Area Under The Curves
The area of the region enclosed by the parabolas $y=x^2-5 x$ and $y=7 x-x^2$ is ________.
Explanation:
$\begin{aligned} y=x^2-5 x, y=7 x-x^2 & \Rightarrow \quad x^2-5 x=7 x-x^2 \\ & \Rightarrow \quad x=0, x=6 \end{aligned}$
$\text { Area }=\int_\limits0^6\left[\left(7 x-x^2\right)-(x-5 x)\right] d x$
$=\int_\limits0^6\left(12 x-2 x^2\right) d x=6 x-\left.\frac{2 x^3}{3}\right|_0 ^6$
$=216-144=72 \text { sq. unit }$
But answer is 198 as per NTA.
Explanation:
Equation of $P Q$
$ \begin{aligned} & y+b=\frac{a^2-b^2}{a+b}\left(x-b^2\right) \\\\ & \Rightarrow y=(a-b) x+a b \end{aligned} $
$\begin{aligned} & S_1=\int\limits_{-b}^a\left((a-b) x+a b-x^2\right) d x \\\\ & =\left(\frac{(a-b)}{2} x^2+a b x-\frac{x^3}{3}\right)_{-b}^a \\\\ & =\frac{1}{6}(a+b)^3\end{aligned}$
$S_2=\frac{1}{2}\left|\begin{array}{ccc}-b & b^2 & 1 \\ 0 & 0 & 1 \\ a & a^2 & 1\end{array}\right|=\frac{1}{2} a b(a+b)$
$\begin{aligned} \frac{S_1}{S_2} & =\frac{\frac{1}{6}(a+b)^3}{\frac{1}{2} \cdot a b(a+b)} \\\\ & =\frac{1}{3} \frac{(a+b)^2}{a b}=\frac{1}{3}\left(\frac{a}{b}+\frac{b}{a}+2\right)\end{aligned}$
$\begin{aligned} & \because \frac{b}{a}+\frac{a}{b} \geq 2 \\\\ & \Rightarrow\left(\frac{S_1}{S_2}\right)_{\min }=\frac{4}{3}=\frac{m}{n} \\\\ & \Rightarrow m+n=7\end{aligned}$
Explanation:
$ 2 y^2=k x $ .........(ii)
Point of intersection of (i) and (ii)
$ \begin{aligned} & k y^2=2\left(y-\frac{2 y^2}{k}\right) \\\\ & \Rightarrow y=0, k y=2\left(1-\frac{2 y}{k}\right) \end{aligned} $
$\begin{aligned} & k y+\frac{4 y}{k}=2 \\\\ & y=\frac{2}{k+\frac{4}{k}}=\frac{2 k}{k^2+4}\end{aligned}$
$\begin{aligned} & A=\int\limits_6^{\frac{2 k}{k^2+4}}\left(\left(y-\frac{k y^2}{2}\right)-\frac{2 y^2}{k}\right) d y \\\\ & A=\left[\frac{y^2}{2}-\left(\frac{k}{2}+\frac{2}{k}\right) \frac{y^3}{3}\right]_0^{\frac{2 k}{k^2+4}}\end{aligned}$
$\begin{aligned} & =\left(\frac{2 k}{k^2+4}\right)^2\left[\frac{1}{2}-\frac{k^2+4}{2 k}\left(\frac{1}{3}\right)\left(\frac{2 k}{k^2+4}\right)\right] \\\\ & =\frac{1}{6} \times 4 \times\left(\frac{1}{k+\frac{4}{k}}\right)^2\end{aligned}$
$\begin{aligned} & \text { A.M. } \geq \text { G.M. } \\\\ & \frac{\left(k+\frac{4}{k}\right)}{2} \geq 2 \\\\ & k+\frac{4}{k} \geq 4\end{aligned}$
$\therefore$ Area is maximum when $k=\frac{4}{k}$
$ \begin{aligned} & \therefore k^2=4 \\\\ & k= \pm 2 \\\\ & k_1=2, k_2=-2 \\\\ & \therefore k_1^2+k_2^2=(+2)^2+(-2)^2 \\\\ & =4+4 \\\\ & =08 \end{aligned} $
The area of the region enclosed by the parabola $(y-2)^2=x-1$, the line $x-2 y+4=0$ and the positive coordinate axes is _________.
