Application of Integrals

Given is $y=x^3$ and the ines $x=-2\&x=1$

For $y=x^3$

$\begin{array}{l}area\left(OAB\right)={\displaystyle \underset{0}{\overset{1}{\int }}ydx}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}{x}^{3}dx}\\ ={\left[\frac{x^4}{4}\right]}_0^1=\frac{1}{4}\\ area\left(ODC\right)={\displaystyle \underset{-2}{\overset{0}{\int }}ydx}\\ ={\displaystyle \underset{-2}{\overset{0}{\int }}{x}^{3}dx}\\ =\left|{\left[\frac{x^4}{4}\right]}_{-2}^0\right|=\left|\left[\frac{0^4}{4}-\frac{{\left(-2\right)}^4}{4}\right]\right|=4\end{array}$

$\therefore$ Total area of the bounded region $=\frac{1}{4}+4$

$=\frac{17}{4}uni{t}^2$

The given equation of curve $y^2\le 4x$ i.e,  $y^2=4x$ - (1) is a parabola and

$4x^2+4y^2\le 9\Rightarrow x^2+y^2\le \frac{9}{4}\Rightarrow x^2+y^2={\left(\frac{3}{2}\right)}^2$ - (2) is a circle

With centre (0,0)and radius 

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

$\begin{array}{l}{x}^2+4x=\frac{9}{4}\\ =x^2+4x-\frac{9}{4}=0\\ =4x^2+16x-9=0\end{array}$

The given equation of the lines are

$\begin{array}{l}2x+y=4------(1)\\ 3x-2y=6-----(2)\\ x-3y+5=0----(3)\end{array}$

$\therefore$ Area of $?ABC=area(PCBQ)-area(?APC)-area(?AQB)$

$\begin{array}{l}={\displaystyle \underset{1}{\overset{4}{\int }}{y}_{(3)}dx-}{\displaystyle \underset{1}{\overset{2}{\int }}{y}_{(1)}dx-{\displaystyle \underset{2}{\overset{4}{\int }}{y}_{(2)}dx-}}\\ ={\displaystyle \underset{1}{\overset{4}{\int }}\frac{\left(x+5\right)}{3}dx-{\displaystyle \underset{1}{\overset{2}{\int }}\left(4-2x\right)dx-{\displaystyle \underset{2}{\overset{4}{\int }}\left(\frac{3x-6}{2}dx\right)}}}\end{array}$

$\therefore$ The point of intersection of the circle and the parabola is . $A\left(\frac{1}{2},\right)\&C\left(\frac{1}{2},-\right)$

Taking in first quadrant

Area of $(OABO)=area(OADO)+area(BADB)$

The given vertices of the triangle are A(2,0),B(4,5)and C(6,3)

So, equation of line AB is $y-0=\frac{5-0}{4-2}\left(x-2\right)$

$=y=\frac{5}{2}\left(x-2\right)$

Similarly equation of BC is

$\begin{array}{l}y-5=\frac{3-5}{6-4}\left(x-4\right)\\ =y=5+\left(-x+4\right)\\ y=9-x\end{array}$

And equation of AC is $y-0=\frac{3-0}{6-2}\left(x-2\right)$

= $y=\frac{3}{4}\left(x-2\right)$

$\therefore$ Area of $?ABC$

= $area(?ABE)+area(BCDE)-area(?ACD)$

$\begin{array}{l}={\displaystyle \underset{2}{\overset{4}{\int }}{y}_{AB}dx+}{\displaystyle \underset{4}{\overset{6}{\int }}{y}_{BC}dx-{\displaystyle \underset{2}{\overset{6}{\int }}{y}_{BC}dx}}\\ ={\displaystyle \underset{2}{\overset{4}{\int }}\frac{5}{2}\left(x-2\right)dx+{\displaystyle \underset{4}{\overset{6}{\int }}\left(9-x\right)dx-{\displaystyle \underset{2}{\overset{6}{\int }}\frac{3}{4}}}}\left(x-2\right)dx\\ =\frac{5}{2}{\left[\frac{x^2}{2}-2x\right]}_2^4+{\left[9x-\frac{x^2}{2}\right]}_4^6-\frac{3}{4}{\left[\frac{x^2}{2}-2x\right]}_2^6\end{array}$

$\begin{array}{l}=\frac{5}{2}\left[\left(\frac{4^2}{2}-2.4\right)-\left(\frac{2^2}{2}-2.2\right)\right]+\left[\left(9.6-\frac{6^2}{2}\right)-\left(9.4-\frac{4^2}{2}\right)\right]-\frac{3}{4}\left[\left(\frac{6^2}{2}-2.6\right)-\left(\frac{2^2}{2}-2.2\right)\right]\\ =\frac{5}{2}\left[\left(8-8\right)-\left(2-4\right)\right]+\left[54-18-36+8\right]-\frac{3}{4}\left[18-12-2+4\right]\\ =5+8-6=7uni{t}^2\end{array}$

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

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)

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