Maths

The given equation of the curve is $y=x^2$ --------(1)

and that of the line is $y=x$ ---------(2)

Solving eq (1) and (2)for x and y

$\begin{array}{l}x=x^2\\ =x^2-x=0\\ =x\left(x-1\right)=0\\ =x=0\&x=1\end{array}$

Where, $x=0,y=0^2=0$

And when $x=1,y=1^2=1$

$\therefore$ The point of intersection of the parabola $y=x^2$ and the line $y=x$

Is O(0,0) and B (1,1)

Hence, area between the curve and the line is

$area\left(DCAO\right)=area\left(?OAB\right)-area\left(OABO\right)$

$\begin{array}{l}={\displaystyle \underset{0}{\overset{1}{\int }}{y}_{line}}dx-{\displaystyle \underset{0}{\overset{1}{\int }}{y}_{curve}}dx\\ ={\displaystyle \underset{0}{\overset{1}{\int }}xdx-{\displaystyle \underset{0}{\overset{1}{\int }}{x}^{2}dx}}\\ ={\left[\frac{x^2}{2}\right]}_0^1-{\left[\frac{x^3}{3}\right]}_0^1\end{array}$

$\begin{array}{l}=\frac{1}{2}-\frac{1}{3}\\ =\frac{3-2}{6}=\frac{1}{6}uni{t}^2\end{array}$

(i) Given of curve is $y=x^2\Rightarrow x^2=y$ and the equation are $x=1\&x=2.$

$\therefore$ Area enclosed

$\begin{array}{l}={\displaystyle \underset{x=1}{\overset{x=2}{\int }}y}dx\\ ={\displaystyle \underset{1}{\overset{2}{\int }}{x}^2}dx\\ ={\left[\frac{x^3}{3}\right]}_1^2=\frac{2^3}{3}-\frac{1^3}{3}\\ =\frac{8-1}{3}=\frac{7}{3}uni{t}^2\end{array}$

(ii) Given equation of curve is $y=x^4$ and the lines are $x=1\&x=5$

So, area enclosed

$\begin{array}{l}={\displaystyle \underset{1}{\overset{5}{\int }}y}dx\\ ={\displaystyle \underset{1}{\overset{5}{\int }}{x}^4}dx\\ ={\left[\frac{x^5}{5}\right]}_1^5=\left(\frac{5^5-1^5}{5}\right)\\ =\frac{\left(3125-1\right)}{5}\\ =\frac{3124}{5}=624.8uni{t}^2\end{array}$

The given equation of the curve is $y^2=4x$ - (1) and

the line is $y=2x$ - (2)

Solving (1) and (2) for x and y

$\begin{array}{l}{\left(2x\right)}^2=4x\\ =4x^2=4x\\ =x^2-x=0\\ =x\left(x-1\right)=0\end{array}$

So,  $x=0\&x=1$

for $x=0$ we get $y=2*0=0$

for $x=1$ , we get $y=2*1=2$

so, the point of intersection are (0,0)and (1,2)

area (DCAO)=area (DCABO)-area ( $?OAB$ )

The equation of circle is $x^2+y^2=4$ which has centre at (0,0) & radius,

$\pi =2$

And the line $x+y=2=y=2-x$

The smaller area of circle is given by

Area (ABCA) area (BOAB) – area (BOA)              

The given equation of the sides of triangle is

$y=2x+1$ --------------------(1)

$y=3x+1$ -------------------(2)

$x=4$ -------------------------(3)

Solving eqn (1) and (2) for x & y we get

$\begin{array}{l}3x+1=2x+1\\ =3x-2x=1-1\\ =x=0\\ \&y=2*0+1=1\end{array}$

$\therefore$ The point of inersection of line (1)and (2)is A (0,1)

Putting x=4 in eq (1) and (2)we get,

$\begin{array}{l}y=2*4+1=8+1=9\\ \&y=3*4+1=12+1=13\end{array}$

$\therefore$ The point of intersection of line (1)and (3) is B(4,9) and C (4,13)

Hence the required area enclosed ABC

$\begin{array}{l}={\displaystyle \underset{0}{\overset{4}{\int }}{y}_{line(2)}dx-}{\displaystyle \underset{0}{\overset{4}{\int }}{y}_{line(1)}dx}\\ ={\displaystyle \underset{0}{\overset{4}{\int }}\left[3x+1\right]}dx-{\displaystyle \underset{0}{\overset{4}{\int }}\left[2x+1\right]}dx\\ ={\left[\frac{3x^2}{2}+x\right]}_0^4-{\left[\frac{2x^2}{2}+x\right]}_0^4\\ =\left[\left(\frac{3}{2}{\left(4\right)}^2+4\right)-\left(\frac{3*0^2}{2}+0\right)\right]-\left[\left(4^2+4\right)-\left(0^2+\right)\right]\\ =24+4-20=8uni{t}^2\end{array}$

Let A (-1,0),B(1,3) and C (3,2) be the vertices of a triangle ABC

So, equation of line AB is $y-0=\frac{3-0}{1-\left(-1\right)}\left(x-\left(-1\right)\right)$

$=y=\frac{3}{2}\left(x+1\right)$ -------------(1)

