Area Under The Curves
The area of the region
$\left\{(x, y):|x-1| \leq y \leq \sqrt{5-x^{2}}\right\}$ is equal to :
The area enclosed by the curves $y=\log _{e}\left(x+\mathrm{e}^{2}\right), x=\log _{e}\left(\frac{2}{y}\right)$ and $x=\log _{\mathrm{e}} 2$, above the line $y=1$ is:
The area of the region enclosed by $y \leq 4 x^{2}, x^{2} \leq 9 y$ and $y \leq 4$, is equal to :
Consider a curve $y=y(x)$ in the first quadrant as shown in the figure. Let the area $\mathrm{A}_{1}$ is twice the area $\mathrm{A}_{2}$. Then the normal to the curve perpendicular to the line $2 x-12 y=15$ does NOT pass through the point.
The area of the smaller region enclosed by the curves $y^{2}=8 x+4$ and $x^{2}+y^{2}+4 \sqrt{3} x-4=0$ is equal to
The area bounded by the curves $y=\left|x^{2}-1\right|$ and $y=1$ is
The odd natural number a, such that the area of the region bounded by y = 1, y = 3, x = 0, x = ya is ${{364} \over 3}$, is equal to :
The area of the region given by
$A=\left\{(x, y): x^{2} \leq y \leq \min \{x+2,4-3 x\}\right\}$ is :
Let the locus of the centre $(\alpha, \beta), \beta>0$, of the circle which touches the circle $x^{2}+(y-1)^{2}=1$ externally and also touches the $x$-axis be $\mathrm{L}$. Then the area bounded by $\mathrm{L}$ and the line $y=4$ is:
The area enclosed by y2 = 8x and y = $\sqrt2$ x that lies outside the triangle formed by y = $\sqrt2$ x, x = 1, y = 2$\sqrt2$, is equal to:
The area of the bounded region enclosed by the curve
$y = 3 - \left| {x - {1 \over 2}} \right| - |x + 1|$ and the x-axis is :
The area of the region S = {(x, y) : y2 $\le$ 8x, y $\ge$ $\sqrt2$x, x $\ge$ 1} is
The area of the region bounded by y2 = 8x and y2 = 16(3 $-$ x) is equal to:
The area bounded by the curve y = |x2 $-$ 9| and the line y = 3 is :
The area of the region enclosed between the parabolas y2 = 2x $-$ 1 and y2 = 4x $-$ 3 is
Let the area enclosed by the x-axis, and the tangent and normal drawn to the curve $4{x^3} - 3x{y^2} + 6{x^2} - 5xy - 8{y^2} + 9x + 14 = 0$ at the point ($-$2, 3) be A. Then 8A is equal to ______________.
Explanation:
$ \begin{aligned} & \text { Area }=\frac{1}{2}\left(\frac{31}{2}-4\right) \times 3=\frac{85}{4} \\\\ & 8 \mathrm{~A}=170 \end{aligned} $
If for some $\alpha$ > 0, the area of the region $\{ (x,y):|x + \alpha | \le y \le 2 - |x|\} $ is equal to ${3 \over 2}$, then the area of the region $\{ (x,y):0 \le y \le x + 2\alpha ,\,|x| \le 1\} $ is equal to ____________.
Explanation:
Point A is the intersection of y = x + $\alpha$ and y = 2 $-$ x lines.
$\therefore$ $y = 2 - y + \alpha $
$ \Rightarrow 2y = 2 + \alpha $
$ \Rightarrow y = {{2 + \alpha } \over 2}$
and $x = 2 - {{2 + \alpha } \over 2} = {{2 - \alpha } \over 2}$
$\therefore$ Point $A = \left( {{{2 - \alpha } \over 2},{{2 + \alpha } \over 2}} \right)$
$\therefore$ AN = y-coordinate of point $A = {{2 + \alpha } \over 2}$
Point B is the intersection of $y = 2 + x$ and $y = - x - \alpha $ lines.
