Mathematics M.C.Q (1 Marks)

Hence required area $= \int_1^2 {\,\log \,x\,dx}   =  (x\,\log x - x)_1^2$

$=  2\log 2 - 1 = (\log 4 - 1)\,\,sq. \,unit$.

We put ${x^2} = t \Rightarrow dx = \frac{{dt}}{{2x}}$ as

$x = 0 \Rightarrow t = 0$ and $x = a \Rightarrow t = {a^2}$,

then it reduces to

$\frac{1}{2}\int_0^{{a^2}} {{e^t}dt = \frac{1}{2}[{e^t}]_0^{{a^2}} = \frac{{{e^{{a^2}}} - 1}}{2}}\,\,sq. \,unit$.

$= \int_0^\pi {y\,\,dx + \left| {\int_\pi ^{2\pi } {y\,dx} } \right| = 4\pi \,sq.} \,unit$

$\left( {\frac{{4a}}{{{m^2}}},\frac{{4a}}{m}} \right)$ and the area enclosed by the two curves is given by 

$\int_0^{4a/{m^2}} {(\sqrt {4ax} - mx)} \,dx$.

$\therefore \,\,\,\int_0^{4a/{m^2}} {(\sqrt {4ax} - mx)} \,dx = \frac{{{a^2}}}{3}$

==> $\frac{8}{3}\frac{{{a^2}}}{{{m^3}}} = \frac{{{a^2}}}{3} $

$\Rightarrow {m^3} = 8 \Rightarrow m = 2$. 

$\Rightarrow {y^2} = 2y \Rightarrow y = 0,\,2$

$\therefore \,$ Required area $ = \int_0^2 {({y^2} - 2y)dy = \left( {\frac{{{y^3}}}{3} - {y^2}} \right)_0^2}$

$={ \frac{4}{3}}  \,\, sq. \,unit$.

$\therefore$ Required area $ = \frac{1}{2} \times AC \times BC$

$ = \frac{1}{2} \times 2\sqrt 2 \times 2\sqrt 2 = 4\,\, sq. \,unit$.

This curve passes through $(1, 2)$, 

$\therefore 2 = a + b$ ...$(i)$

and area bounded by this curve and line $x = 4$ and $x-$ axis is $8$ sq. unit,

then $\int_{\,0}^{\,4} {(a\sqrt x + bx)\,} dx = 8$

==> $\frac{{2a}}{3}[{x^{3/2}}]_0^4 + \frac{b}{2}[{x^2}]_0^4 = 8$, 

==> $\frac{{2a}}{3}.8 + 8b = 8$

==> $2a + 3b = 3$ ...$(ii)$

From equation $(i)$ and $(ii),$ we get

$a = 3,\,b = - 1$.

Now check from options, only $(b)$ satisfies the above condition.

$\Rightarrow y = \frac{{3x + 10}}{{x - 2}}$

Required area is $\int_3^4 {\,\,y\,dx = \int_3^4 {\frac{{3x + 10}}{{x - 2}}\,\,dx} } $

$= [3x + 16\log (x - 2)]_3^4 = 3 + 16\log 2$ sq. unit.

$ = 16 \times 4 - \int_0^{16} {\sqrt x } dx = 64 - \left[ {\frac{{{x^{3/2}}}}{{3/2}}} \right]_0^{16} = \frac{{64}}{3}.$

$ = \left[ {\frac{{{2^x}}}{{\log 2}} - {x^2} + \frac{{{x^3}}}{3}} \right]_0^2$

$ = \frac{4}{{\log 2}} - 4 + \frac{8}{3} - \frac{1}{{\log 2}}$

$ = \frac{3}{{\log 2}} - \frac{4}{3}$.