Equation of line BC is $y-3=\frac{2-3}{3-1}\left(x-1\right)$

$=y=\frac{-1}{2}\left(x-1\right)+3=\frac{-1}{2}x+\frac{1}{2}+3=\frac{-x}{2}+\frac{7}{2}$ ---------------(2)

Equation of line AC is $y-0=\frac{2-0}{3-\left(-1\right)}\left[x-\left(-1\right)\right]$

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

$\therefore$ Area of $?$ ABC= area ( $?ABE$ ) $+area\left(BCDE\right)$ $-area\left(?ACD\right)$

$\begin{array}{l}={\displaystyle \underset{-1}{\overset{1}{\int }}{y}_{eq(1)}dx+{\displaystyle \underset{1}{\overset{3}{\int }}{y}_{eq(2)}dx-{\displaystyle \underset{-1}{\overset{3}{\int }}{y}_{eq3}dx}}}\\ ={\displaystyle \underset{-1}{\overset{1}{\int }}\frac{3}{2}\left(x+1\right)dx{\displaystyle \underset{1}{\overset{3}{\int }}\left(\frac{-x}{2}+\frac{7}{2}\right)dx-}}{\displaystyle \underset{-1}{\overset{1}{\int }}\frac{1}{2}\left(x+1\right)}\\ =\frac{3}{2}{\left[\frac{x^2}{2}+x\right]}_{-1}^1+\frac{1}{2}{\left[-\frac{x^2}{2}+7x\right]}_1^{3}dx-\frac{1}{2}{\left[\frac{x^2}{2}+x\right]}_{-1}^3\end{array}$

$\begin{array}{l}=\frac{3}{2}\left[\left(\frac{1^2}{2}+1^2\right)-\left(\frac{{\left(-1\right)}^2}{2}+\left(\pi \right)\right)\right]+\frac{1}{2}\left[\left(\frac{3^2}{2}+7*3\right)-\left(\frac{-1^2}{2}+7*1\right)\right]-\frac{1}{2}\left[\left(\frac{3^2}{2}+3\right)-\left(\frac{{\left(-1\right)}^2}{2}+\left(-1\right)\right)\right]\\ =\frac{3}{2}\left[\frac{1}{2}+1-\frac{1}{2}+1\right]+\frac{1}{2}\left[\frac{-9}{2}+21+\frac{1}{2}-7\right]-\frac{1}{2}\left[\frac{9}{2}+3-\frac{1}{2}+1\right]\\ =\frac{3}{2}\left[2\right]+\frac{1}{2}\left[10\right]-\frac{1}{2}\left[8\right]=3+5-4=8-4=4uni{t}^2\end{array}$

The equation of the curve is $y=x^2+2\Rightarrow x^2=y-2$ - (1) and

lines are

$y=x$ - (2)

$x=0$ - (3)

$x=3$ - (4)

Equation (1)is a parabola with vertex (0,2)

Equation (2)is a straight line passing origin with shape = $tan\theta =1=\theta =45^{?}$

$\therefore$ The required area enclosed OBCDO = area (ODCAO)-area (OBAO)

The equation of the given circle is

$=x^2+y^2=1$ - (1)

$=\; (x-1)2^{}+y^2=1$ - (1) - (2)

Equation (1) is a circle with centre 0 (0,0) and radius 1. Equation (2) is a circle with centre c (1,0) and radius 1.

Solving (1) and (2)

The equation given circle is

$\begin{array}{l}4x^2+4y^2=9\\ \Rightarrow x^2+y^2=\frac{9}{4}\\ \Rightarrow x^2+y^2={\left(\frac{3}{2}\right)}^2\end{array}$

i.e, centre (0,0), radius $r=\frac{3}{2}$

since $x^2=4y$ intersect the circle

we can put $x^2=4y$ in $x^2+y^2={\left(\frac{3}{2}\right)}^2$

$\begin{array}{l}\left(4y\right)+y^2=\frac{9}{4}\\ \Rightarrow y^2+4y-\frac{9}{4}=0\\ \Rightarrow 4y^2+16y-9=0\\ \Rightarrow 4y^2+18y-2y-9=0\end{array}$

$\begin{array}{l}\Rightarrow 2y\left(2y+9\right)-1\left(2y+9\right)=0\\ \Rightarrow \left(2y+9\right)\left(2y-1\right)=0\\ \Rightarrow y=\frac{-9}{2}\&y=\frac{1}{2}\\ \end{array}$

$y=\frac{-9}{2},x^2=4*\left(\frac{-9}{2}\right)=-18$ which is not possible or $x^2$ cannot be (-)ve

As $y=3$ intersect $y^2=4x$ at Athen,

$\begin{array}{l}{3}^2=4x\\ \Rightarrow x=\frac{9}{4}\end{array}$

$\therefore$ A has coordinate $\left(\frac{a}{4},3\right)$

Hence, area of curve = ${\displaystyle \underset{y=0}{\overset{y=3}{\int }}x}dy$

$\begin{array}{l}={\displaystyle \underset{0}{\overset{3}{\int }}\frac{y^2}{4}dy={\left[\frac{y3}{4*3}\right]}_0^3}\\ =\frac{3^3}{12}-\frac{0}{12}\\ =\frac{27}{12}=\frac{9}{4}uni{t}^2\end{array}$

$\therefore$ Option (B) is correct