Part (B) is the correct answer.
Reason: The given curves are y = cos x and \(y=sinx;0\le x\le \frac { \pi }{ 2 } \)
The curves intersect, where
\(cosx=sinx\Rightarrow tanx=1\Rightarrow x=\frac { \pi }{ 4 } .\)
And \(y=cos\frac { \pi }{ 4 } =sin\frac { \pi }{ 4 } =\frac { 1 }{ \sqrt { 2 } } \)

Therefore, Reqd, area = ar (OPBO) = ar (OPA) + ar (APBA)
\(=\overset { \frac { 1 }{ \sqrt { 2 } } }{ \underset { 0 }{ \int { } } } { sin }^{ -1 }ydy+\overset { 1 }{ \underset { \frac { 1 }{ \sqrt { 2 } } }{ \int { } } } { cos }^{ -1 }ydy\)
\(=\sqrt { 2 } -1\)

The given curve is x² = 4y
which is an upward parabola with vertex (0,0).
The given line is x = 4y - 2
Solving (1) and (2):

\( { (4y-2) }^{ 2 }=4y\)
\(\Rightarrow { 16y }^{ 2 }-16y+4=4y\)
\(\Rightarrow { 16y }^{ 2 }-20y+4=0\Rightarrow { 4y }^{ 2 }-5y+1=0\)
\(\Rightarrow (4y-1)(y-1)=0\Rightarrow y=\frac { 1 }{ 4 } ,1\)
\(When \ y=\frac { 1 }{ 4 } ,thenx=4\left( \frac { 1 }{ 4 } \right) -2\)
\(=1-2=-1\)
\(When \ y=1,thenx=4(1)-2=4-2=2\)
\(Thus(2)meets(1)at\)
\(A\left( -1,\frac { 1 }{ 4 } \right) andB(2,1).\)
\(\therefore Reqd.area=ar(ALOMBDA)-ar(LMBOAL)\)
\(=\overset { 2 }{ \underset { -1 }{ \int { } } } \frac { x+2 }{ 4 } dx-\overset { 2 }{ \underset { -1 }{ \int { } } } \frac { { x }^{ 2 } }{ 4 } dx\)
\(=\frac { 1 }{ 4 } \left[ \frac { { x }^{ 2 } }{ 2 } +2x \right] ^{ 2 }_{ -1 }-\frac { 1 }{ 4 } \left[ \frac { { x }^{ 3 } }{ 3 } \right] _{ -1 }^{ 2 }\)

\(=\frac { 1 }{ 4 } \left[ \left( 2+4 \right) -\left( \frac { 1 }{ 2 } -2 \right) \right] -\frac { 1 }{ 12 } \left[ 8-\left( -1 \right) \right] \)
\(=\frac { 1 }{ 4 } \left[ 6+\frac { 3 }{ 2 } \right] -\frac { 1 }{ 12 } [9]=\frac { 15 }{ 8 } -\frac { 3 }{ 4 } \)
\(=\frac { 15-6 }{ 8 } =\frac { 9 }{ 8 } sq.units.\)

\(\text { Line is } y=3 x+2\)
\(\text { When } y=0, x=-\frac{2}{3}\)
\(\text { Area } =-\int_{-1}^{-2 / 3}(3 x+2) d x+\int_{-2 / 3}^{1}(3 x+2) d x\)
\(=-\left[\frac{3 x^{2}}{2}+2 x\right]_{-1}^{-2 / 3}+\left[\frac{3 x^{2}}{2}+2 x\right]_{-2 / 3}^{1} \)
\(=-\left(\frac{2}{3}-\frac{4}{3}\right)+\left(\frac{3}{2}-2\right)+\left(\frac{3}{2}+2\right)-\left(\frac{2}{3}-\frac{4}{3}\right) \)
\(=\frac{2}{3}-\frac{1}{2}+\frac{7}{2}+\frac{2}{3}=3+\frac{4}{3}=\frac{13}{3} \mathrm{sq} \text { units } \)

Eliminating y from the equations, we get

\(3 x-\frac{3}{2} x^{2}+12=0 \Rightarrow x^{2}-2 x-8=0 \)
\(\Rightarrow (x-4)(x+2)=0 \Rightarrow x=-2,4 \)
\(\text { Area }=\int_{-2}^{4}\left\{\frac{3 x+12}{2}-\frac{3}{4} x^{2}\right\} d x \)
\(=\left[\frac{3 x^{2}}{4}+6 x-\frac{x^{3}}{4}\right]_{-2}^{4} \)
\(=[(12+24-16)-(3-12+2)] \text { sq units } \)
\(=[20+7]=27 \text { sq units } \)