$ = 2\,\left[ {\frac{x}{2}\sqrt {4 - {x^2}} + 2{{\sin }^{ - 1}}\frac{x}{2}} \right]_1^2$

$ = 2\,\left[ {2.\frac{\pi }{2} - \left[ {\frac{{\sqrt 3 }}{2} - 2.\frac{\pi }{6}} \right]} \right]$

$ = 2\,\left[ {\pi - \left[ {\frac{{\sqrt 3 }}{2} - \frac{\pi }{3}} \right]} \right]$

$ = \frac{{8\pi }}{3} - \sqrt 3 $.

$ = \frac{1}{2}[x]_0^{\pi /2} - \frac{1}{4}[\sin 2x]_0^{\pi /2} = \frac{\pi }{4}$.

There are two points on the curve $(a, a),(- a,- a)$

The equation of the line at $(a,a)$ is,

$y - a = {\left( {\frac{{dy}}{{dx}}} \right)_{(a,\,a)}}(x - a)$

$ = {\left( {\frac{{ - {a^2}}}{{{x^2}}}} \right)_{(a,\,a)}}(x - a)$

$\Rightarrow$ $y - a = - (x - a)$ 

therefore, equation of the tangent at $(a,a)$ is $x + y = 2\,a$.

The interception of line $x + y = 2a$ with $x-$ axis is $2a$ and with $y-$ axis is $2a.$

$\therefore $ Required area $=  \frac{1}{2} \times 2a \times 2a = 2{a^2}$.

$\therefore $ $x = 0$, $4$

Required area $ = \int_0^4 {(4x - {x^2})dx = \left[ {\frac{{4{x^2}}}{2} - \frac{{{x^3}}}{3}} \right]_0^4} $

$ = 32 - \frac{{64}}{3} = \frac{{32}}{3}\,\, sq. \,unit$

we get $x = - 1,\,4$.

Curve does not intersect $x -$ axis between $x = - 1$ and $x = 4$.

$\therefore$ Area $ = \int_{ - 1}^4 {(4 + 3x - {x^2})dx = \frac{{125}}{6}} $.

$2 \times 2\sqrt a \times \frac{2}{3}\left| {{x^{3/2}}} \right|_0^a $

$= \frac{8}{3}{a^2}\,\, sq. \,unit$

the curve is symmetrical about $y -$ axis as well as $x - $ axis.

$\therefore$ Whole area of given ellipse

$ = 4({\rm{area\,\, }}\,{\rm{of}}\,BCO) = 4 \times \int_0^a {y\,dx = 4\int_0^a {\frac{b}{a}\sqrt {{a^2} - {x^2}} } dx} $

$ = 4ab\int_0^{\pi /2} {\left( {\frac{{1 + \cos 2\theta }}{2}} \right)\,d\theta } $, {Putting 

$x = a\sin \theta $}

$ = 2ab\left( {\int_0^{\pi /2} {\,\,d\theta + \int_0^{\pi /2} {\,\,\,\cos 2\theta \,d\theta } } } \right)$

$ = [\theta ]_0^{\pi /2} + \left[ {\frac{{\sin 2\theta }}{2}} \right]_0^{\pi /2} = \pi ab\,\,\, sq. \,unit$

$ = 4\sqrt a \times \frac{2}{3}[{x^{3/2}}]_0^a = \frac{{8\sqrt a }}{3}.a\sqrt a = \frac{8}{3}{a^2}$.

$ = 2.\frac{1}{2}\left[ {x\sqrt {9 - {x^2}} + 9{{\sin }^{ - 1}}\frac{x}{3}} \right]_1^3 $

$= \left[ {9\frac{\pi }{2} - \sqrt 8 - 9{{\sin }^{ - 1}}\left( {\frac{1}{3}} \right)} \right]$

$ = \left[ {9\left( {\frac{\pi }{2} - {{\sin }^{ - 1}}\left( {\frac{1}{3}} \right)} \right) - \sqrt 8 } \right]$

$ = \left[ {9{{\cos }^{ - 1}}\left( {\frac{1}{3}} \right) - \sqrt 8 } \right]$

$= [9{\sec ^{ - 1}}(3) - \sqrt 8 ]$.