Explanation:
Solving the equations
$\begin{array}{r} (y-2)^2=x-1 \text { and } x-2 y+4=0 \\ x=2(y-2) \end{array}$
$\begin{aligned} & \frac{x^2}{4}=x-1 \\ & x^2-4 x+4=0 \\ & (x-2)^2=0 \\ & x=2 \end{aligned}$
Exclose area (w.r.t. y-axis) $=\int_\limits0^3 x d y-\text { Area of } \Delta$.
$\begin{aligned} & =\int_\limits0^3\left((y-2)^2+1\right) d y-\frac{1}{2} \times 1 \times 2 \\ & =\int_\limits0^3\left(y^2-4 y+5\right) d y-1 \\ & =\left[\frac{y^3}{3}-2 y^2+5 y\right]_0^3-1 \\ & =9-18+15-1=5 \end{aligned}$
Let the area of the region $\left\{(x, y): 0 \leq x \leq 3,0 \leq y \leq \min \left\{x^2+2,2 x+2\right\}\right\}$ be A. Then $12 \mathrm{~A}$ is equal to __________.
Explanation:
$\begin{aligned} & y=2 x+2 \\ & A=\int_\limits0^2\left(x^2+2\right) d x+\int_\limits2^3(2 x+2) d x \\ & A=\frac{41}{3} \\ & 12 A=41 \times 4=164 \end{aligned}$
The area (in sq. units) of the part of the circle $x^2+y^2=169$ which is below the line $5 x-y=13$ is $\frac{\pi \alpha}{2 \beta}-\frac{65}{2}+\frac{\alpha}{\beta} \sin ^{-1}\left(\frac{12}{13}\right)$, where $\alpha, \beta$ are coprime numbers. Then $\alpha+\beta$ is equal to __________.
Explanation:
$\begin{aligned} & \text { Area }=\int_\limits{-13}^{12} \sqrt{169-y^2} d y-\frac{1}{2} \times 25 \times 5 \\ & =\frac{\pi}{2} \times \frac{169}{2}-\frac{65}{2}+\frac{169}{2} \sin ^{-1} \frac{12}{13} \\ & \therefore \alpha+\beta=171 \end{aligned}$
If the points of intersection of two distinct conics $x^2+y^2=4 b$ and $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ lie on the curve $y^2=3 x^2$, then $3 \sqrt{3}$ times the area of the rectangle formed by the intersection points is _________.
Explanation:
Putting $y^2=3 x^2$ in both the conics
We get $x^2=b$ and $\frac{b}{16}+\frac{3}{b}=1$
$\Rightarrow \mathrm{b}=4,12 \quad(\mathrm{b}=4$ is rejected because curves coincide)
$\therefore \mathrm{b}=12$
Hence points of intersection are
$( \pm \sqrt{12}, \pm 6) \Rightarrow \text { area of rectangle }=432$
If the area of the region $\left\{(x, y): 0 \leq y \leq \min \left\{2 x, 6 x-x^2\right\}\right\}$ is $\mathrm{A}$, then $12 \mathrm{~A}$ is equal to ________.
