Application of Integrals

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

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

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

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

(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 ( $?ABE$ ) $+area\left(BCDE\right)$ $-area\left(?ACD\right)$

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

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