$\Rightarrow {x^2} = 8x \Rightarrow x = 0$,8

$\therefore$ Required area =$\int_0^8 {(2\sqrt 2 \sqrt x - x)dx} $

$ = \left[ {\frac{{4\sqrt 2 }}{3}{x^{3/2}} - \frac{{{x^2}}}{2}} \right]_0^8 $

$= \frac{{128}}{3} - \frac{{64}}{2} = \frac{{32}}{3}\,\, sq. \,unit$

${y^2} = 4ax$ and the chord $y = 2ax$ is obtained by solving these equations simultaneously.

${y^2} = 4ax,y = 2ax \Rightarrow {(2ax)^2} = 4ax$

==> $x[4{a^2}x - 4a] = 0$

$ \Rightarrow 4ax[ax - 1] = 0$

==> $x = 0$ or $x = \frac{1}{a}$

Also $x = 0 \Rightarrow y = 0$ and $x = \frac{1}{a}$

==>$y = \pm 2$

Hence the required points are $(0,0)$ and $\left[ {\frac{1}{a},2} \right]$.

Now required area $ = \int_0^{1/a} {[\sqrt {4ax} - 2ax} ]dx = \frac{1}{{3a}}\,\, sq. \,unit$.

$ = \int_{ - 1}^0 { - {x^2}dx + \int_0^1 {{x^2}dx} } $

$=\left( {\frac{{ - {x^3}}}{3}} \right)_{ - 1}^0$$ + \left( {\frac{{{x^3}}}{3}} \right)_0^1$

$ = \left| {\,\frac{{ - 1}}{3}\,} \right| + \left| {\,\frac{1}{3}\,} \right| = \frac{2}{3}$.

==> $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1$,

which is equation of an ellipse. 

Remember area enclosed by ellipse is $\pi \,ab$

$i.e.,$ $\pi \;2.3 = 6\pi $.

$\therefore$ Area $O A B=\int_{0}^{2} y d x$

$=\int_{0}^{2} \sqrt{4-x^{2}} d x$

$=\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{0}^{2}$

$=2\left(\frac{\pi}{2}\right)=\pi $ sq. units

Thus, the correct answer $A.$

On solving, we get $x = 0,x = 1$

Therefore, required area $= \int_0^1 {({x^3} - \sqrt x )} \,dx$

$ = \left[ {\frac{{{x^4}}}{4} - \frac{{2x\sqrt x }}{3}} \right]_0^1 $

$= \left[ {\frac{1}{4} - \frac{2}{3}} \right] = \frac{5}{{12}}$,

(Area can not be negative).

On solving, we get $x = 0$, $x = 1$

Therefore, required area $A = \int_0^1 {({x^2} - x)} dx$

$ = \left[ {\frac{{{x^3}}}{3} - \frac{{{x^2}}}{2}} \right]_0^1$

$ = \frac{1}{3} - \frac{1}{2} = \frac{1}{6}\,\, sq. \,unit$.

$\therefore$ Required area $ = 4\int_0^1 {(1 - {x^2})\,dx} $

$ = 4\left[ {x - \frac{{{x^3}}}{3}} \right]_0^1 = \frac{8}{3}$.

$=2\int_0^1 {(x - {x^2})\,dx = 2 \times \frac{1}{6} = \frac{1}{3}} $.

We know that area bounded by the curves 

$ = \int_{{x_1}}^{{x_2}} {ydx = \int_0^{\pi /4} {\cos xdx - \int_0^{\pi /4} {\,\,\sin x\,dx} } } $

$ = [\sin x]_0^{\pi /4} - [ - \cos x]_0^{\pi /4}$

$ = \left( {\sin \frac{\pi }{4} - \sin 0} \right) + \left( {\cos \frac{\pi }{4} - \cos 0} \right) $

$= \left( {\frac{1}{{\sqrt 2 }} - 0} \right) + \left( {\frac{1}{{\sqrt 2 }} - 1} \right)$

$ = \sqrt 2 - 1$.