Given equation of curve is \(y=x|x|=\left\{\begin{array}{ll} x^{2}, & \text { if } x \geq 0 \\ -x^{2}, & \text { if } x< 0 \end{array}\right. \)
The graph of above curve between the ordinates x = - 3 and x = 3 is given below

Clearly, shaded portion is the required region. 
∴ Required Area
= Area of region OALO + Area of region OBMO
\(=\left|\int_{-3}^{0} y_{2} d x\right|+\int_{0}^{3} y_{1} d x  \)
\(=\left|\int_{-3}^{0}\left(-x^{2}\right) d x\right|+\int_{0}^{3} x^{2} d x=\left|\left[-\frac{x^{3}}{3}\right]_{-3}^{0}\right|+\left[\frac{x^{3}}{3}\right]_{0}^{3}  \)
\(=\left|\left[\frac{(-3)^{3}}{3}\right]\right|+\left[\frac{27}{3}-0\right]=|-9|+[9-0 \)
= 9 + 9 = 18 sq.units
Hence,the required area is 18 sq units.

\(= \frac{a^{2}}{12}(4 \pi-3 \sqrt{3}) \text { sq units }\)

\(\frac{16}{3} \text { sq units }\)

Given curve is \(y=\sqrt{3 x+4} \)
On squaring both sides, we get
\(y^{2}=3 x+4 \)
\(\Rightarrow y^{2}=3\left(x+\frac{4}{3}\right) \Rightarrow y^{2}=3\left[x-\left(\frac{-4}{3}\right)\right] \)
which is the equation of the parabola of the form
y² = 4 a x. whose vertex is \(\left(-\frac{4}{3}, 0\right)\) and symmetrical about x-axis.
As  \(y=\sqrt{3 x+4} \) is a positive square root, so we take a upper part of the parabola \(y^{2}=3 x+4 \)
upper part of the parabola \(y^{2}=3 x+4 \)
The area of the region bounded by the curve
\(y=\sqrt{3 x+4}\) between  x = 0, x = 4 and the x-axis,is the area shown in the figure given below
 
∴ Required area \(= \int_{0}^{4} y d x=\int_{0}^{4}(\sqrt{3 x+4}) d x  \)
\(=\int_{0}^{4}(3 x+4)^{1 / 2} d x=\left[\frac{(3 x+4)^{3 / 2}}{\frac{3}{2} \cdot 3}\right]_{0}  \)
\(\begin{array}{l} =\frac{2}{9}\left[(12+4)^{3 / 2}-(4)^{3 / 2}\right] \quad\left[\because \int(a x+b)^{n} d x=\frac{(a x+b)^{n+1}}{a(n+1)}\right] \end{array}  \)
\(=\frac{2}{9}\left[(16)^{3 / 2}-(4)^{3 / 2}\right]  \)
\(=\frac{2}{9}\left[\left(2^{4}\right)^{3 / 2}-\left(2^{2}\right)^{3 / 2}\right]  \)
\(=\frac{2^{4}}{9}\left[(2)^{6}-(2)^{3}\right]=\frac{2}{9}(64-8)  \)
\(=\frac{2}{9} \times 56=\frac{112}{9} \text { sq units } \)
Hence, the required is \( \frac{112}{9} \text { sq units }.\)

On the graph it represents the area bounded by the curve \(y=\left| x-2 \right| \), x-axis and between the ordinates at x = -1 and x = 3.

Given curve is  4x² + y² = 36 
\(\begin{array}{ll} \therefore & \frac{4 x^2}{36}+\frac{y^2}{36}=1 \end{array}\)
\(\begin{array}{ll} \therefore & \frac{x^2}{9}+\frac{y^2}{36}=1 \end{array}\)              ...(i)
We know that the standard equation of ellipse is
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)                      ...(ii)
On comparing Eqs. (i) and (ii), we get
a² = 9 and b² = 36
\(\Rightarrow\) a = 3 and b = 6
Here, we see that a < b, so the vertical ellipse will be formed.