Explanation:
We have
$\begin{aligned} & A=\frac{1}{2} \times 4 \times 8+\int_\limits4^6\left(6 x-x^2\right) d x \\ & A=\frac{76}{3} \\ & 12 A=304 \end{aligned}$
Explanation:
$\begin{aligned} & A=\int_\limits0^1\left[\left(8-4 y^2\right)-\left(-2 y^2\right)\right] d y+ \\ & \int_\limits1^{3 / 2}\left[\left(8-4 y^2\right)-(2 y-4)\right] d y \\ & =\left[8 y-\frac{2 y^3}{3}\right]_0^1+\left[12 y-y^2-\frac{4 y^3}{3}\right]_1^{3 / 2}=\frac{107}{12}=\frac{m}{n} \\ & \therefore m+n=119 \end{aligned}$
The area of the region $\left\{(x, y): x^{2} \leq y \leq\left|x^{2}-4\right|, y \geq 1\right\}$ is
The area of the region enclosed by the curve $f(x)=\max \{\sin x, \cos x\},-\pi \leq x \leq \pi$ and the $x$-axis is
The area of the region enclosed by the curve $y=x^{3}$ and its tangent at the point $(-1,-1)$ is :
Area of the region $\left\{(x, y): x^{2}+(y-2)^{2} \leq 4, x^{2} \geq 2 y\right\}$ is
The area of the region $\left\{(x, y): x^{2} \leq y \leq 8-x^{2}, y \leq 7\right\}$ is :
The area bounded by the curves $y=|x-1|+|x-2|$ and $y=3$ is equal to :
The area of the region given by $\{ (x,y):xy \le 8,1 \le y \le {x^2}\} $ is :
The area of the region $A = \left\{ {(x,y):\left| {\cos x - \sin x} \right| \le y \le \sin x,0 \le x \le {\pi \over 2}} \right\}$ is
Let $\Delta$ be the area of the region $\left\{ {(x,y) \in {R^2}:{x^2} + {y^2} \le 21,{y^2} \le 4x,x \ge 1} \right\}$. Then ${1 \over 2}\left( {\Delta - 21{{\sin }^{ - 1}}{2 \over {\sqrt 7 }}} \right)$ is equal to
Let $[x]$ denote the greatest integer $\le x$. Consider the function $f(x) = \max \left\{ {{x^2},1 + [x]} \right\}$. Then the value of the integral $\int\limits_0^2 {f(x)dx} $ is
Let $A=\left\{(x, y) \in \mathbb{R}^{2}: y \geq 0,2 x \leq y \leq \sqrt{4-(x-1)^{2}}\right\}$ and
$
B=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: 0 \leq y \leq \min \left\{2 x, \sqrt{4-(x-1)^{2}}\right\}\right\} \text {. }
$.
Then the ratio of the area of A to the area of B is
The area enclosed by the curves ${y^2} + 4x = 4$ and $y - 2x = 2$ is :
Explanation:
$ \begin{aligned} & y^2=\frac{3 x}{2}, x+y=3, y=0 \\\\ & 2 y^2=3(3-y) \\\\ & 2 y^2+3 y-9=0 \\\\ & 2 y^2-3 y+6 y-9=0 \\\\ & (2 y-3)(y+2)=0 ; y=3 / 2 \\\\ & \text { Area }\left(\int_0^{\frac{3}{2}}\left(x_R-x_2\right) d y\right)-A_1 \\\\ & =\int_0^{\frac{3}{2}}\left((3-y)-\frac{2 y^2}{3}\right) d y-\frac{\pi}{8}(2) \\\\ & A=\left(3 y-\frac{y^2}{2}-\frac{2 y^3}{9}\right)_0^{\frac{3}{2}}-\frac{\pi}{4} \\\\ & 4 \mathrm{~A}+\pi=4\left[\frac{9}{2}-\frac{9}{8}-\frac{3}{4}\right]=\frac{21}{2}=10.50 \\\\ & \therefore 4(4 \mathrm{~A}+\pi)=42 \end{aligned} $
If A is the area in the first quadrant enclosed by the curve $\mathrm{C: 2 x^{2}-y+1=0}$, the tangent to $\mathrm{C}$ at the point $(1,3)$ and the line $\mathrm{x}+\mathrm{y}=1$, then the value of $60 \mathrm{~A}$ is _________.
