Application of Integrals

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

Given equation of the circle is

∴ option (A) is correct

The given equation of the curve is

$\begin{array}{l}{y}^2=4x\\ \Rightarrow y=\pm \surd 3\end{array}$

$\begin{array}{l}\mathrm{\to \; y}=+\surd 2x\end{array}$ in Ist quadrant

So, area of curve enclosed by $y^2=4x$

And $x=3=2*\; area$ area (AOCA)

Given curve is $x^2=4y$ and the equation of line is $x=4y-2$

The point of intersection of the curve and the line can be determine as follows.

Put, $x=4y-2\Rightarrow x+2=4y\Rightarrow y\frac{x+2}{4}$

In $x^2=4y$ to determine value of x

i.e, $x^2=4*\frac{\left(x+2\right)}{4}=x+2$

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

$\Rightarrow x=2$ and $x=-1$

$x=2$ , we have ${\left(2\right)}^2=4y\Rightarrow 4-4y\Rightarrow y=1$

And at $x=-1$ we have ${\left(-1\right)}^2=4y\Rightarrow y=\frac{1}{4}$

So, the coordinates A and B are (2,1) and ( $-1,\frac{1}{4}$ )

$\therefore$ The required area before the line & the curve is area $B{D}^{\iota }AB$ = area of trapezium (BNMAB)- area under curve BDA

$\begin{array}{l}={\displaystyle \underset{-1}{\overset{2}{\int }}{y}_{line}}dx-{\displaystyle \underset{-1}{\overset{2}{\int }}{y}_{curve}}dx\\ ={\displaystyle \underset{-1}{\overset{2}{\int }}\frac{x+2}{4}da-{\displaystyle \underset{-1}{\overset{2}{\int }}\frac{x^2}{4}}}\\ =\frac{1}{4}{\displaystyle \underset{-1}{\overset{2}{\int }}xda+\frac{2}{4}{\displaystyle \underset{-1}{\overset{2}{\int }}dx-\frac{1}{4}{\displaystyle \underset{-1}{\overset{2}{\int }}{x}^{2}dx}}}\end{array}$

$\begin{array}{l}=\frac{1}{4}{\left[\frac{x^2}{2}\right]}_{-1}^2+\frac{2}{4}{\left[x\right]}_{-1}^2-\frac{1}{4}{\left[\frac{x^3}{3}\right]}_{-1}^2\\ =\frac{1}{8}\left[2^2-{\left(-1\right)}^2\right]+\frac{1}{2}\left[2-\left(-1\right)\right]-\frac{1}{12}\left[2^3-{\left(-1\right)}^3\right]\end{array}$

$\begin{array}{l}=\frac{3}{8}+\frac{3}{2}-\frac{9}{12}=\frac{3}{8}+\frac{3}{2}-\frac{3}{4}=\frac{3+4*3-2*3}{8}\\ =\frac{3+12-6}{8}=\frac{9}{8}uni{t}^2\end{array}$

Given that equation of

curve $y=x^2$

line $y=\left|x\right|=\left\{x,ifx\ge 0\&-x,ifx\angle 0\right\}$

Since the line passes through A&B in Ist and IInd quadrants

the equation must satisfy

$\Rightarrow y=x^2$

$\Rightarrow x=x^2$ for Ist quadrant and

$\Rightarrow -x=x^2$ for IInd t quadrant

So, $\Rightarrow x^2-x=0$ and $x^2+x=0$

$\Rightarrow x\left(x-1\right)=0$ and $x\left(x+1\right)=0$

$\Rightarrow x=0,1$ $x=0,-1$

$x=0,y=0$

$x=1,y=1$ i.e, A has coordinate (1,1)

$x=-1,y=1$ i.e, B has coordinate (1,1)

Now, area of AODA = area (AOM)-area (ADOM)

$={\displaystyle \underset{0}{\overset{1}{\int }}{y}_{line}dx-{\displaystyle \underset{0}{\overset{1}{\int }}{y}_{curve}dx}}$

$={\displaystyle \underset{0}{\overset{1}{\int }}xdx-{\displaystyle \underset{0}{\overset{1}{\int }}{x}^2}dx={\left[\frac{x^2}{2}\right]}_0^1-}{\left[\frac{a^3}{3}\right]}_0^1$

$=\frac{1}{2}-\frac{1}{3}=\frac{3-2}{6}=\frac{1}{6}unit{s}^2$

$\therefore$ The required area of the region bounded by curve $y=x^2$ and line $y=\left|x\right|$ is $\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}unit{s}^2$