Area Under The Curves

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

$ \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 $

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

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

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

$ A=\int_{-1}^{\frac{1}{2}}\left(8 x^{3}-1\right) d x+\int_{\frac{1}{2}}^{1} 8 x^{3}-1 $

$ \begin{aligned} A= & \left(\frac{8 x^{4}}{4}-x\right)_{-1}^{1 / 2}+\left(8 \frac{x^{4}}{4}-x\right)_{1 / 2}^{1} \\\\ & =\left|\left(2 x^{4}-x\right)_{-1}^{1 / 2}\right|+\left(2 x^{4}-x\right)_{1 / 2}^{1} \\\\ & =\left|\left(2 \cdot \frac{1}{16}-\frac{1}{2}\right)-(2+1)\right| +\left[(2-1)-\left(\frac{2}{16}-\frac{1}{2}\right)\right] \\\\ & =\left|\frac{1}{8}-\frac{1}{2}-3\right|+1-\frac{1}{8}+\frac{1}{2} \\\\ & =\left|\frac{1-4-24}{8}\right|+\frac{8-1+4}{8} \\\\ & =\frac{27}{8}+\frac{11}{8}=\frac{38}{8}=\frac{19}{4} \end{aligned} $

Let the value of $ a=1 $

then area between curves $ x^{2}=y $ and $ x+y=2 $ is $ k $

$ \begin{aligned} & y=2-x \\\\ \Rightarrow \quad & x^{2}=2-x \\\\ & x^{2}+x-2=0 \\\\ & x^{2}+2 x-x-2=0 \\\\ & x=-2, x=1 \end{aligned} $

$ \text { Area }=\int_{-2}^{1}\left(x^{2}-2+x\right) d x $

$ =\left[\frac{x^{3}}{3}-2 x+\frac{x^{2}}{2}\right]_{-2}^{1} $

$ =\left(\frac{1}{3}-2+\frac{1}{2}\right)-\left(\frac{-8}{3}+4+2\right) $

$ =\frac{5}{6}-2+\frac{8}{3}-6 $

$ =\frac{16+5}{6}-8 $

$ =\frac{21}{6}-8 $

$ =\frac{7}{2}-8=\frac{-9}{2} $

Area $ =\frac{9}{2}=k $

We have,

$ \begin{array}{ll} & y^{2}=4 x+4 \text { and } y^{2}=5 x-20 \\\\ & \Rightarrow 4 x+4=5 x-20 \\\\ & x=24 \Rightarrow y^{2}=4(x+1) \\\\ & y^{2}=4 \times 25 \Rightarrow y= \pm 10 \end{array} $

Shaded area

$ \begin{array}{l} =2\left[\int_{-1}^{24} 2 \sqrt{x+1}-\int_{4}^{24} \sqrt{5} \sqrt{x-4}\right] \\\\ =2\left[2\left[\frac{(x+1)^{3 / 2}}{3 / 2}\right)_{-1}^{24}-\sqrt{5}\left(\frac{(x-4)^{3 / 2}}{3 / 2}\right)_{4}^{24}\right] \\\\ =2\left[\frac{4}{3}(25)^{3 / 2}-\frac{2 \sqrt{5}}{3}(\sqrt{20})^{3}\right] \end{array} $

$ =\frac{2}{3}[4 \times 125-2 \times \sqrt{5} \times 20 \times 2 \sqrt{5}] $

$ =\frac{2}{3}[500-400]=\frac{200}{3} $ sq. units

We have,

$ \begin{aligned} & C_1: y^2=4(x+7) \\\\ & C_2: y^2=5(2-x)=-5(x-2) \end{aligned} $

$\begin{aligned} & \therefore \text { Required area }=\left|\int\left(x_1-x_2\right) d y\right| \\\\ & =\left|\int_{-2 \sqrt{5}}^{2 \sqrt{5}}\left[\left(2-\frac{y^2}{5}\right)-\left(\frac{y^2}{4}-7\right)\right] d y\right| \\\\ & =\left|\int_{-2 \sqrt{5}}^{2 \sqrt{5}}\left(-\frac{9 y^2}{20}+9\right) d y\right| \\\\ & =\left|\left[-\frac{3 y^3}{20}+9 y\right]_{-2 \sqrt{5}}^{2 \sqrt{5}}\right|=24 \sqrt{5}\end{aligned}$