Explanation:
$ y=2 x^2+1 $
Tangent at (1, 3) : $y-3=4(x-1)$
$ y=4 x-1 $
$\mathrm{A}=\int\limits_0^1\left(2 \mathrm{x}^2+1\right) \mathrm{dx}-$ area of $(\Delta \mathrm{QOT})-$ area of $(\Delta \mathrm{PQR})+$ area of $(\Delta \mathrm{QRS})$
$ \mathrm{A}=\left(\frac{2}{3}+1\right)-\frac{1}{2}-\frac{9}{8}+\frac{9}{40}=\frac{16}{60} $
$ \therefore $ $ 60 A=16 $
If the area of the region $\left\{(x, \mathrm{y}):\left|x^{2}-2\right| \leq y \leq x\right\}$ is $\mathrm{A}$, then $6 \mathrm{A}+16 \sqrt{2}$ is equal to __________.
Explanation:
On solving, $\left|x^2-2\right|=y$ and $y=x$, we get $(1,1)$ and $(2,2)$
$ \begin{aligned} & A=\int\limits_1^{\sqrt{2}}\left(x-\left(2-x^2\right)\right) d x+\int\limits_{\sqrt{2}}^2\left(x-\left(x^2-2\right)\right) d x \\\\ & A=\left[\frac{x^2}{2}-2 x+\frac{x^3}{3}\right]_1^{\sqrt{2}}+\left[\frac{x^2}{2}-\frac{x^3}{3}+2 x\right]_{\sqrt{2}}^2 \end{aligned} $
$ \begin{aligned} A=\left(1-2 \sqrt{2}+\frac{2 \sqrt{2}}{3}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)+ & \left(2-\frac{8}{3}+4\right) \\ & -\left(1-\frac{2 \sqrt{2}}{3}+2 \sqrt{2}\right) \end{aligned} $
$ \begin{aligned} & A=-4 \sqrt{2}+\frac{4 \sqrt{2}}{3}+\frac{7}{6}+\frac{10}{3} \\\\ & A=\frac{-8 \sqrt{2}}{3}+\frac{9}{2} \Rightarrow A=\frac{-16 \sqrt{2}+27}{6} \\\\ & 6 A=-16 \sqrt{2}+27 \\\\ & 6 A+16 \sqrt{2}=27 \end{aligned} $
Let $y = p(x)$ be the parabola passing through the points $( - 1,0),(0,1)$ and $(1,0)$. If the area of the region $\{ (x,y):{(x + 1)^2} + {(y - 1)^2} \le 1,y \le p(x)\} $ is A, then $12(\pi - 4A)$ is equal to ___________.
Explanation:
Now, to find the area of the region
$ \left\{(x, y) ;(x+1)^2+(y-1)^2 \leq 1, y \leq p(x)\right\} $
Thus, area of shaded region is
$ \begin{aligned} & A =\int\limits_{-1}^0\left\{\left(1-x^2\right)-\left(1-\sqrt{1-(x+1)^2}\right\} d x\right. \\\\ & =\int\limits_{-1}^0\left\{-x^2+\sqrt{1-(x+1)^2}\right\} d x \\\\ & =\int\limits_{-1}^0-x^2 d x+\int_{-1}^0 \sqrt{1-(x+1)^2} d x \\\\ & =-\left(\frac{x^3}{3}\right)_{-1}^0+\left[\frac{\sqrt{1-(x+1)^2}}{2}+\frac{1}{2} \sin ^{-1}(x+1)\right]_{-1}^0 \\\\ & =-\frac{1}{3}+\left[\frac{1}{2}+\frac{\pi}{4}-\frac{1}{2}\right] \\\\ & A =\frac{\pi}{4}-\frac{1}{3} \end{aligned} $
Now, $12(\pi-4 A)$
$ \begin{aligned} & =12\left[\pi-4\left(\frac{\pi}{4}-\frac{1}{3}\right)\right] \\\\ & =12\left(\pi-\pi+\frac{4}{3}\right)=16 \end{aligned} $
Let the area enclosed by the lines $x+y=2, \mathrm{y}=0, x=0$ and the curve $f(x)=\min \left\{x^{2}+\frac{3}{4}, 1+[x]\right\}$ where $[x]$ denotes the greatest integer $\leq x$, be $\mathrm{A}$. Then the value of $12 \mathrm{~A}$ is _____________.
