Area Under The Curves

y = $\sqrt x $

y = x $-$ 2

$\therefore\,\,\,$ $\sqrt x $ = x $-$ 2

$ \Rightarrow $$\,\,\,$ x = x² $-$ 4x + 4

x² $-$ 5x + 4 = 0

x² $-$ 4x $-$ x + 4 = 0

$ \Rightarrow $$\,\,\,$ x(x $-$ 4) $-$ (x $-$ 4) = 0

$ \Rightarrow $$\,\,\,$ (x $-$ 4) (x $-$ 1) = 0

$\therefore\,\,\,$ x = 4, 1

and y = 2, $-$ 1

$\therefore\,\,\,$ Their point of intersection (4, 2) and (1, $-$ 1)



Required area is shown in the shaded figure.

$\therefore\,\,\,$ Required area

= $\int\limits_0^2 {\sqrt x \,dx + \int\limits_2^4 {\left( {\sqrt x - x + 2} \right)\,dx} } $

= $\int\limits_0^4 {\sqrt x \,dx + \int\limits_2^4 {\left( {2 - x} \right)\,dx} } $

= $\left[ {{2 \over 3}{x^{{3 \over 2}}}} \right]_0^4 + \left[ {2x - {{{x^2}} \over 2}} \right]_2^4$

= ${2 \over 3}\left( 8 \right)$ + 2$\left( {4 - 2} \right)$ $-$ ${1 \over 2}$ (16 $-$ 4)

= ${{16} \over 3}$ + 4 $-$ 6

= ${{10} \over 3}$

The given equation ${x^2} + {y^2} = 4$ is equation of circle of radius 2 centred at origin and equation ${y^2} = 3x$ is the equation of parabola.

${x^2} + {y^2} = 4$ ..... (1)

${y^2} = 3x$ ..... (2)

Substituting Eq. (2) in Eq. (1), we get

${x^2} + 3x - 4 = 0$

$ \Rightarrow {x^2} + 4x - x - 4 = 0$

$ \Rightarrow x(x + 4) - 1(x + 4) = 0$

$ \Rightarrow (x - 1)(x + 4) = 0$

$ \Rightarrow (x - 1) = 0$ and $(x + 4) = 0$

Therefore, x = 1, $-$4. Considering x = 1, then from Eq. (2), we get $y = \sqrt 3 , - \sqrt 3 $.

Therefore $(1,\sqrt 3 )$ and $(1, - \sqrt 3 )$ are the points of intersection of parabola and circle.

The required area (A) is the area of the shaded region shown in the figure. Therefore,

$A = 2\left[ {\int\limits_0^1 {{y_2}dx + \int\limits_1^2 {{y_1}dx} } } \right]$

From Eq. (1), we get ${y_1} = \sqrt {4 - {x^2}} $

From Eq. (2), we get ${y_2} = \sqrt {3x} $

Therefore, $A = 2\left[ {\int\limits_0^1 {\sqrt {3x} dx + \int\limits_1^2 {\sqrt {4 - {x^2}} dx} } } \right]$

$ = 2\left[ {\int\limits_0^1 {\sqrt 3 {x^{1/2}}dx + \int\limits_1^2 {\sqrt {{2^2} - {x^2}} dx} } } \right]$

Using standard integral : $\int {{x^n}dx = {{{x^{n - 1}}} \over {n + 1}}} $, we have

$\int {\sqrt {{a^2} - {x^2}} dx = {x \over 2}\sqrt {{a^2} - {x^2}} + {{{a^2}} \over 2}{{\tan }^{ - 1}}{x \over {\sqrt {{a^2} - {x^2}} }}} $

Therefore,

$A = 2\left[ {\left. {\left( {\sqrt 3 {{{x^{3/2}}} \over {3/2}}} \right)} \right|_0^1 + \left. {\left( {{x \over 2}\sqrt {4 - {x^2}} + {4 \over 2}{{\tan }^{ - 1}}{x \over {\sqrt {4 - {x^2}} }}} \right)} \right|_1^2} \right]$

$ = 2\left[ {\sqrt 3 .{1 \over {3/2}} - 0 + {2 \over 2}\sqrt {4 - 4} + 2{{\tan }^{ - 1}}{2 \over {\sqrt {4 - 4} }} - {1 \over 2}\sqrt {4 - 1} - 2{{\tan }^{ - 1}}{1 \over {\sqrt {4 - 1} }}} \right]$

$ = 2\left[ {{{2\sqrt 3 } \over 3} + {{\tan }^{ - 1}}(\infty ) - {{\sqrt 3 } \over 2} - 2{{\tan }^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)} \right]$

