(2)^3\right\}-\left\{8(-2)-\frac{2}{3}(-2)^3\right\} \\ \quad=\left(16-\frac{16}{3}\right)-\left(-16+\frac{16}{3}\right) \\ \quad=16-\frac{16}{3}+16-\frac{16}{3} \\ \quad=32-\frac{32}{3}=\frac{96-32}{3}=\frac{64}{3} \end{array} \end{aligned} $

$\therefore$ The area of the region bounded by the curves is $\frac{64}{3}$ sq. units.

$ y=x^2-5 x+4=(x-1)(x-4) $

$ \begin{aligned} A=A_1+A_2=\int_0^1\left(x^2\right. & -5 x+4) d x +\int_1^2-\left(x^2-5 x+4\right) \\ =\left[\frac{x^3}{3}-\frac{5 x^2}{2}+4 x\right]_0^1 & +\left[\frac{-x^3}{3}+\frac{5 x^2}{2}-4 x\right]^2_1 \end{aligned} $

$ \begin{aligned} & =\left(\frac{1}{3}-\frac{5}{2}+4\right)+\left(-\frac{8}{3}+10-8+\frac{1}{3}-\frac{5}{2}+4\right) \\ & =\left(\frac{3}{2}+\frac{1}{3}\right)+\left(6-\frac{7}{3}-\frac{5}{2}\right) \\ & =\frac{9+2}{6}+\left(\frac{36-14-15}{6}\right) \\ & =\frac{11}{6}+\frac{7}{6}=\frac{18}{6}=3 \end{aligned} $

$ \begin{aligned} &\text { Required area (A) }\\ &=\int_0^{\frac{\pi}{4}}(\cos x-\sin x) d x+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x-\cos x) d x \end{aligned} $

$ \begin{aligned} = & (\sin x+\cos x)_0^{\frac{\pi}{4}}+(-\cos x-\sin x)_{\frac{\pi}{4}}^{\frac{\pi}{2}} \\ = & {\left[\left(\sin \frac{\pi}{4}+\cos \frac{\pi}{4}\right)-(\sin 0+\cos 0)\right] }-\left[\left(\cos \frac{\pi}{2}+\sin \frac{\pi}{2}\right)-\left(\cos \frac{\pi}{4}+\sin \frac{\pi}{4}\right)\right] \\ = & (\sqrt{2}-1)-(1-\sqrt{2}) \\ = & \sqrt{2}-1-1+\sqrt{2} \\ = & 2 \sqrt{2}-2=2(\sqrt{2}-1) \text { sq. unit } \end{aligned} $

$ \text { } y=\sqrt{4-x^2} \text { and } y^2=3 x $

$ \begin{aligned} &\begin{array}{ll} \Rightarrow & 3 x=4-x^2 \\ \Rightarrow & x^2+3 x-4=0 \\ \Rightarrow & (x+4)(x-1)=0 \\ & x=1,-4 \Rightarrow y=\sqrt{3} \end{array}\\ &\therefore \text { Required area, }\\ &\begin{aligned} & =\frac{1}{3} \int_0^{\sqrt{3}} y^2 d y+\int_{\sqrt{3}}^2 \sqrt{4-y^2} d y \\ & =\frac{1}{3}\left[\frac{y^3}{3}\right]_0^{\sqrt{3}}+\left[\frac{y}{2} \sqrt{4-y^2}+\frac{4}{2} \sin ^{-1} \frac{y}{2}\right]_{\sqrt{3}}^2 \\ & =\frac{1}{9}(3 \sqrt{3})+2 \frac{\pi}{2}-\frac{\sqrt{3}}{2}-2\left(\frac{\pi}{3}\right) \\ & =\frac{\sqrt{3}}{3}-\frac{\sqrt{3}}{2}+\frac{\pi}{3}=\frac{\pi}{3}-\frac{1}{2 \sqrt{3}} \end{aligned} \end{aligned} $

$ \text { Area of enclosed region, } $

$ \begin{aligned} & =2\left(\frac{1}{2} \times 1 \times 1\right)+(x) \\ & =1+1=2 \end{aligned} $

Here, $x^4=x^{\frac{1}{3}} \Rightarrow x^{12}=x$

$ \Rightarrow x\left(x^{11}-1\right)=0 $

So, $x=0$ or $x^{11}=1$ which gives $x=1$

$ \begin{aligned} \therefore \quad A & =\int_0^1\left(x^{\frac{1}{3}}-x^4\right) d x=\left[\frac{3}{4} x^{\frac{4}{3}}-\frac{1}{5} x^5\right]_0^1 \\ & =\left(\frac{3}{4}-\frac{1}{5}\right)=\frac{15-4}{20}=\frac{11}{20} \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

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