Now, required area = 4(Area of region OAB in first quadrant)
\(\begin{aligned} =4\left|\int_0^3 y d x\right| \end{aligned}\)
\(\begin{aligned} =\left|4 \int_0^3 2 \sqrt{9-x^2} d x\right| \end{aligned}\)
\(\begin{aligned} \left[\because \frac{y^2}{36}=1-\frac{x^2}{9} \Rightarrow y=2 \sqrt{9-x^2}\right] \end{aligned}\)
\(\begin{aligned} =8 \int_0^3 \sqrt{9-x^2} d x \end{aligned}\)
\(\begin{aligned} =8\left|\left[\frac{x}{2} \sqrt{9-x^2}+\frac{9}{2} \sin ^{-1}\left(\frac{x}{3}\right)\right]_0^3\right| \end{aligned}\)
\(\left[\because \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x}{a}\right)\right]\)
\(=8\left\{\left[\frac{3}{2} \sqrt{9-9}+\frac{9}{2} \sin ^{-1} \frac{3}{3}-0-\frac{9}{2} \sin ^{-1} 0\right]\right\}\)
\(\begin{aligned} =8\left|\left[0+\frac{9}{2} \times \frac{\pi}{2}-0\right]\right| \end{aligned}\)
\(\begin{aligned} =8 \times \frac{9 \pi}{4} \end{aligned}\)
\(=18 \pi\)
Hence, the required area is 18\(\pi\) sq units.

The given curves are
y = x² (parabola)
y = x (line)
These intersect at
O(0, 0) and A(1, 1).
 
The area bounded by the curves = Shaded area
 \(=\int _{ 0 }^{ 1 }{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) } dx\)
\(=\int _{ 0 }^{ 1 }{ \left( x-{ x }^{ 2 } \right) } dx\)
\(=\left[ \frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] _{ 0 }^{ 1 }\)
\(=\frac { 1 }{ 2 } -\frac { 1 }{ 3 } =\frac { 1 }{ 6 } sq.units\)

Eqns. of PQ, QR and PR are:
PQ : y = \(\frac{5}{2}\)(x-2)
QR : y = 9-x
PR : y = \(\frac{3}{4}\) (x-2)
Req. Area = \(=\int _{ 2 }^{ 4 }{ \frac { 5 }{ 2 } \left( x-2 \right) dx+\int _{ 4 }^{ 6 }{ \left( 9-x \right) dx-\int _{ 2 }^{ 6 }{ \frac { 3 }{ 4 } \left( x-2 \right) dx } } } \)
​​​\(=\left[ \frac { 5 }{ 4 } { \left( x-2 \right) }^{ 2 } \right] _{ 2 }^{ 4 }-\frac { 1 }{ 2 } \left[ \left( 9-x \right) ^{ 2 } \right] _{ 4 }^{ 6 }-\frac { 3 }{ 8 } \left[ \left( x-2 \right) ^{ 2 } \right] _{ 2 }\)
= 5 + 8 - 6 = 7 sq.units.

(i) (a) : Here, teacher explained about cosine curve.
(ii) (c) : Required area \(\int_{0}^{\pi / 2} \cos x d x\) 
\(=[\sin x]_{0}^{\pi / 2}=\sin \frac{\pi}{2}-\sin 0=1-0=1 \text { sq. unit }\) 
(iii) (b) : Required area = \(\left|\int_{\pi / 2}^{3 \pi / 2} \cos x d x\right|=\left|[\sin x]_{\pi / 2}^{3 \pi / 2}\right|\) 
\(=\left|\sin \frac{3 \pi}{2}-\sin \frac{\pi}{2}\right|=|-1-1|=|-2|\) 
= 2 sq. units [Since, area can't be negative]
(iv) (a) : Required area = \(\int_{3 \pi / 2}^{2 \pi} \cos x d x=[\sin x]_{3 \pi / 2}^{2 \pi}\) 
\(=\sin 2 \pi-\sin \frac{3 \pi}{2}=0-(-1)=1 \text { sq. unit }\) 
(v) (d) : Required area
\(=\int_{0}^{\pi / 2} \cos x d x+\left|\int_{\pi / 2}^{3 \pi / 2} \cos x d x\right|+\int_{3 \pi / 2}^{2 \pi} \cos x d x\) 
= 1 + 2 + 1 = 4 sq. units.