Now, $\tan {\pi \over 2} = \infty $ and $\tan {\pi \over 6} = {1 \over {\sqrt 3 }}$. Therefore, the area of the smaller portion enclosed between the two curves is obtained as follows:

$A = 2\left[ {{{2\sqrt 3 } \over 3} - {{\sqrt 3 } \over 2} + 2{{\tan }^{ - 1}}\left( {\tan {\pi \over 2}} \right) - 2{{\tan }^{ - 1}}\left( {\tan {\pi \over 6}} \right)} \right]$

$ = 2\left[ {\sqrt 3 {1 \over 6} + 2{\pi \over 2} - 2{\pi \over 6}} \right]$

$ = 2\left[ {\sqrt 3 + {1 \over 6} + 2{{2\pi } \over 6}} \right] = 2\left[ {{{\sqrt 3 } \over 6} + {{4\pi } \over 6}} \right]$

$ = {{\sqrt 3 } \over 3} + {{4\pi } \over 3} = \left( {{1 \over {\sqrt 3 }} + {{4\pi } \over 3}} \right)$ sq. units

Area of shaded region

= $\int\limits_0^1 {\left( {1 + \sqrt x } \right)dx} + \int\limits_1^2 {\left( {3 - x} \right)dx} - \int\limits_0^2 {{{{x^2}} \over 4}dx} $

= $\left[ x \right]_0^1 + \left[ {{{{x^{{3 \over 2}}}} \over {{3 \over 2}}}} \right]_0^1$ + $3\left[ x \right]_1^2 - \left[ {{{{x^2}} \over 2}} \right]_1^2 - \left[ {{{{x^3}} \over {12}}} \right]_0^2$

= ${5 \over 2}$ sq. unit

Required Area

= A₁ + A₂

= $\left| {\int\limits_1^3 {\left( {1 - x} \right)} dx} \right| + \left| {\int\limits_3^4 {\left( {{x^2} - 5x + 4} \right)dx} } \right|$

= $\left| {\left[ {x - {{{x^2}} \over 2}} \right]_1^3} \right| + \left| {\left[ {{{{x^3}} \over 3} - {5 \over 2}{x^2} + 4x} \right]_3^4} \right|$

= $\left| {\left[ {\left( {3 - {9 \over 2}} \right) - \left( {1 - {1 \over 2}} \right)} \right]} \right| + \left| {\left[ {\left( {{{64} \over 3} - 40 + 16} \right) - \left( {9 - {{45} \over 2} + 12} \right)} \right]} \right|$

= $\left| {\left( {2 - 4} \right)} \right| + \left| {\left( {{{ - 8} \over 3} + {3 \over 2}} \right)} \right|$

= 2 + ${7 \over 6}$

= ${{19} \over 6}$ sq. unit.

Points of intersection of the two curves are $\left( {0,0} \right),\left( {2,2} \right)$ and $\left( {2, - 2} \right)$

Area $=$ Area $(OAB)-$ area under parabola ($0$ to $2$ )

$ = {{\pi \times {{\left( 2 \right)}^2}} \over 4} - \int\limits_0^2 {\sqrt 2 \sqrt x } \,dx$

$ = \pi - {8 \over 3}$

Required area


$ = $ Area of $ABCD$ $-$ $ar$ $(ABOCD)$


$ = {1 \over 4}\left[ {{{{y^2}} \over 2} + y} \right]_{ - 1/2}^1\,\, - {1 \over 2}\left[ {{{{y^3}} \over 3}} \right]_{ - 1}^1$

$ = {1 \over 4}\left[ {{3 \over 2} + {3 \over 8}} \right] - {9 \over {48}}$

$ = {{15} \over {32}} - {9 \over {48}} = {{27} \over {96}} = {9 \over {32}}$

Given curves are ${x^2} + {y^2} = 1$ and ${y^2} = 1 - x.$

Intersection points are $x = 0,1$

Area of shaded portion is the required area.

So, Required Area $=$ Area of semi-circle $+$ Area bounded by parabola

$ = {{\pi {r^2}} \over 2} + 2\int\limits_0^1 {\sqrt {1 - x} dx} $

$ = {\pi \over 2} + 2\int\limits_0^1 {\sqrt {1 - x} \,dx} $

( As radius of circle $=1$ )

$ = {\pi \over 2} + 2\left[ {{{{{\left( {1 - x} \right)}^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over { - {\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \right]_0^1$

$ = {\pi \over 2} - {4 \over 3}\left( { - 1} \right) = {\pi \over 2} + {4 \over 3}$ Sq. unit

Given curves are

$y = \sqrt x $ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

and $2y - x + 3 = 0\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

On solving both we get $y=-1,3$



Required area $ = \int\limits_0^3 {\left\{ {\left( {2y + 3} \right) - {y^2}} \right\}} dy$

$\left. {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {y^2} + 3y - {{{y^3}} \over 3}} \right|_0^3 = 9.$

Given curves ${x^2} = {y \over 4}$ and ${x^2} = 9y$ are the parabolas whose equations can be written as $y = 4{x^2}$ and $y = {1 \over 9}{x^2}.$

Also, given $y=2.$

Now, shaded portion shows the required area which is symmetric.