$ \begin{aligned} &\text { As }(\alpha, \beta) \text { is the stationary point of }\\ &\begin{aligned} & & y & =2 x-x^2, \frac{d y}{d x}=0 \\ & & & 2-2 x \\ \Rightarrow & & x & =0 \\ \Rightarrow & & \alpha & =1 \end{aligned} \end{aligned} $

$ \begin{aligned} & \therefore \text { Required area }=\int_0^1\left(y_2-y_1\right) d x \\ & \left.=\int_0^1\left(2^x-2 x+x^2\right)\right) d x=\left[\frac{2^x}{\log 2}-x^2+\frac{x^3}{3}\right]_0^1 \\ & =\left[\frac{2^1}{\log 2}-1+\frac{1}{3}-\frac{1}{\log 2}\right] \end{aligned} $

$ \begin{aligned} & =\frac{2}{\log 2}-1+\frac{1}{3}-\frac{1}{\log 2} \\ & =\frac{1}{\log 2}-\frac{2}{3} \\ & \Rightarrow \frac{3-\log 4}{3 \log 2} \end{aligned} $

$ \begin{aligned} & x=y^2, x=3-2 y^2 \\ & y=\sqrt{x} \\ & y=\sqrt{\frac{3-x}{2}} \end{aligned} $

Vertices $(0,0)$ and $(3,0)$

intersect at ( 1,1 ) and (1, -1)

Area of shaded region

$=2 \times($ Area of $P Q M)$

$ \begin{aligned} &\begin{aligned} & =2\left\{\left[\int_0^1 \sqrt{x} d x\right]+\left[\int_1^3 \sqrt{\frac{3-x}{2}} d x\right]\right\} \\ & \left.=2\left[\left(\frac{2}{3} x^{3 / 2}\right)_0^1+\left(\frac{-1}{\sqrt{2}} \cdot \frac{2}{3}(3-x)^{3 / 2}\right)\right)_1^3\right] \\ & =2\left[\frac{2}{3}-\left(0-\frac{1}{\sqrt{2}} \cdot \frac{2}{3} 2^{3 / 2}\right)\right] \\ & =2\left(\frac{2}{3}+\frac{4}{3}\right)=2 \times \frac{6}{3}=4 \end{aligned}\\ &\text { Area of required region }=4 \mathrm{sq} \text { units } \end{aligned} $

Given curves are

$ \begin{array}{r} 3 x^2-y^2-2 x y+4 x+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \text { and } 3 x^2-y^2-2 x y+6 x+2 y=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

Eqs. (i) and (ii) can be written as

$(3 x+y+1)(x-y+1)=0$, and $(3 x+y)(x-y+2)=0$, respectively.

Thus, the curve (i) represents the pair of lines

$ \begin{array}{r} L_1: 3 x+y+1=0 \\ L_2: x-y+1=0 \end{array} $

and the curve (ii) represents the pair of lines

$ \begin{array}{r} L_3: 3 x+y=0 \\ L_4: x-y+2=0 \end{array} $

Here, $L_1 \| L_3$ and $L_2 \| L_4$

So, the region closed by the given curves will be a parallelogram.

Now, required area $=$ Area of parallelogram $A B C D$ $=2 \times($ area of $\triangle A B C)$

$ \begin{aligned} & =2 \times \frac{1}{2}\left|\begin{array}{ccc} -\frac{1}{2} & \frac{1}{2} & 1 \\ -\frac{1}{4} & \frac{3}{4} & 1 \\ -\frac{1}{2} & \frac{3}{2} & 1 \end{array}\right| \\ & =\left[-\frac{1}{2}\left(\frac{3}{4}-\frac{3}{2}\right)-\frac{1}{2}\left(-\frac{1}{4}+\frac{1}{2}\right)+1\left(-\frac{3}{8}+\frac{3}{8}\right)\right] \\ & =-\frac{1}{2}\left(-\frac{3}{4}\right)-\frac{1}{2}\left(\frac{1}{4}\right)+0 \\ & =\frac{3}{8}-\frac{1}{8}=\frac{2}{8}=\frac{1}{4} \end{aligned} $

To find the area under the curve $ y = |\sin x - \cos x| $ from $ x = 0 $ to $ x = \frac{\pi}{2} $, and above the $ X $-axis, we proceed as follows:

The expression $|\sin x - \cos x|$ changes based on $x$. Solving $|\sin x - \cos x| = 0$:

$ \begin{align*} \sin x - \cos x &= 0 \\ \sin x &= \cos x \\ \tan x &= 1 \\ x &= \frac{\pi}{4} \end{align*} $

This implies that $ \sin x - \cos x $ is negative from $ 0 $ to $ \frac{\pi}{4} $ and positive from $ \frac{\pi}{4} $ to $ \frac{\pi}{2} $.

