Ncert Solutions Maths class 12th

Given equation of the curve is $\left|x\right|+\left|y\right|=1$ , which can be break down into each quadrant .

For Ist quadrant,  $\left|x\right|=x,\left|y\right|=y$

i.e.,  $x+y=1$ - (1)

Similarly for IInd, IIIRd nad IVth quadrant

$-x+y=1$ - (2)

$-x-y=1$ - (3)

$x-y=1$ - (4)

We draw the above focus lines on a graph and find the area enclosed which is a square.

$\therefore$ Required area $\left(?ABCD\right)=4*area\left(?AOB\right)$ .

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

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

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

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

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

When $x=-1,y={\left(-1\right)}^2=1$

And $x=2,y=2^2=4$

$\therefore$ The point of intersection of the parabola and the lines $A(-1,1)\&B(2,4)$

Hence the required area enclosed region is $ea(AOBA)=area(DABCD)-area(DAOBCD)$

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

$\begin{array}{l}=\left[\left(\frac{2^2}{2}2.2\right)-\left(\frac{{\left(-1\right)}^2}{2}+2\left(-1\right)\right)\right]-\left[\frac{2^3}{3}-\frac{{\left(-1\right)}^3}{3}\right]\\ =\left[2+4-\frac{1}{2}+2\right]-\left[\frac{8+1}{3}\right]=\frac{15}{2}-3=\frac{9}{2}uni{t}^2\\ \end{array}$

The Given equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

And the equation of the line in $\frac{x}{a}+\frac{y}{b}=1$

With x and y intersept a and b

So, required area of the enclosed region is

$area(BCAB)=area(OBCAO)-area(?OBA)$

Given equation of the ellipse is $\frac{x^2}{9}+\frac{y^2}{4}=1$ Which as major axis aling x- axis and that of the line is $\frac{x}{3}+\frac{y}{2}=1$ which has x and y intercepts at 3 and 2respectively.

Required area of enclosed region is area $area(BCAB)=area(OBAO)?area(ABOA)$

The given equation of parabola is $4y=3x^2\Rightarrow x^2=\frac{4}{3}y$ ------------(1)

And the line is $2y=3x+12\Rightarrow y=\frac{3}{2}x+6$ ----------------------(2)

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

$\begin{array}{l}{x}^2=\frac{4}{3}\left(\frac{3}{2}x+6\right)=2x+8\\ \Rightarrow x^2-2x-8=0\\ \Rightarrow x^2-4x+2x-8=0\\ \Rightarrow x\left(x-4\right)+2\left(x-4\right)=0\\ \Rightarrow \left(x-4\right)\left(x+2\right)=0\\ \Rightarrow x=4\&x=-2\end{array}$

At, $x=4,y=\frac{3}{2}*4+6=6+6=12$

And $x=-2,y=\frac{3}{2}*\left(-1\right)+6=-3+6=3$

Thus, the point of intersection of (1)&(2)are $A(4,12)\&B(-2,3)$

$\therefore$ Area of the enclosed region (BOAB)

=area (CBAD) – area (OADC)

$\begin{array}{l}={\displaystyle \underset{-2}{\overset{4}{\int }}{y}_{line}dx-}{\displaystyle \underset{-2}{\overset{4}{\int }}{y}_{curve}dx}\\ ={\displaystyle \underset{-2}{\overset{4}{\int }}\left(\frac{3}{2}x+6\right)dx-{\displaystyle \underset{-2}{\overset{4}{\int }}\frac{3x^2}{4}dx}}\\ ={\left[\frac{3}{2}\frac{x^2}{2}+6x\right]}_{-2}^4-{\left[\frac{3}{4}*\frac{x^3}{3}\right]}_{-2}^4\\ =\left[\left(\frac{3}{4}*4^2+6*4\right)-\left(\frac{3}{4}{\left(-2\right)}^2+6*\left(-2\right)\right)\right]-\frac{1}{4}\left[4^3-{\left(-2\right)}^3\right]\\ =\left[12+24-3+12\right]-\frac{1}{4}\left[64+8\right]\\ =45-18=27uni{t}^2\end{array}$

The equation of the parabola is $y^2=4a^2$ -----------(1)

and that of line is $y=mx$ ------(2)