$\therefore$ Area $ = 2\int\limits_0^2 {\left( {\sqrt {9y} - \sqrt {{y \over 4}} } \right)} dy$

Area $ = 2\int\limits_0^2 {\left( {3\sqrt y - {{\sqrt y } \over 2}} \right)} dy$

$ = 2\left[ {{2 \over 3} \times 3.{y^{{3 \over 2}}} - {1 \over 2} \times {2 \over 3}.{y^{{3 \over 2}}}} \right]_0^2$

$ = 2\left[ {2{y^{{3 \over 2}}} - {1 \over 3}{y^{{3 \over 2}}}} \right] = \left. {2 \times {5 \over 3}{y^{{3 \over 2}}}} \right|_0^2$

$ = 2.{5 \over 3}2\sqrt 2 = {{20\sqrt 2 } \over 3}$

Area of required region $AOCB$

$ = \int\limits_0^1 {xdx} + \int\limits_1^e {{1 \over x}dx = {1 \over 2}} + 1 = {3 \over 2}$ sq. units

$\therefore$ Required area

$ = \left[ {\int\limits_0^{{\pi \over 4}} {\left( {\cos \,x - \sin x} \right)dx + } } \right.$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\int\limits_{{\pi \over 4}}^{{5\pi \over 4}} {\left( {\sin x - \cos x} \right)dx + } $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\int\limits_{{5\pi \over 4}}^{{3\pi \over 2}} {\left( {\cos x - \sin x} \right)dx } $

$ = 4\sqrt 2 - 2$

The given parabola is ${\left( {y - 2} \right)^2} = x - 1$

Vertex $\left( {1,2} \right)$ and it meets $x$-axis at $\left( {5,0} \right)$

Also it gives ${y^2} - 4y - x + 5 = 0$

So, that equation of tangent to the parabola at $\left( {2,3} \right)$ is

$y.3 - 2\left( {y + 3} \right) - {1 \over 2}\left( {x + 2} \right) + 5 = 0$

or $x - 2y + 4 = 0$

which meets $x$-axis at $\left( { - 4,0} \right).$

In the figure shaded area is the required area.

Let us draw $PD$ perpendicular to $y$-axis.



Then required area

$ = Ar\,\,\Delta BOA + Ar\,\,\left( {OCPD} \right)\, - \,Ar\left( {\Delta APD} \right)$

$ = {1 \over 2} \times 4 \times 2 + \int_0^3 {xdy - {1 \over 2}} \times 2 \times 1$

$ = 3 + \int_0^3 {{{\left( {y - 2} \right)}^2} + 1\,dy} $

$ = 3 + \left[ {{{{{\left( {y - 2} \right)}^3}} \over 3} + y} \right]_0^3$

$ = 3 + \left[ {{1 \over 3} + 3 + {8 \over 3}} \right] = 3 + 6 = 9$ Sq. units

$x + 2{y^2} = 0 \Rightarrow {y^2} = - {x \over 2}$

$\left[ {} \right.$ Left handed parabola with vertex at $\left( {0,0} \right)$ $\left. {} \right]$

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

$\left[ {} \right.$ Left handed parabola with vertex at $\left( {1,0} \right)$ $\left. {} \right]$

Solving the two equations we get the points of intersection

as $\left( { - 2,1} \right),\left( { - 2, - 1} \right)$



The required area is $ACBDA,$ given by

$ = \left| {\int\limits_{ - 1}^1 {\left( {1 - 3{y^2} - 2{y^2}} \right)dy} } \right|$

$ = \left| {\left[ {y - {{5{y^3}} \over 3}} \right]_{ - 1}^1} \right|$

$ = \left| {\left( {1 - {5 \over 3}} \right) - \left( { - 1 + {5 \over 3}} \right)} \right|$

$ = 2 \times {2 \over 3} = {4 \over 3}\,\,$ sq. units.