Thus, the function can be defined piecewise as:

$ |\sin x - \cos x| = \begin{cases} \cos x - \sin x & \text{for } 0 \leq x < \frac{\pi}{4} \\ \sin x - \cos x & \text{for } \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \end{cases} $

Integrating from $ x = 0 $ to $ x = \frac{\pi}{4} $:

$ \int_0^{\pi/4} (\cos x - \sin x) \, dx = \int_0^{\pi/4} \cos x \, dx - \int_0^{\pi/4} \sin x \, dx $

Calculating each integral:

$ \int_0^{\pi/4} \cos x \, dx = [\sin x]_0^{\pi/4} = \sin\frac{\pi}{4} - \sin 0 = \frac{\sqrt{2}}{2} $

$ \int_0^{\pi/4} \sin x \, dx = [-\cos x]_0^{\pi/4} = -\cos\frac{\pi}{4} + \cos 0 = -\frac{\sqrt{2}}{2} + 1 $

Thus, the area from $ x = 0 $ to $ x = \frac{\pi}{4} $ is:

$ \int_0^{\pi/4} (\cos x - \sin x) \, dx = \frac{\sqrt{2}}{2} - \left( -\frac{\sqrt{2}}{2} + 1 \right) = \sqrt{2} - 1 $

Integrating from $ x = \frac{\pi}{4} $ to $ x = \frac{\pi}{2} $:

$ \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx = \int_{\pi/4}^{\pi/2} \sin x \, dx - \int_{\pi/4}^{\pi/2} \cos x \, dx $

Calculating each integral:

$ \int_{\pi/4}^{\pi/2} \sin x \, dx = [-\cos x]_{\pi/4}^{\pi/2} = -\cos\frac{\pi}{2} + \cos\frac{\pi}{4} = 0 + \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} $

$ \int_{\pi/4}^{\pi/2} \cos x \, dx = [\sin x]_{\pi/4}^{\pi/2} = \sin\frac{\pi}{2} - \sin\frac{\pi}{4} = 1 - \frac{\sqrt{2}}{2} $

Thus, the area from $ x = \frac{\pi}{4} $ to $ x = \frac{\pi}{2} $ is:

$ \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx = \frac{\sqrt{2}}{2} - \left( 1 - \frac{\sqrt{2}}{2} \right) = \sqrt{2} - 1 $

Total Area:

Adding the two parts:

$ (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2(\sqrt{2} - 1) $

Thus, the total area under the curve is $ 2(\sqrt{2} - 1) $ square units.

$ \text { We have, } $

$ y^2=8(x+2) \text { and } y^2=4(1-x) $

The intersecting point of the above 2 parabola are $(-1, \pm 2 \sqrt{2})$

$\therefore$ Required area

$ \begin{aligned} & =\left\lvert\, 2\left[\int_2^{2 \sqrt{2}}\left(1-\frac{y^2}{4}\right) d y+\int_{2 \sqrt{2}}^4\left(\frac{y^2}{8}-2\right) d y\right]\right. \\ & =\left\lvert\, 2\left[\left[y-\frac{y^3}{12}\right]_2^{2 \sqrt{2}}+\left[\frac{y^3}{24}-2 y\right]_{2 \sqrt{2}}^4\right]\right. \\ & =\left\lvert\, 2\left[\left(2 \sqrt{2}-\frac{16 \sqrt{2}}{12}\right)-\left(2-\frac{8}{12}\right)+\left(\frac{64}{24}-8\right)\right.\right. \\ & \left.-\left(\frac{16 \sqrt{2}}{24}-4 \sqrt{2}\right)\right] \end{aligned} $

$ \begin{aligned} & =\left\lvert\, 2\left[\frac{2 \sqrt{2}}{3}-\frac{4}{3}-\frac{16}{3}+\frac{10 \sqrt{2}}{3}\right]\right. \\ & =\left|2\left[\frac{12 \sqrt{2}}{3}-\frac{20}{3}\right]\right| \\ & =\left|\frac{8}{3}[3 \sqrt{2}-5]\right|=\frac{8}{3}(5-3 \sqrt{2}) \end{aligned} $

Given the problem of finding the area of a region bounded by a circle and a parabola, let's explore how to calculate it.