Explanation:
$ \text { Required area }=\left[\int\limits_0^{\frac{1}{2}}\left(x^2+\frac{3}{4}\right) d x\right]+\left[\frac{1}{2}\left(\frac{3}{2}+\frac{1}{2}\right) \times 1\right] $
$ \begin{aligned} & =\left[\frac{x^3}{3}+\frac{3 x}{4}\right]_0^{\frac{1}{2}}+1 \\\\ & =\frac{1}{24}+\frac{3}{8}-0+1=\frac{1+9+24}{24}=\frac{34}{24}=\frac{17}{12} \end{aligned} $
$ \text { So, } 12 \mathrm{~A}=12 \times \frac{17}{12}=17 $
If the area of the region $S=\left\{(x, y): 2 y-y^{2} \leq x^{2} \leq 2 y, x \geq y\right\}$ is equal to $\frac{n+2}{n+1}-\frac{\pi}{n-1}$, then the natural number $n$ is equal to ___________.
Explanation:
$ S=\left\{(x, y): 2 y-y^2 \leq x^2 \leq 2 y, x \geq y\right\} $
Here, we have three curves
$ \begin{aligned} &2 y-y^2 =x^2 ..........(i)\\\\ &x^2 =2 y ..........(2)\\\\ &\text {and}~~ x = y ...........(3) \end{aligned} $
From Eq. (i),
$ x^2+y^2-2 y=0 $
$\Rightarrow x^2+(y-1)^2=1$ is a circle with centre $(0,1)$ and radius is 1.
Intersection points of Equations (i) and (iii) are,
$ \begin{array}{rlrl} & 2 x-x^2 =x^2 \\\\ &\Rightarrow 2 x^2 =2 x \\\\ &\Rightarrow x(x-1) =0 \\\\ &\Rightarrow x =0,1 \\\\ & \text { When, } x=0 \text {, then } y=0 \\\\ & \text { and when } x=1 \text {, then } y=1 \end{array} $
$\therefore$ Required points are $(0,0)$ and $(1,1)$.
Also, intersection points of (ii) and (iii)
$ \begin{aligned} & \quad x^2=2 x \\\\ & \Rightarrow x(x-2)=0 \\\\ & \Rightarrow x=0,2 \\\\ & \text { When, } x=0 \text {, then } y=0 \\\\ & \text { and when } x=2 \text {, then } y=2 \\\\ & \therefore \text { Required points are }(0,0) \text { and }(2,2) \end{aligned} $
$\therefore$ Required area $=$ Area of triangle - Area of part I - Area of part II
$ =\frac{1}{2} \times 2 \times 2-\int\limits_0^2 \frac{x^2}{2}-\left(\frac{\pi}{4}-\frac{1}{2}\right) $
$ \begin{aligned} & =2-\frac{1}{2}\left(\frac{x^3}{3}\right)_0^2-\frac{\pi}{4}+\frac{1}{2} \\\\ & =2-\frac{1}{2} \times \frac{8}{3}-\frac{\pi}{4}+\frac{1}{2} \\\\ & =\frac{7}{6}-\frac{\pi}{4} \\\\ & =\frac{5+2}{5+1}-\frac{\pi}{5-1} \\\\ & \therefore n =5 \end{aligned} $
Let $A$ be the area bounded by the curve $y=x|x-3|$, the $x$-axis and the ordinates $x=-1$ and $x=2$. Then $12 A$ is equal to ____________.
Explanation:
$=\int\limits_{-1}^{2}\left|3 x-x^{2}\right|$
$A=\int\limits_{-1}^{0} x^{2}-3 x d x+\int\limits_{0}^{2} 3 x-x^{2} d x$
$\left.\left.=\frac{x^{3}}{3}-\frac{3 x^{2}}{2}\right]_{-1}^{0}+\frac{3 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{2}$
$=0-\left(\frac{-1}{3}-\frac{3}{2}\right)+\left(6-\frac{8}{3}\right)-0$
$=\frac{31}{6}$
$\therefore 12 A=62$
$\left\{(x, y):|2 x-1| \leq y \leq\left|x^{2}-x\right|, 0 \leq x \leq 1\right\}$ be $\mathrm{A}$.
