Area Under The Curves
Let the area of the region bounded by the curve $y=\max \{\sin x, \cos x\}$, lines $x=0, x=\frac{3 \pi}{2}$, and the $x$-axis be A . Then, $\mathrm{A}+\mathrm{A}^2$ is equal to $\_\_\_\_$。
Explanation:
$ \begin{aligned} & f(x)=\max (\sin x, \cos x) \\ & =\left\{\begin{array}{c} \cos x, x \in\left(0, \frac{\pi}{4}\right) \\ \sin x, x \in\left(\frac{\pi}{4}, \frac{5 \pi}{4}\right) \\ \cos x, x \in\left(\frac{5 \pi}{4}, \frac{3 \pi}{2}\right) \end{array}\right. \end{aligned} $
$ \begin{aligned} & \Rightarrow \int_0^{\frac{3 \pi}{4}}|f(x)| d x=\int_0^{\frac{\pi}{4}}(\cos x) d x+\int_{\frac{\pi}{4}}^{\frac{5 \pi}{4}}(\sin x) d x +\int_{\frac{5 \pi}{4}}^{\frac{3 \pi}{2}}(\cos x) d x \\ & =3 \Rightarrow \mathrm{~A}=3 \\ & A+A^2=12 \end{aligned} $
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a function such that $f(x)+3 f\left(\frac{\pi}{2}-x\right)=\sin x, x \in \mathbf{R}$. Let the maximum value of $f$ on $\mathbf{R}$ be $\alpha$. If the area of the region bounded by the curves $g(x)=x^2$ and $h(x)=\beta x^3, \beta>0$, is $\alpha^2$, then $30 \beta^3$ is equal to $\_\_\_\_$ .
Explanation:
$f(x)+3f\left(\frac{\pi}{2}-x\right)=\sin x \qquad \ldots(1)$
Now replace $x$ by $\frac{\pi}{2}-x$ in (1). This is a standard NCERT step to get a second equation in the same two terms.
$\Rightarrow f\left(\frac{\pi}{2}-x\right)+3f(x)=\cos x \qquad \ldots(2)$
Now we solve the two linear equations (1) and (2) to find $f(x)$.
Multiply (1) by $3$:
$3f(x)+9f\left(\frac{\pi}{2}-x\right)=3\sin x \qquad \ldots(3)$
Subtract (2) from (3):
$\big(3f(x)-3f(x)\big)+\big(9-1\big)f\left(\frac{\pi}{2}-x\right)=3\sin x-\cos x$
$8f\left(\frac{\pi}{2}-x\right)=3\sin x-\cos x$
Now put $x \rightarrow \frac{\pi}{2}-x$ in this result to get $f(x)$:
$8f(x)=3\cos x-\sin x$
$f(x)=\frac{1}{8}(3\cos x-\sin x)$
To find the maximum value, write $3\cos x-\sin x$ in the form $R\cos(x+\phi)$, where $R=\sqrt{3^2+(-1)^2}=\sqrt{10}$. So the maximum value of $3\cos x-\sin x$ is $\sqrt{10}$.
$f_{\max}=\frac{\sqrt{10}}{8}$
$\mathrm{y}=\mathrm{g}(\mathrm{x}) \& \mathrm{y}=\mathrm{h}(\mathrm{x})$ intersect as shown in the figure
The curves are $g(x)=x^2$ and $h(x)=\beta x^3$ with $\beta>0$. Their points of intersection come from $x^2=\beta x^3$:
$x^2=\beta x^3 \Rightarrow x^2(1-\beta x)=0 \Rightarrow x=0 \text{ and } x=\frac{1}{\beta}$
For $0
$\begin{aligned} & \therefore \text { Area bounded }=\Delta=\left|\int_0^{\frac{1}{\beta}}\left(\beta \mathrm{x}^3-\mathrm{x}^2\right) \mathrm{dx}\right| \\ & =\left|\left[\frac{\beta x^4}{4}-\frac{x^3}{3}\right]_0^{1/\beta}\right| \\ & =\left|\left(\frac{\beta}{4}\cdot\frac{1}{\beta^4}-\frac{1}{3}\cdot\frac{1}{\beta^3}\right)\right| \\ & =\left|\left(\frac{1}{4\beta^3}-\frac{1}{3\beta^3}\right)\right|=\frac{1}{12 \beta^3}=\alpha^2 \text { (given) } \\ & \Rightarrow \frac{1}{12\beta^3}=\left(\frac{\sqrt{10}}{8}\right)^2=\frac{10}{64}=\frac{5}{32} \\ & \Rightarrow \beta^3=\frac{1}{12}\cdot\frac{32}{5}=\frac{8}{15} \\ & \Rightarrow 30 \beta^3=30\cdot\frac{8}{15}=16\end{aligned}$
If the area of the region bounded by $16x^2 - 9y^2 = 144$ and $8x - 3y = 24$ is $A$, then $3(A + 6 \log_e(3))$ is equal to ________.