The equation of the circle is:

$ x^2 + y^2 = 2ax $

This can be rewritten by completing the square for $x$:

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

The equation of the parabola is:

$ y^2 = ax $

Finding Points of Intersection:

To find the intersection points between the circle and the parabola, we set:

$ x^2 + ax = 2ax $

$ x^2 - ax = 0 $

$ x(x-a) = 0 $

Thus, $x = 0$ or $x = a$.

For $x = 0$, substituting in the parabola $y^2 = ax$, we get $y = 0$.

For $x = a$, substituting in the parabola $y^2 = ax$, we get $y = \pm a$.

Thus, the points of intersection are $(0, 0)$, $(a, a)$, and $(a, -a)$.

Calculating the Area:

We need to integrate the area between the circle and the parabola from $x = 0$ to $x = a$:

$ \text{Required Area} = \int_0^a \left( \sqrt{a^2 - (x-a)^2} - \sqrt{a} \sqrt{x} \right) \, dx $

This evaluates to:

$ = \left[ \frac{x-a}{2} \sqrt{a^2-(x-a)^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x-a}{a}\right) - \sqrt{a} \times \frac{2}{3} x^{\frac{3}{2}} \right]_0^a $

$ = \left[ -\sqrt{a} \cdot a^{\frac{3}{2}} \times \frac{2}{3} - \frac{a^2}{2} \cdot \sin^{-1}(-1) \right] $

$ = -\frac{2}{3} a^2 + \frac{a^2}{2} \times \frac{\pi}{2} $

Therefore, the area of the smaller region lying above the $X$-axis is:

$ a^2 \left(\frac{\pi}{4} - \frac{2}{3}\right) $

$\begin{aligned} & (y)=x^3-19 x+30 \\ & =(x-2)(x-3)(x+5)\end{aligned}$

$\begin{aligned} & \therefore \text { Required area } \\ & \begin{aligned} &=\int_{-5}^2\left(x^3-19 x+30\right) d x+\left|\int_2^3 x^3-19 x+30 d x\right| \\ &= {\left[\frac{x^4}{4}-\frac{19 x^2}{2}+30 x\right]_{-5}^2 } \\ &+\left|\left[\frac{x^4}{4}-\frac{19 x^2}{2}+30 x\right]_2^3\right| \\ &=(4-38+60)-\left(\frac{625}{4}-\frac{475}{2}-150\right) \\ &+\left|\left(\frac{81}{4}-\frac{171}{2}+90\right)-(4-38+60)\right| \\ &=26+\frac{925}{4}+\left|\frac{99}{4}-26\right|=\frac{1029}{4}+\frac{5}{4} \\ &= \frac{1034}{4}=\frac{517}{2}\end{aligned}\end{aligned}$

Given, $ y=1+4 x-x^{2} $

$ \Rightarrow \quad \frac{d y}{d x}=4-2 x=0 \Rightarrow x=2 $ [point of maxima]

$ \Rightarrow \quad y_{\max }=1+4 \times 2-4=5 $

when, $ x=0, y=1 $

$ \therefore $ Graph according to question and above information

$ \Rightarrow $ If $ y=m x $ bisects the area bounded

$ \Rightarrow A_{1}=A_{2} $

$ \Rightarrow \quad \int_{0}^{3 / 2}\left(1+4 x-x^{2}\right) d x=2 \int_{0}^{3 / 2} m x d x $

$ \Rightarrow\left[x+2 x^{2}-\frac{x^{3}}{3}\right]_{0}^{3 / 2}=\left[m x^{2}\right]_{0}^{3 / 2} $

$ \Rightarrow \frac{3}{2}+2 \times \frac{9}{4}-\frac{27}{8} \times \frac{1}{3}=m \times \frac{9}{4} $