$\therefore$ $y = 2 - y - \alpha $
$ \Rightarrow 2y = 2 - \alpha $
$ \Rightarrow y = {{2 - \alpha } \over 2}$
$\therefore$ $x = {{2 - \alpha } \over 2} - 2 = {{2 - \alpha - 4} \over 2} = {{ - \alpha - 2} \over 2}$
$\therefore$ Point $B = \left( {{{ - \alpha - 2} \over 2},{{2 - \alpha } \over 2}} \right)$
$\therefore$ $BM = y - $coordinate of point $B = {{2 - \alpha } \over 2}$
Area of the common region $BRAE$
$ = \Delta CDE - \left( {\Delta BCR + \Delta ARD} \right)$
$ = {1 \over 2} \times 4 \times 2 - \left( {{1 \over 2}( - \alpha + 2) \times BM + {1 \over 2} \times (2 + \alpha ) \times AN} \right)$
$ = 4 - \left( {{1 \over 2} \times (2 - \alpha ) \times {{(2 - \alpha )} \over 2} + {1 \over 2} \times (2 + \alpha ) \times {{2 + \alpha } \over 2}} \right)$
$ = 4 - \left[ {{{{{(2 - \alpha )}^2}} \over 4} + {{{{(2 + \alpha )}^2}} \over 4}} \right]$
Given, $4 - \left[ {{{{{(2 - \alpha )}^2}} \over 4} + {{{{(2 + \alpha )}^2}} \over 4}} \right] = {3 \over 2}$
$ \Rightarrow {{{{(2 - \alpha )}^2}} \over 4} + {{{{(2 + \alpha )}^2}} \over 4} = {5 \over 2}$
$ \Rightarrow {(2 - \alpha )^2} + {(2 + \alpha )^2} = 10$
$ \Rightarrow 4 + {\alpha ^2} - 4\alpha + 4 + {\alpha ^2} + 4\alpha = 10$
$ \Rightarrow 2{\alpha ^2} + 8 = 10$
$ \Rightarrow 2{\alpha ^2} = 2$
$ \Rightarrow {\alpha ^2} = 1$
$ \Rightarrow \alpha = \, \pm \,1$
Given that $\alpha > 0$ so accepted value of $\alpha = + \,1$.
Now, $0 \le y \le x + 2\alpha $ and $|x| \le 1$
$ \Rightarrow 0 \le y \le x + 2$ and $ - 1 \le x \le 1$
Area of $ABCD = {1 \over 2}(1 + 3) \times (1 - ( - 1))$
$ = {1 \over 2} \times 4 \times 2$
$ = 4$ sq. unit
For real numbers a, b (a > b > 0), let
Area $\left\{ {(x,y):{x^2} + {y^2} \le {a^2}\,and\,{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} \ge 1} \right\} = 30\pi $
and
Area $\left\{ {(x,y):{x^2} + {y^2} \le {b^2}\,and\,{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} \le 1} \right\} = 18\pi $
Then, the value of (a $-$ b)2 is equal to ___________.
Explanation:
and $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \geq 1$ is exterior of ellipse
$\therefore$ Area $=\pi a^{2}-\pi a b=30 \pi$
Similarly $x^{2}+y^{2} \geq b^{2}$ and $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leq 1$ gives
$\pi a b-\pi b^{2}=18 \pi$
By (1) and (2), $\frac{a}{b}=\frac{5}{3} \Rightarrow a=\frac{5 b}{3}$
$\Rightarrow \pi \cdot \frac{25 b^{2}}{9}-\pi \cdot \frac{5 b^{2}}{3}=30 \pi$
$\Rightarrow\left(\frac{25}{9}-\frac{5}{3}\right) b^{2}=30$
$\Rightarrow \frac{10}{9} b^{2}=30 \Rightarrow b^{2}=27$
and $a^{2}=\frac{25}{9} \cdot 27=75$
$(a-b)^{2}=(5 \sqrt{3}-3 \sqrt{3})^{2}=3 \cdot 4=12$
If the area of the region $\left\{ {(x,y):{x^{{2 \over 3}}} + {y^{{2 \over 3}}} \le 1,\,x + y \ge 0,\,y \ge 0} \right\}$ is A, then ${{256A} \over \pi }$ is equal to __________.
