Area Under The Curves

${A_1} + {A_2} = xy - 8$ & ${A_1} = 2{A_2}$

${A_1} + {{{A_1}} \over 2} = xy - 8$

${A_1} = {2 \over 3}(xy - 8)$

$\int\limits_4^x {f(x)dx = {2 \over 3}(xf(x) - 8)} $

Differentiate w.r.t. x

$f(x) = {2 \over 3}\{ xf'(x) + f(x)\} $

${2 \over 3}xf'(x) = {1 \over 3}f(x)$

$2\int {{{f'(x)} \over {f(x)}}dx = \int {{{dx} \over x}} } $

$2\ln f(x) = \ln x + \ln c$

${f^2}(x) = cx$

Which passes through (4, 2)

$4 = c \times 4 \Rightarrow c = 1$

Equation of required curve

${y^2} = x$

Equation of normal having slope ($-$6) is

$y = - 6x - 2\left( {{1 \over 4}} \right)( - 6) - {1 \over 4}{( - 6)^3}$

$y = - 6x + 57$

Which does not pass through $(10,\, - 4)$

$\cos \theta = {{2\sqrt 3 } \over 4} = {{\sqrt 3 } \over 2} \Rightarrow \theta = 30^\circ $

Area of the required region

$ = {2 \over 3}\left( {4 \times {1 \over 2}} \right) + {4^2} \times {\pi \over 6} - {1 \over 2} \times 4 \times 2\sqrt 3 $

$ = {4 \over 3} + {{8\pi } \over 3} - 4\sqrt 3 = {1 \over 3}\left\{ {4 - 12\sqrt 3 + 8\pi } \right\}$

Area $ = 2\int\limits_0^{\sqrt 2 } {(1 - |{x^2} - 1|)dx} $

$2\left[ {\int\limits_0^1 {\left( {1 - (1 - {x^2})} \right)dx + \int\limits_1^{\sqrt 2 } {\left( {2 - {x^2}} \right)dx} } } \right]$

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

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

$\mathrm{a}$ is a odd natural number and

$\left| {\int\limits_1^3 {{y^a}dy} } \right| = {{364} \over 3}$

$ \Rightarrow \left| {{1 \over {a + 1}}\left( {{y^{a + 1}}} \right)_1^3} \right| = {{364} \over 3}$

$ \Rightarrow {{{3^{a + 1}} - 1} \over {a + 1}} = \, \pm \,{{364} \over 3}$

Solving with $( - )$ sign,

${{{3^{a + 1}} - 1} \over {a + 1}} = {{364} \over 3} \Rightarrow (a = 5)$

Solving with $( + )$ sign,

${{{3^{a + 1}} - 1} \over {a + 1}} = {{ - 364} \over 3}$, No a exist

$\therefore$ $(a = 5)$

$A\left( {2,2\sqrt 2 } \right),B\left( {1,2\sqrt 2 } \right),C\left( {1,\sqrt 2 } \right)$

Area $ = \int\limits_0^{4\sqrt 2 } {\left( {{y \over {\sqrt 2 }} - {{{y^2}} \over 8}} \right)dy - } $ area ($\Delta$BAC)

$ = \left[ {{{{y^2}} \over {2\sqrt 2 }} - {{{y^3}} \over {24}}} \right]_0^{4\sqrt 2 } - {1 \over 2} \times AB \times BC$

$ = 8\sqrt 2 - {{32 \times 4\sqrt 2 } \over {24}} - {1 \over 2} \times 1 \times \sqrt 2 $

$ = 8\sqrt 2 - {{16\sqrt 2 } \over 3} - {{\sqrt 2 } \over 2}$

$ = {{\sqrt 2 } \over 6}(48 - 32 - 3) = {{13\sqrt 2 } \over 6}$

Required area

$ = \int\limits_1^4 {\left( {\sqrt {8x} - \sqrt 2 x} \right)dx} $

$ = \left. {{{2\sqrt 8 } \over 3}{x^{{3 \over 2}}} - {{{x^2}} \over {\sqrt 2 }}} \right|_1^4$

$ = {{16\sqrt 3 } \over 3} - {{16} \over {\sqrt 2 }} - {{2\sqrt 8 } \over 3} + {1 \over {\sqrt 2 }}$

$ = {{11\sqrt 2 } \over 6}$ sq. units

${c_1}:{y^2} = 8x$

${c_2}:{y^2} = 16(3 - x)$

Solving c₁ and c₂

$48 - 16x = 8x$

$x = 2$

$\therefore$ $y = \pm \,4$

$\therefore$ Area of shaded region

$ = 2\int\limits_0^4 {\left\{ {\left( {{{48 - {y^2}} \over {16}}} \right) - \left( {{{{y^2}} \over 8}} \right)} \right\}dy} $

$ = {1 \over 8}\left[ {48y - {y^3}} \right]_0^4 = 16$

$y = 3$ and $y = |{x^2} - 9|$

Intersect in first quadrant at $x = \sqrt 6 $ and $x = \sqrt {12} $

Required area

$ = 2\left[ {{2 \over 3}\left( {6 \times \sqrt 6 } \right) + \int\limits_{\sqrt 6 }^3 {\left( {3 - \left( {9 - {x^2}} \right)} \right)dx + \int\limits_3^{\sqrt {12} } {\left( {3 - \left( {{x^2} - 9} \right)} \right)dx} } } \right]$

$ = 2\left[ {4\sqrt 6 + \left. {\left( {{{{x^3}} \over 3} - 6x} \right)} \right|_{\sqrt 6 }^3 + \left. {\left( {12x - {{{x^3}} \over 3}} \right)} \right|_3^{\sqrt {12} }} \right]$

$ = 2\left[ {4\sqrt 6 + \left( {4\sqrt 6 - 9} \right) + \left( {8\sqrt {12} - 27} \right)} \right]$

$ = 2\left[ {8\sqrt 6 + 16\sqrt 3 - 36} \right] = 8\left[ {2\sqrt 6 + 4\sqrt 3 - 9} \right]$

Area of the shaded region

$ = 2\int_0^1 {\left( {{{{y^2} + 3} \over 4} - {{{y^2} + 1} \over 2}} \right)dy} $

$ = 2\int_0^1 {\left( {{1 \over 4} - {{{y^2}} \over 4}} \right)dy} $

$ = 2\left[ {{1 \over 4} - {1 \over {12}}} \right] = {1 \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