$i.e.,$ $\frac{{{\pi ^3}}}{4}$.

Also area bounded by curve $y = \sin x$ and $x -$ axis is $2$ sq. unit. 

Hence required area is $\frac{{{\pi ^3}}}{4} - 2 = \frac{{{\pi ^3} - 8}}{4}$.

$y = 2 - {x^2}$.....$(ii)$

$\therefore $ By equation $(i)$ and $(ii),$ we get,  $x = \pm 1$

$\therefore $ $y = \pm 1$

$\therefore $ Required area $ = 2\left[ {\int_0^1 {(2 - {x^2})dx - \int_0^1 {{x^2}dx} } } \right]$

$ = 2\,\left[ {2x - \frac{{2{x^3}}}{3}} \right]_0^1 = 4\left[ {x - \frac{{{x^3}}}{3}} \right]_0^1 = 4\left( {\frac{2}{3}} \right) = \frac{8}{3}$.

Since, $\sin x \ge \cos x$ on the interval $\left[ {\frac{\pi }{4},\frac{{5\pi }}{4}} \right]$

$\therefore $ Area of one such region $ = \int_{\pi /4}^{5\pi /4} {(\sin x - \cos x)\;dx} $

$ = 2\sqrt 2 sq.\;unit$

The area of region bounded by the curves

$x\left(y-e^x\right)=\sin x$ or $y=\frac{\sin x}{x}+e^x$ and

$2 x y=2 \sin x+x^3$

or $\quad y=\frac{\sin x}{x}+\frac{x^2}{2}$

between lines $x=1$ and $x=2$ is

$A=\int \limits_1^2\left\{\left(\frac{\sin x}{x}+e^x\right)-\left(\frac{\sin x}{x}+\frac{x^2}{2}\right)\right\} d x$

$=\int \limits_1^2\left(e^x-\frac{x^2}{2}\right) d x=\left|\left[e^x-\frac{x^3}{6}\right]_1^2\right|=e^2-e-\frac{7}{6}$

Area of region defincd by

$A_1=\left\{\begin{array}{r}\left.(x, y): x \geq 0, y \geq 0,2 x+2 y-x^2\right\} \\ -y^2>1>x+y\end{array}\right\}$

Similarly, $\left|A_2\right|$ is the area of region definced by

$\left.A_2= \forall (x, y) \leq x \geq 0, y \geq 0, x+y>1>x^2+y^2\right\}$

Here, $\left|A_1\right|=\left|A_2\right|$

and area $\left|A_3\right|$ is the area of refion defined by

$A_3=\left\{(x, y): x \geq 0, y \geq 0, x+y>1>x^3+y^3\right\}$

then $\left|A_3\right|>\left|A_2\right|=\left|A_1\right|$

We have,

$y=x^2+x+10$

$x^2+x+\frac{1}{4}=y-10+\frac{1}{4}$

$\left(x+\frac{1}{2}\right)^2=\left(y-\frac{39}{4}\right)$

Latusrectum of parabola is $1 .$

$\therefore$ Length of chord is also

$1$

Area is maximum when chord is latusrectum.