Then $(6 \mathrm{~A}+11)^{2}$ is equal to
Explanation:
$ \begin{aligned} & x-x^{2}=2 x-1 \\\\ & x^{2}+x-1=0 \\\\ & x=\frac{-1+\sqrt{5}}{2} \end{aligned} $
Area $=2($ area of $B C E)$
$A=2 \int_{\frac{1}{2}}^{\frac{\sqrt{5}-1}{2}}\left(x-x^{2}\right)-(2 x-1) d x$
$=2 \int_{\frac{1}{2}}^{\frac{\sqrt{5}-1}{2}} (1-x-x^{2}) d x=\left.2\left(x-\frac{x^{2}}{2}-\frac{x^{3}}{3}\right)\right|_{\frac{1}{2}} ^{]_{5}^{\frac{\sqrt{5}-1}{2}}}$
$ \begin{aligned} =2\left\{\left(\frac{\sqrt{5}-1}{2}-\frac{1}{2}\right)-\left\{\left(\frac{\sqrt{5}-1}{2}\right)^{2}\right.\right. & \left.-\frac{1}{4}\right\} \frac{1}{2} \\\\ & \left.-\left[\left(\frac{\sqrt{5}-1}{2}\right)^{3}-\frac{1}{8}\right] \frac{1}{3}\right\} \end{aligned} $
$ A=\frac{-11+5 \sqrt{5}}{6} $
$\Rightarrow(6 A+11)^{2}=125$
Let for $x \in \mathbb{R}$,
$ f(x)=\frac{x+|x|}{2} \text { and } g(x)=\left\{\begin{array}{cc} x, & x<0 \\ x^{2}, & x \geq 0 \end{array}\right. \text {. } $
Then area bounded by the curve $y=(f \circ g)(x)$ and the lines $y=0,2 y-x=15$ is equal to __________.
Explanation:
$g(x)=\left[\begin{array}{cc}x^{2} & x \geq 0 \\ x & x<0\end{array}\right.$
$f(x)=f[g(x)]=\left[\begin{array}{cl}g(x), & g(x) \geq 0 \\ 0, & g(x)<0\end{array}\right.$
$\operatorname{fog}(x)=\left[\begin{array}{cc}x^{2}, & x \geq 0 \\ 0, & x<0\end{array}\right.$
$2 y-x=15$
$\mathrm{A}=\int_{0}^{3}\left(\frac{\mathrm{x}+15}{2}-\mathrm{x}^{2}\right) \mathrm{dx}+\frac{1}{2} \times \frac{15}{2} \times 15$
= $\frac{x^{2}}{4}+\frac{15 x}{2}-\left.\frac{x^{3}}{3}\right|_{0} ^{3}+\frac{225}{4}$
$=\frac{9}{4}+\frac{45}{2}-9+\frac{225}{4}=\frac{99-36+225}{4}$
$=\frac{288}{4}=72$
$\left\{(x, y): y \geq x^2, y \geq(1-x)^2, y \leq 2 x(1-x)\right\}$.
Then $540 \mathrm{~A}$ is equal to :
Explanation:
$y = {x^2}$ and $y = 2x(1 - x)$
$ \Rightarrow {x^2} = 2x - 2{x^2}$
$ \Rightarrow x = 0,x = {2 \over 3}$
Now,
$y = {(1 - x)^2}$ and $y = 2x(1 - x)$
$ \Rightarrow 1 + {x^2} - 2x = 2x - 2{x^2}$
$ \Rightarrow 3{x^2} - 4x + 1 = 0$
$x = 1,x = {1 \over 3}$
$\therefore$ $A = \int\limits_{{1 \over 3}}^{{2 \over 3}} {(2x - 2{x^2})dx - \left\{ {\int\limits_{{1 \over 3}}^{{1 \over 2}} {{{(1 - x)}^2}dx + \int\limits_{{1 \over 2}}^{{2 \over 3}} {{x^2}dx} } } \right\}} $
= $\left( {{x^2} - {{2{x^3}} \over 3}} \right)_{{1 \over 3}}^{{2 \over 3}} - \left\{ {\left( {{{{{\left( {x - 1} \right)}^3}} \over 3}} \right)_{{1 \over 3}}^{{1 \over 2}} + \left( {{{{x^3}} \over 3}} \right)_{{1 \over 2}}^{{2 \over 3}}} \right\}$
$ = {5 \over {108}}$
$\therefore$ $540A = 25$
Let $\alpha$ be the area of the larger region bounded by the curve $y^{2}=8 x$ and the lines $y=x$ and $x=2$, which lies in the first quadrant. Then the value of $3 \alpha$ is equal to ___________.