The Point of intersection of(1)and (2) is given by

$\begin{array}{l}{\left(mx\right)}^2=4ax\\ \Rightarrow m^2{x}^2-4ax=0\\ \Rightarrow x\left(m^{2}x-4a\right)=0\\ \Rightarrow x=0\&x=\frac{4a}{m^2}\end{array}$

For, $x=0,y^2=4a*0\to y=0$ i.e, O(0,0)

For, $x=\frac{4a}{m^2},y^2=4a*\frac{4a}{m^2}\to y=\frac{4a}{m}$ (in first quadrant)

i.e, $A\left(\frac{4a}{m^2},\frac{4a}{m}\right)$

Hence, the required area enclosed by the curve and the lines is

$area(DACO)=area(OCABO)-area(?OAB)$

The given equation of the curve is $y=sinx$

$\therefore$ The required area bounded by the curve

$\begin{array}{l}={\displaystyle \underset{0}{\overset{\pi }{\int }}y}dx+{\displaystyle \underset{\pi }{\overset{2\pi }{\int }}y}dx\\ ={\displaystyle \underset{0}{\overset{\pi }{\int }}sinxdx+{\displaystyle \underset{\pi }{\overset{2\pi }{\int }}sinxdx}}\\ =\left|{\left[-cosx\right]}_0^{\pi }\right|+\left|{\left[-cosx\right]}_0^{\pi }\right|\\ =|-\left[cosx-cos\theta \right]|+|-\left[cos2\pi -cos\pi \right]|\\ =|-\left[-1-1\right]|+|-\left[1+1\right]|\\ =\left|2\right|+|-2|=2+2=4uni{t}^2\end{array}$

Given equation of lines is $y=|x+3|$ -------(1)

The point (x,y)satisfying (1)are

Hence plotting the above in graph we get     

Now, ${\displaystyle \underset{-6}{\overset{0}{\int }}|x+3|dx=}{\displaystyle \underset{-6}{\overset{-3}{\int }}|x+3|dx+{\displaystyle \underset{-3}{\overset{0}{\int }}|x+3|dx}}$

We know that, $\begin{array}{l}{\displaystyle \underset{-6}{\overset{0}{\int }}|x+3|dx=}{\displaystyle \underset{-6}{\overset{-3}{\int }}|x+3|dx+{\displaystyle \underset{-3}{\overset{0}{\int }}|x+3|dx}}\\ y=|x+3|=\left\{\begin{array}{l}x+3,if,x+3\ge 0\Rightarrow x\ge -3\\ -\left(x+3\right),if,x+3\angle 0\Rightarrow x\angle -3\end{array}\right\}\end{array}$

So, ${\displaystyle \underset{-6}{\overset{0}{\int }}|x+3|dx=}{\displaystyle \underset{-6}{\overset{-3}{\int }}-\left(x+3\right)dx+}{\displaystyle \underset{-3}{\overset{0}{\int }}\left(x+3\right)dx}$

$\begin{array}{l}=-{\left[\frac{x^2}{2}+3x\right]}_{-6}^{-3}+{\left[\frac{x^2}{2}+3x\right]}_{-3}^0\\ =-\left\{\left[\frac{{\left(-3\right)}^2}{2}+3\left(-3\right)\right]-\left[\frac{{\left(-6\right)}^2}{2}+3\left(-6\right)\right]\right\}+\left\{\left[\frac{0^2}{2}+3*0\right]-\left[\frac{{\left(-3\right)}^2}{2}+3*\left(-3\right)\right]\right\}\\ =-\left\{\frac{9}{2}-9-\frac{36}{2}+18\right\}+\left\{-\frac{9}{2}+9\right\}\\ =-\frac{9}{2}+9+18-18-\frac{9}{2}+9\\ =9uni{t}^2\end{array}$

Given curve is $y=4x^2\Rightarrow x^2=\frac{1}{4}y$ - (1)

$x=0$  i.e, y-axis and y=4 and y=1

Hence, the required area in Ist quadrant i.e, area ABCD = ${\displaystyle \underset{y=1}{\overset{y=4}{\int }}xdy}$

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

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

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

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