Explanation:
$\therefore$ Area of shaded region
$ = \int\limits_{ - {{\left( {{1 \over 2}} \right)}^{{3 \over 2}}}}^0 {\left( {{{\left( {1 - {x^{{2 \over 3}}}} \right)}^{{3 \over 2}}} + x} \right)dx + \int\limits_0^1 {{{\left( {1 - {x^{{2 \over 3}}}} \right)}^{{3 \over 2}}}dx} } $
$ = \int\limits_{ - {{\left( {{1 \over 2}} \right)}^{{3 \over 2}}}}^0 {{{\left( {1 - {x^{{2 \over 3}}}} \right)}^{{3 \over 2}}}dx + \int\limits_{ - {{\left( {{1 \over 2}} \right)}^{{3 \over 2}}}}^0 {xdx} } $
Let $x = {\sin ^3}\theta $
$\therefore$ $dx = 3{\sin ^2}\theta \cos \theta d\theta $
$ = \int\limits_{ - {\pi \over 4}}^{{\pi \over 2}} {3{{\sin }^2}\theta {{\cos }^4}\theta d\theta + \left( {0 + {1 \over {16}}} \right)} $
$ = {{9\pi } \over {64}} + {1 \over {16}} - {1 \over {16}} = {{36\pi } \over {256}} = A$
$\therefore$ ${{256A} \over \pi } = 36$
${A_1} = \left\{ {(x,y):|x| \le {y^2},|x| + 2y \le 8} \right\}$ and
${A_2} = \left\{ {(x,y):|x| + |y| \le k} \right\}$. If 27 (Area A1) = 5 (Area A2), then k is equal to :
Explanation:
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 area (in sq. units) of the region enclosed between the parabola y2 = 2x and the line x + y = 4 is __________.
Explanation:
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
Let S be the region bounded by the curves y = x3 and y2 = x. The curve y = 2|x| divides S into two regions of areas R1, R2. If max {R1, R2} = R2, then ${{{R_2}} \over {{R_1}}}$ is equal to ______________.
Explanation:
$C_{1}: y=x^{3}$
$C_{2}: y^{2}=x$
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, $\frac{R_{1}+R_{2}}{R_{1}}=\frac{5 / 12}{1 / 48} \Rightarrow 1+\frac{R_{2}}{R_{1}}=20$
$\Rightarrow \frac{R_{2}}{R_{1}}=19$
$R = \left\{ {(x,y):\max \{ 0,{{\log }_e}x\} \le y \le {2^x},{1 \over 2} \le x \le 2} \right\}$ is ,
$\alpha {({\log _e}2)^{ - 1}} + \beta ({\log _e}2) + \gamma $, then the value of ${(\alpha + \beta - 2\lambda )^2}$ is equal to :
Explanation:
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
Explanation:
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
Explanation:
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$
Explanation:
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)$
Explanation:
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
$f(x) = \left\{ \matrix{ \min \,\{ (x + 6),{x^2}\}, - 3 \le x \le 0 \hfill \cr \max \,\{ \sqrt x ,{x^2}\} ,\,0 \le x \le 1. \hfill \cr} \right.$
If the area bounded by y = f(x) and x-axis is A, then the value of 6A is equal to ___________.
Explanation:
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
Explanation:
Explanation:
$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$
by the curves y = x2 – 1 and y = 1 – x2 is equal to :
A = {(x, y) : |x| + |y| $ \le $ 1, 2y2 $ \ge $ |x|}
For point of intersection
A = {(x, y) : (x – 1)[x] $ \le $ y $ \le $ 2$\sqrt x $, 0 $ \le $ x $ \le $ 2}, where [t]
denotes the greatest integer function, is :
{ (x, y) : 0 $ \le $ y $ \le $ x2 + 1, 0 $ \le $ y $ \le $ x + 1,
${1 \over 2}$ $ \le $ x $ \le $ 2 } is :
$A = \int\limits_{{1 \over 2}}^1 {({x^2} + 1)dx + } $$\int\limits_1^2 {} $${(x + 1)dx}$
${{\left| x \right|} \over 2} + {{\left| y \right|} \over 3} = 1$ and inside the ellipse ${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$ is :
and $g(x) = \left( {x - {1 \over 2}} \right)^2,x \in R$
Then the area (in sq. units) of the region bounded by the curves, y = ƒ(x) and y = g(x) between the lines, 2x = 1 and 2x = $\sqrt 3 $, is :
{(x,y) $ \in $ R2 : x2 $ \le $ y $ \le $ 3 – 2x}, is :
$\Delta $OQR = ${1 \over 2}$, then 'a' satisfies the equation :
C1 : y2
= ax, C2 : x2
= ay (a > 0)
{(x, y) $ \in $ R2 | 4x2 $ \le $ y $ \le $ 8x + 12} is :