Explanation:
${A_1} = \int\limits_0^2 {2\sqrt 2 \sqrt x - xdx} $
$ = \left. {2\sqrt 2 \times {2 \over 3}\,.\,{x^{{3 \over 2}}} - {{{x^2}} \over 2}} \right]_0^2$
$ = {{4\sqrt 2 } \over 3} \times 2\sqrt 2 - 2$
$ = {{16} \over 3} - 2 = {{10} \over 3}$
${A_2} = \left. {\int\limits_2^8 {2\sqrt 2 \sqrt x - xdx = {{4\sqrt 2 } \over 3}\,.\,{x^{{3 \over 2}}} - {{{x^2}} \over 2}} } \right]_2^8$
$ = {{4\sqrt 2 } \over 3}(16\sqrt 2 - 2\sqrt 2 ) - 30$
$ = {{112} \over 3} - 30 = {{22} \over 3}$
${A_2} > {A_1} \Rightarrow 3\alpha = 22$
If the area enclosed by the parabolas $\mathrm{P_1:2y=5x^2}$ and $\mathrm{P_2:x^2-y+6=0}$ is equal to the area enclosed by $\mathrm{P_1}$ and $\mathrm{y=\alpha x,\alpha > 0}$, then $\alpha^3$ is equal to ____________.
Explanation:
$ \begin{aligned} & \text { Area between } P_{1} \text { and } P_{2} \quad \text { [Say } \left.A_{1}\right] \\\\ & =\int\limits_{-2}^{2}\left(x^{2}+6\right)-\frac{5}{2} x^{2} d x \\\\ & =2 \int\limits_{0}^{2}\left(6-\frac{3}{2} x^{2}\right) d x=2\left[6 x-\frac{x^{3}}{2}\right]_{0}^{2}=16 \end{aligned} $
$ \begin{aligned} & a x=\frac{5}{2} x^2 \Rightarrow x=0, \frac{2 a}{5} \\\\ & \text { Area between } P_1 \text { and } y=a x \quad \text { [Say } A_2 \text { ] } \\\\ & =\int\limits_0^{\frac{2 \alpha}{5}} a x-\frac{5}{2} x^2 d x \\\\ & \left.=\frac{a x^2}{2}-\frac{5}{6} x^3\right]_0^{\frac{2 a}{5}}: \frac{2 a^3}{75} \\\\ & A_1=A_2 \Rightarrow \frac{2 a^3}{75}=16 \\\\ & a^3=600 \end{aligned} $
If the area of the region bounded by the curves $y^2-2y=-x,x+y=0$ is A, then 8 A is equal to __________
Explanation:
$ \begin{aligned} & y^{2}-2 y=-x \\\\ & x+y=0 \end{aligned} $
Area $=\int_{0}^{3}\left(2 y-y^{2}\right)-(-y) d y$
$ =\int_{0}^{3}\left(3 y-y^{2}\right) d y $
$ \begin{aligned} & \left.=\frac{3 y^{2}}{2}-\frac{y^{3}}{3}\right]_{0}^{3} \\\\ & =\frac{27}{2}-9 \\\\ & =\frac{27-18}{2}=\frac{9}{2}=A \end{aligned} $
$8 A=\frac{9}{2} \times 8=36$ sq. units
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