The area enclosed between the curves

${y^2} = x$ and $y = \left| x \right|$

From the figure, area lies between

${y^2} = x$ and $y = x$



$\therefore$ Required area $ = \int_0^1 {\left( {{y_2} - {y_1}} \right)} dx$

$ = \int_0^1 {\left( {\sqrt x - x} \right)dx = \left[ {{{{x^{3/2}}} \over {3/2}} - {{{x^2}} \over 2}} \right]} _0^1$

$\therefore$ Required area $ = {2 \over 3}\left[ {{x^{3/2}}} \right]_0^1 - {1 \over 2}\left[ {{x^2}} \right]_0^1$

$ = {2 \over 3} - {1 \over 2} = {1 \over 6}$

Intersection points of ${x^2} = 4y$ and ${y^2} = 4x$ are $\left( {0,0} \right)$ and $\left( {4,4} \right).$ The graph is as shown in the figure.



By symmetry, we observe

${S_1} = {S_3} = \int\limits_0^4 {ydx = \int\limits_0^4 {{{{x^2}} \over 4}dx} } $

$ = \left[ {{{{x^3}} \over {12}}} \right]_0^4 = {{16} \over 3}$ sq. unit

Also ${S_2} = \int\limits_0^4 {\left( {2\sqrt x - {{{x^2}} \over 4}} \right)} dx$

$\,\,\,\,\,\,\,\,\,\,\,\,$ $ = \left[ {{{2{x^{{3 \over 2}}}} \over {{3 \over 2}}} - {{{x^3}} \over {12}}} \right]_0^4$

$\,\,\,\,\,\,\,\,\,\,\,\,$ $ = {4 \over 3} \times 8 - {{16} \over 3} = {{16} \over 3}$

$\therefore$ ${S_1}:{S_2}:{S_3} = 1:1:1$

The graph of the curve $y = {\log _e}\left( {x + e} \right)$ is as shown in the fig.



Required area

$A = \int\limits_{1 - e}^0 {ydx} = \int\limits_{1 - e}^0 {{{\log }_e}} \left( {x + e} \right)dx$

put $x + e = t \Rightarrow dx = dt$

also At $x = 1 - e,t = 1$

At $x = 0,\,\,t = e$

$\therefore$ $A = \int\limits_1^e {{{\log }_e}} \,tdt = \left[ {t\,{{\log }_e}t - t_1^e} \right]$

$e - e - 0 + 1 = 1$

Hence the required area is $1$ square unit.

Given that

$\int\limits_{\pi /4}^\beta {f\left( x \right)} dx = \beta \sin \beta + {\pi \over 4}\cos \,\beta + \sqrt 2 \beta $

Differentiating $w.r.t$ $\beta $

$f\left( \beta \right) = \beta \cos \beta + \sin \beta - {\pi \over 4}\sin \beta + \sqrt 2 $

$f\left( {{\pi \over 2}} \right) = \left( {1 - {\pi \over 4}} \right)\sin {\pi \over 2} + \sqrt 2 $

$ = 1 - {\pi \over 4} + \sqrt 2 $

The required area is shown by shaded region



$A = \int\limits_1^3 {\left| {x - 2} \right|dx = 2\int\limits_2^3 {\left( {x - 2} \right)} } dx$

$ = 2\left[ {{{{x^2}} \over 2} - 2x} \right]_2^3 = 1$

$A = \int\limits_{ - 1}^0 {\left\{ {\left( {3 + x} \right) - \left( { - x + 1} \right)} \right\}dx + } $

$\,\,\,\,\,\,\,\,\,\,\,\,$ $\int\limits_0^1 {\left\{ {\left( {3 - x} \right) - \left( { - x + 1} \right)} \right\}dx + } $

$\,\,\,\,\,\,\,\,\,\,\,\,$ $\int\limits_1^2 {\left\{ {\left( {3 - x} \right) - \left( {x - 1} \right)} \right\}dx} $

$ = \int\limits_{ - 1}^0 {\left( {2 + 2x} \right)dx + \int\limits_0^1 {2dx + \int\limits_1^2 {\left( {4 - 2x} \right)dx} } } $

$ = \left[ {2x - {x^2}} \right]_{ - 1}^0 + \left[ {2x} \right]_0^1 + \left[ {4x - {x^2}} \right]_1^2$

$ = 0 - \left( { - 2 + 1} \right) + \left( {2 - 0} \right) + \left( {8 - 4} \right) - \left( {4 - 1} \right)$

$ = 1 + 2 + 4 - 3 = 4$ sq. units

First we draw each curve as separate graph



NOTE : Graph of $y = \left| {f\left( x \right)} \right|$ can be obtained from the graph of the curve $y = f\left( x \right)$ by drawing the mirror image of the portion of the graph below $x$-axis, with respect to $x$-axis.

Clearly the bounded area is as shown in the following figure.



Required area $ = 4\int\limits_0^1 {\left( { - \ln x} \right)} dx$

$ = - 4\left[ {x\,\ln x + - x_0^1} \right] = 4\,\,$ sq. units