Area Under The Curves

Required area (above x-axis)

${A_1} = 2\int\limits_0^4 {\left( {{{8 - x} \over 2} - \sqrt x } \right)dx} $

$ = 2\left( {16 - {{16} \over 4} - {8 \over {3/2}}} \right) = {{40} \over 3}$

and ${A_2} = 4\left( {{1 \over 2}\,.\,{k^2}} \right) = 2{k^2}$

$\therefore$ $27\,.\,{{40} \over 3} = 5\,.\,(2{k^2})$

$\Rightarrow$ k = 6

The required area $ = \int_{ - 4}^2 {\left( {4 - y - {{{y^2}} \over 2}} \right)dy} $

$ = \left[ {4y - {{{y^2}} \over 2} - {{{y^3}} \over 6}} \right]_{ - 4}^2$

$ = 18$ square units

$C_{1}: y=x^{3}$

$C_{2}: y^{2}=x$

and $C_{3}=y=2|x|$

$C_{1}$ and $C_{2}$ intersect at $(1,1)$

$C_{2}$ and $C_{3}$ intersect at $\left(\frac{1}{4}, \frac{1}{2}\right)$

Clearly $R_{1}=\int_{0}^{1 / 4}(\sqrt{x}-2 x) d x=\frac{2}{3}\left(\frac{1}{8}\right)-\frac{1}{16}=\frac{1}{48}$

and $R_{1}+R_{2}=\int_{0}^{1}\left(\sqrt{x}-x^{3}\right) d x=\frac{2}{3}-\frac{1}{4}=\frac{5}{12}$

So, $\frac{R_{1}+R_{2}}{R_{1}}=\frac{5 / 12}{1 / 48} \Rightarrow 1+\frac{R_{2}}{R_{1}}=20$

$\Rightarrow \frac{R_{2}}{R_{1}}=19$

According to the question,

${1 \over 2}\int_0^{3/2} {(1 + 4x - {x^2})dx = \int_0^{3/2} {mx\,dx} } $

$ \Rightarrow {1 \over 2}\left[ {\left( {x + 2{x^2} - {{{x^3}} \over 3}} \right)} \right]_0^{3/2} = {m \over 2}[x]_0^{3/2} \Rightarrow {3 \over 2} + {9 \over 2} - {9 \over 8} = {{9m} \over 4}$

$\Rightarrow$ m = 39/18 $\Rightarrow$ 12m = 26

f'(x) = 6x² $-$ 6x $-$ 12 = 6(x $-$ 2) (x + 1)

Point = (2, $-$20) & ($-$1, 7)


$A = \int\limits_{ - 1}^0 {(2{x^3} - 3{x^2} - 12x)dx + \int\limits_0^2 {(12x + 3{x^2} - 2{x^3})\,dx} } $

$A = \left( {{{{x^4}} \over 2} - {x^3} - 6{x^2}} \right)_{ - 1}^0 + \left( {6{x^2} + {x^3} - {{{x^4}} \over 2}} \right)_0^2$

4A = 114

For A & B

3x² = 6x + 24 $\Rightarrow$ x² $-$ 2x $-$ 8 = 0

$\Rightarrow$ x = $-$2, 4

Area $ = \int\limits_{ - 2}^4 {\left( {{3 \over 2}x + 6 - {3 \over 4}{x^2}} \right)dx} $

$ = \left[ {{{3{x^2}} \over 4} + 6x - {{{x^3}} \over 4}} \right]_{ - 2}^4 = 27$

Required area (shaded)

$ = 2\left[ {\int\limits_0^2 {\left( {{{4 - {y^2}} \over 4}} \right)dy - \int\limits_0^1 {\left( {{{1 - {x^2}} \over 2}} \right)dx} } } \right]$

$ = 2\left[ {{4 \over 3} - {1 \over 3}} \right] = (2)$

E : x² + 4y² = 5

Tangent at P : x + 4y = 5

Required area

$ = \int\limits_1^{\sqrt 5 } {\left( {{{5 - x} \over 4} - {{\sqrt {5 - {x^2}} } \over 2}} \right)dx} $

$ = \left[ {{{5x} \over 4} - {{{x^2}} \over 8} - {x \over 4}\sqrt {5 - {x^2}} - {5 \over 2}{{\sin }^{ - 1}}{x \over {\sqrt 5 }}} \right]_1^{\sqrt 5 }$

$ = {5 \over 4}\sqrt 5 - {5 \over 4} - {5 \over 4}{\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)$

If we assume $\alpha$, $\beta$, $\gamma$, $\in$ Q (Not given in question) then $\alpha$ = ${5 \over 4}$, $\beta$ = $-$${5 \over 4}$ & $\gamma$ = $-$${5 \over 4}$

|$\alpha$ + $\beta$ + $\gamma$| = 1.25

Area is $\int\limits_{ - 3}^{ - 2} {(x + 6)dx + \int\limits_{ - 2}^0 {{x^2}dx + \int\limits_0^1 {\sqrt {x}dx = A} } } $

$ = {7 \over 2} + \left[ {{{{x^3}} \over 3}} \right]_{ - 2}^0 + \left[ {{2 \over 3}{x^{3/2}}} \right]_0^1$

$ = {7 \over 2} + {8 \over 3} + {2 \over 3} = {{41} \over 6}$

So, 6A = 41

Question is incomplete it should be area bounded by y = || x $-$ 1 | $-$ 2 | and y = 2.

$A = \int\limits_{{\pi \over 4}}^{{{5\pi } \over 4}} {(\sin x - \cos x)dx} $

$= [ - \cos x - \sin x]_{\pi /4}^{5\pi /4}$

$ = - \left[ {\left( {\cos {{5\pi } \over 4} + \sin {5\pi \over 4}} \right) - \left( {\cos {\pi \over 4} + \sin {\pi \over 4}} \right)} \right]$

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

$ = {4 \over {\sqrt 2 }} = 2\sqrt 2 $

$ \Rightarrow {A^4} = {\left( {2\sqrt 2 } \right)^4} = 64$