Area of shaded region $=$

Area of rectangle $O A B C-2 \int \limits_{-1 / 2}^0$ parabola

$=10-2 \int \limits_{\frac{-1}{2}}^0\left(x^2+x+10\right) d x$

$=10-2\left[\frac{x^3}{3}+\frac{x^2}{2}+10\right]_{-1}^2$

$=10-2\left[\frac{-1}{24}+\frac{1}{8}-\frac{10}{2}\right]=\frac{1}{6}$

Area of region $O B A O$

$=\operatorname{area}(T)=2 \int_0^a \sqrt{4 e x} d x$

$=4 \sqrt{a}\left[\frac{2}{3} x^{3 / 2}\right]_0^a=4 \sqrt{a}\left[\frac{2}{8} a^{3 / 2}-0\right]=\frac{8}{3} a^2$

Area of rectangle $=(P Q R S)$

$=R S \times R Q=2 y(a-x)$

$\quad A=2 y\left(a-\frac{y^2}{4 a}\right) \quad\left[\because x=\frac{y^2}{4 a}\right]$

$\frac{d A}{d y}=2 a-\frac{3 y^2}{2 a}$

For maxima or minima $\frac{d A}{d y}=0$

$2 a-\frac{3 y^2}{2 a}=0 \Rightarrow y=\frac{2 a}{\sqrt{3}}$

$\therefore$ Area of rectangle

$P Q R S=2 \times \frac{2 a}{\sqrt{3}}\left(a-\frac{a}{3}\right)=\frac{8 a^2}{3 \sqrt{3}}$

$\frac{\text { Arca of rectangle } P Q R S}{\text { Area of rectangle }(T)}$

$=\frac{8 a^2 / 3 \sqrt{3}}{8 a^2 / 3}=\frac{1}{\sqrt{3}}$

We have,

$y=\frac{1}{4}\left|4-x^2\right| \text { and } y=7-|x|$

Graph of given curves

Area of shaded region

$=2\left[\int \limits_0^4(7-x) d x-\int \limits_0^4 \frac{1}{4}\left|4-x^2\right| d x\right]$

$=2\left[\int \limits_0^4(7-x) d x-\frac{1}{4}\left[\int \limits_0^2\left(4-x^2\right) d x\right.\right.$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\left.\left.+\int \limits_2^4\left(x^2-4\right) d x\right]\right]$

$=2\left[\left[7 x-\frac{x^2}{2}\right]_0^4-\frac{1}{4}\left[4 x-\frac{x^3}{3}\right]_0^2\right.$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\left.-\frac{1}{4}\left[\frac{x^3}{3}-4 x\right]_2^4\right]$

We have, parabola $y^2=4 x+1$ and circle $x^2+y^2 \leq 1$

Area of shaded region,

$A_1=2\left(\int \limits_{-1 / 4}^0 \sqrt{4 x+1} d x+\int_0^1 \sqrt{1-x^2} d x\right)$

$\Rightarrow \quad A_1=2\left[\frac{2}{3}(4 x+1)^{3 / 2}\right]_{-1 / 4}^0+2 \times \frac{\pi}{2}$

$\Rightarrow \quad A_1=\frac{1}{3}+\frac{\pi}{2}$

$\begin{aligned} A_2 =\text { Area of circle }-A_1 \\ =\pi-\left(\frac{1}{3}+\frac{\pi}{2}\right)=\frac{\pi}{2}-\frac{1}{3} \\ \therefore\left|A_1-A_2\right| =\left|\frac{1}{3}+\frac{\pi}{2}-\frac{\pi}{2}+\frac{1}{3}\right|=\frac{2}{3} \end{aligned}$

Given, $A_1=\left\{(x, y): x^2+2 y^2 \leq 1\right\}$

$A_2=\left\{(x, y):\left.\left|x^3+2 \sqrt{2}\right| y\right|^3 \leq 1\right\}$

$A_3=\{(x, y): \max (|x|, \sqrt{2}|y| \leq 1)\}$

Graph of $A_1, A_2$ and $A_3$ are

Clearly from graph $A_1 \subset A_2 \subset A_3$

We have, $y=\left|x^3-4 x^2+3 x\right|$

Area of region bounded by $y \leq f(x), X$-axis and $0 \leq x \leq 3$ is $\int_0^3\left|x^3-4 x^2+3 x\right| d x$