$ \therefore \quad m=\frac{13}{6} $

Given curves are $ y=x^{3} $ and $ y=\sqrt{x} $

Both curves intersects at $ x=0 $ and $ x=1 $

Shaded region is required area

$ \begin{aligned} A & =\left|\int_{0}^{1}\left(x^{3}-\sqrt{x}\right) d x\right|=\left|\left[\frac{x^{4}}{4}-\frac{2 x^{3 / 2}}{3}\right]_{0}^{1}\right| \\ & =\left|\left[\frac{1}{4}-\frac{2}{3}\right]\right| \\ & =\frac{5}{12} \text { sq. unit } \end{aligned} $

Given equation : $4{x^2} - 4px + 5p = 0$

for rational roots, D must be perfect square

$D = 16{p^2} - 4 \times 4 \times 5p = 16p(p - 5)$

So, max. Integral value of $p = 9$ for making D is perfect square

$\therefore$ $q = 9$

Area of shared region

$ = \int\limits_0^9 {{{(x - 9)}^2}dx} $

$ = \left. {{{{{(x - 9)}^3}} \over 3}} \right|_0^9 = {{{9^3}} \over 3} = 243$ sq. units

$ |\cos x-\sin x| \leq y \leq \sin x $

Intersection point of $\cos x-\sin x=\sin x$

$ \Rightarrow \tan x=\frac{1}{2} $

Let $\psi=\tan ^{-1} \frac{1}{2}$

So, $\tan \psi=\frac{1}{2}, \sin \psi=\frac{1}{\sqrt{5}}, \cos \psi=\frac{2}{\sqrt{5}}$

Area $ = \int\limits_{{{\tan }^{ - 1}}{1 \over 2}}^{{\pi \over 4}} {(\sin x - (\cos x - \sin x))dx + \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {(\sin x - (\sin x - \cos x))dx} } $

$ =\int\limits_{{{\tan }^{ - 1}}{1 \over 2}}^{\pi / 4}(2 \sin x-\cos x) d x+\int\limits_{\pi / 4}^{\pi / 2} \cos x d x $

$ =[-2 \cos x-\sin x]_{{{\tan }^{ - 1}}{1 \over 2}}^{\pi / 4}+[\sin x]_{\pi / 4}^{\pi / 2} $

$ =-\sqrt{2}-\frac{1}{\sqrt{2}}+2 \cos ({{{\tan }^{ - 1}}{1 \over 2}}) +\sin ({{{\tan }^{ - 1}}{1 \over 2}})+\left(1-\frac{1}{\sqrt{2}}\right)$

$=-\sqrt{2}-\frac{1}{\sqrt{2}}+2\left(\frac{2}{\sqrt{5}}\right)+\left(\frac{1}{\sqrt{5}}\right)+1-\frac{1}{\sqrt{2}}$

$ = \sqrt 5 - 2\sqrt 2 + 1$

Combining both the graph we get,

Both graph cuts each other at $y=2$ and $y=1$

$\therefore x^2=2$

$\Rightarrow x=\sqrt2$

Two graphs cuts each other at $x=1$ and $x=\sqrt2$

In 0 to 1, $f(x) = \max ({x^2},1 + [x]) = 1$

In 1 to $\sqrt2$, $f(x) = \max ({x^2},1 + [x]) = 2$

In $\sqrt2$ to 2, $f(x) = \max ({x^2},1 + [x]) = {x^2}$

$\therefore$ $\int\limits_0^2 {f(x)dx} $

$ = \int\limits_0^1 {1\,dx + \int\limits_1^{\sqrt 2 } {2\,dx + \int\limits_{\sqrt 2 }^2 {{x^2}\,dx} } } $

$ = \left[ x \right]_0^1 + 2\left[ x \right]_1^{\sqrt 2 } + \left[ {{{{x^3}} \over 3}} \right]_{\sqrt 2 }^2$

$ = 1 + 2(\sqrt 2 - 1) + {8 \over 3} - {{2\sqrt 2 } \over 3}$

$ = {8 \over 3} - 1 + 2\sqrt 2 - {{2\sqrt 2 } \over 3}$

$ = {5 \over 3} + {{6\sqrt 2 - 2\sqrt 2 } \over 3}$

$ = {5 \over 3} + {{4\sqrt 2 } \over 3}$

$ = {{5 + 4\sqrt 2 } \over 3}$

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