$\int_0^1\left(x^3-4 x^2+3 x\right)-\int_1^3\left(x^3-4 x^2+3 x\right) d x$

$=\left[\frac{x^4}{4}-\frac{2 x^3}{3}+\frac{3 x^2}{2}\right]_0^1-\left[\frac{x^4}{4}-\frac{4 x^3}{3}+\frac{\left.3 x^2\right]^3}{2}\right\rfloor_1$

$\left.=\left[\left(\frac{1}{4}-\frac{4}{3}+\frac{3}{2}\right)-(0)\right\rfloor\right]$

We have, $y=2 x-4 x^3$

Given, area of $I$ region = area of $II$ region

$\therefore \quad \int \limits_a^b\left(2 x-4 x^3\right) d x=2(b-a) c$

${\left[x^2-x^4\right\}_a=2(b-a) c }$

$\Rightarrow \quad\left(b^2-b^4\right)-\left(a^2-a^4\right)=2(b-a) c$

$\Rightarrow \quad\left(b^2-a^2\right)-\left(b^4-a^4\right)=2(b-a) c$

$\Rightarrow \quad\left(b^2-a^2\right)\left(1-\left(b^2+a^2\right)\right)=2(b-a) c$

$\Rightarrow \quad(b+a)\left(1-b^2-a^2\right)=2 c$

$\Rightarrow \quad(b+a)\left(1-(b+a)^2+2 a b\right)=2 c$

$\quad\left[\because c=2 x-4 x^3\right]$

$\Rightarrow \quad c=2 x-4 x^3$

$\therefore \quad$ Let $a, b$ and $\alpha$ are roots.

$\therefore a+b+\alpha=0, a b+(a+b) \alpha=-\frac{1}{2}$

$a b \alpha=-\frac{c}{4}$

$a+b=-\alpha, a b=\alpha^2-\frac{1}{2}, a b=\frac{-c}{4 \alpha}$

$1-\alpha^2+2\left(\alpha^2-\frac{1}{2}\right)=-8 \alpha\left(\alpha^2-\frac{1}{2}\right)$

$\Rightarrow \quad 1-\alpha^2+2 \alpha^2-1=8 \alpha^2-4$

$\Rightarrow \quad \alpha^2=4 \Rightarrow \alpha=\frac{2}{\sqrt{7}}$

Given, $y=\cos x$

Equation of line joining

$\left(\frac{-\pi}{4}, \cos \left(\frac{-\pi}{4}\right)\right)$ and $(0,2)$ is

$y-2 =\frac{2-1 / \sqrt{2}}{0+\pi / 4}(x-0)$

$\Rightarrow \quad y-2 =(8-2 \sqrt{2}) x$

$\Rightarrow \quad y =\frac{(8-2 \sqrt{2})}{\pi} x+2$

Equation of line joining $\left(\frac{\pi}{4}, \cos \frac{\pi}{4}\right)$ and $(0,2)$ is

$y-2 =\frac{2-\frac{1}{\sqrt{2}}}{0-\frac{\pi}{4}}(x-0)$

$\Rightarrow \quad y =\left(\frac{-8+2 \sqrt{2}}{\pi}\right) x+2$

Graph of given curve, $y=\cos x$, and line are

Area of shaded region

$=2 \text { area of curve } A P C A$

$=2 \int \limits_0^{\pi / 4}\left[\frac{(-8+2 \sqrt{2})}{\pi} x+2-\cos x\right] d x$

$=2\left[\frac{-8 x^2}{2 \pi}+\frac{2 \sqrt{2} x^2}{2 \pi}+2 x-\sin x\right]_0^{\pi / 4}$

$=2\left[\frac{-4}{\pi}\left(\frac{\pi}{4}\right)^2+\frac{\sqrt{2}}{\pi}\left(\frac{\pi}{4}\right)^2+\frac{2 \pi}{4}-\frac{1}{\sqrt{2}}\right]$

$=2\left[\frac{\pi}{4}\left(\frac{-4}{4}+\frac{\sqrt{2}}{4}+2\right)-\frac{1}{\sqrt{2}}\right]$

$=\left(\frac{4+\sqrt{2}}{8}\right) \pi-\sqrt{2}$

$\int \limits_0^{2 \pi} \min \left\{|x-\pi|, \cos ^{-1}(\cos x)\right\} d x$

Graph of $y=|x-\pi|$ and $y=\cos ^{-1}(\cos x)$

$\int \limits_0^{2 \pi} \min \left\{|x-\pi|, \cos ^{-1}(\cos x) d x=\right.\text { Area of }$

shaded region.

$=\frac{1}{2} \times \pi \times \frac{\pi}{2}+\frac{1}{2} \times \pi \times \frac{\pi}{2}$

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

Given,

$A(1,1), B(3,2), C(2,3)$

Slope of line $l_2=\frac{2-1}{3-1}=\frac{1}{2}$

Line $l_1$ is parallel to $b_2$

$\therefore$ Equation of line $l_1$ is

$\begin{aligned}\Rightarrow & y-3 &=\frac{1}{2}(x-2) \\\Rightarrow & y &=\frac{x}{2}+2 \end{aligned}$

Area of shaded region

$\begin{array}{l}=\int \limits_1^3\left(\text { line } l_1-\text { curve } f(x)\right) d x \\ =\int \limits_1^3\left(\frac{x}{2}+2\right) d x-4 \\=\left[\frac{x^2}{4}+2 x\right]_1^3-4 \\ =\left(\frac{9}{4}+6\right)-\left(\frac{1}{4}+2\right)-4 \\=\frac{9}{4}+6-\frac{1}{4}-2-4=2\end{array}$

We have,

$\begin{array}{l}A=\left\{(x, y): x^2+y^2 \leq 100\right\} \\B=\{(x, y): \sin (x+y) > 0 \end{array}$

$\sin (x+y) > 0$

$\therefore \quad x+y \in(0, \pi) \cup(2 \pi, 3 \pi)$

$\quad x+y=c$ equation of line

Required area = shaded region

$\frac{1}{2}$ area of circle $=\frac{1}{2} \pi(10)^2=50 \pi$ 

We have, $y=x^2$ and $y=1-x^2$

Intersection point of $y=x^2$ and $y=1-x^2$ is $A\left(\frac{1}{\sqrt{2}}, \frac{1}{2}\right)$ and $C\left(\frac{-1}{\sqrt{2}}, \frac{1}{2}\right)$

Area of shaded region

$=2 \int \limits_0^{1 / \sqrt{2}}\left[\left(1-x^2\right)-\left(x^2\right)\right] d x$

$=2 \int \limits_0^{1 / \sqrt{2}}\left(1-2 x^2\right) d x$

$=2\left[x-\frac{2 x^3}{3}\right]_0^{1 / \sqrt{2}}$

$=2\left[\frac{1}{\sqrt{2}}-\frac{1}{3 \sqrt{2}}\right]$

$=\frac{2}{\sqrt{2}}\left[\frac{3-1}{3}\right]=\frac{2 \sqrt{2}}{3}$ sq units

We have, $y=\| x-3|-4|-5$

$y=\left\{\begin{array}{cc}-x-6, & x<-1 \\x-4, & -1 \leq x<3 \\-x+2, & 3 \leq x<7 \\x-12, & x \geq 7\end{array}\right.$

The graph of function are

Area of region bounded by

$y=|| x-3|-4|-5$ and $X$-axis

$=$ Area of $\triangle A P B+$ Area of trapezium

$B P Q C+$ Area of trapezium $Q C D R+$ Area

of $\triangle R S D$

$=\frac{1}{2} \times 5 \times 5+\frac{1}{2} \times 4 \times(5+1)$

$+\frac{1}{2} \times 4 \times(5+1)+\frac{1}{2} \times 5 \times 5$

$=49$