Explanation:
$16x^2-9y^2=144$
$16x^2-(8x-24)^2=144$
$16x^2-64(x-3)^2=144 \rightarrow x^2-4(x-3)^2=9$
$3x^2-24x+45 \rightarrow x^2-8x+15=0$
$x=3,5$
$\mathrm{Area}=\int\limits_{3}^{5}\sqrt{\frac{16x^2-144}{9}}-\frac{1}{2}\cdot 2\cdot \frac{16}{3}$
$=\frac{4}{3}\int\limits_{3}^{5}\sqrt{x^2-9}-\frac{16}{3}$
$=\frac{4}{3}\left(\frac{x}{2}\sqrt{x^2-9}-\frac{9}{2}\log_e\left(x+\sqrt{x^2-9}\right)\right)_{3}^{5}-\frac{16}{3}$
$=\frac{4}{3}\left(\frac{5}{2}\cdot 4-\frac{9}{2}\log_e 9-\frac{3}{2}\cdot 0+\frac{9}{2}\log_e 3\right)-\frac{16}{3}$
$\mathrm{Area}=8-6\log_e 3=A$
$\therefore A+6\ell n3=8$
$\Rightarrow 3(A+6\ell n3)=24$
Explanation:
$0 \leq 9 x \leq y^2 ~\&~ y \geq 3 x-6$
$\begin{aligned} & A=\text { Required Area }=\left[\int_0^1(-3 \sqrt{x}) d x-\int_0^1(3 x-6) d x\right] \\ & A=-\left.3\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)\right|_0 ^1-\left.\left(\frac{3 x^2}{2}-6 x\right)\right|_0 ^1 \\ & A=-2[1-0]\left[\frac{3}{2}-6\right] \\ & A=-2-\frac{3}{2}+6=\frac{5}{2} \text { Sq. unit } \\ & \therefore 6 A=6 \times \frac{5}{2}=15 \end{aligned}$
If the area of the region $\{(x, y):|x-5| \leq y \leq 4 \sqrt{x}\}$ is $A$, then $3 A$ is equal to _________.
Explanation:
$\begin{aligned} & \text { Area }=\int_1^{25} 4 \sqrt{x} d x-\frac{1}{2} \times(5-1) \cdot 4-\frac{1}{2} \times(25-5) \times 20 \\ & =4\left[\frac{2}{3} x^{\frac{3}{2}}\right]_1^{25}-8-200=\frac{368}{3} \\ & \Rightarrow 3 A=368 \end{aligned}$
The area of the region bounded by the curve $y=\max \{|x|, x|x-2|\}$, the $x$-axis and the lines $x=-2$ and $x=4$ is equal to__________
Explanation:
Area =
$\begin{aligned} & \frac{1}{2} \times 2 \times 2+\int_0^1\left(2 x-x^2\right) d x+\int_1^3 x d x+\int_3^4\left(x^2-2 x\right) d x \\ & =2+\left[\frac{2 x^2}{2}-\frac{x^3}{3}\right]_0^1+\left[\frac{x^2}{2}\right]_1^3+\left[\frac{x^3}{3}-\frac{2 x^2}{2}\right]_3^4 \end{aligned}$
= 12 square units
If the area of the region $\left\{(x, y):\left|4-x^2\right| \leq y \leq x^2, y \leq 4, x \geq 0\right\}$ is $\left(\frac{80 \sqrt{2}}{\alpha}-\beta\right), \alpha, \beta \in \mathbf{N}$, then $\alpha+\beta$ is equal to _________.
Explanation:
$\begin{aligned} &\text { Area = }\\ &\int_{\sqrt{2}}^2\left(x^2-\left(4-x^2\right)\right) d x+(2 \sqrt{2}-2) \times 4-\int_2^{2 \sqrt{2}}\left(x^2-4\right) d x \end{aligned}$
$\begin{aligned} & =\left[\frac{2 x^3}{3}-4 x\right]_{\sqrt{2}}^2+8 \sqrt{2}-8-\left[\frac{x^3}{3}-4 x\right]_2^{2 \sqrt{2}} \\ & =\frac{40 \sqrt{2}}{3}-16 \\ & \Rightarrow \alpha=6, \beta=16 \Rightarrow \alpha+\beta=22 \end{aligned}$
If the area of the larger portion bounded between the curves $x^2+y^2=25$ and $\mathrm{y}=|\mathrm{x}-1|$ is $\frac{1}{4}(\mathrm{~b} \pi+\mathrm{c}), \mathrm{b}, \mathrm{c} \in N$, then $\mathrm{b}+\mathrm{c}$ is equal to _________
Explanation:
$\begin{aligned} & \mathrm{x}^2+\mathrm{y}^2=5 \\ & \mathrm{x}^2+(\mathrm{x}-1)^2=25 \Rightarrow \mathrm{x}=4 \\ & \mathrm{x}^2+(-\mathrm{x}+1)^2=5 \Rightarrow \mathrm{x}=-3 \\ & \mathrm{~A}=25 \pi-\int_{-3}^4 \sqrt{25-\mathrm{x}^2} \mathrm{dx}+\frac{1}{2} \times 4 \times 4+\frac{1}{2} \times 3 \times 3 \\ & \mathrm{~A}=25 \pi+\frac{25}{2}-\left[\frac{\mathrm{x}}{2} \sqrt{25-\mathrm{x}^2}+\frac{25}{2} \sin ^{-1} \frac{\mathrm{x}}{5}\right]_{-3}^4 \\ & \mathrm{~A}=25 \pi+\frac{25}{2}-\left[6+\frac{25}{2} \sin ^{-1} \frac{4}{5}+6+\frac{25}{2} \sin ^{-1} \frac{3}{5}\right] \\ & \mathrm{A}=25 \pi+\frac{1}{2}-\frac{25}{2} \cdot \frac{\pi}{2} \\ & \mathrm{~A}=\frac{75 \pi}{4}+\frac{1}{2} \\ & \mathrm{~A}=\frac{1}{4}(75 \pi+2) \\ & \mathrm{b}=75, \mathrm{c}=2 \\ & \mathrm{~b}+\mathrm{c}=75+2=77 \end{aligned}$
Let the area of the region enclosed by the curve $y=\min \{\sin x, \cos x\}$ and the $x$ axis between $x=-\pi$ to $x=\pi$ be $A$. Then $A^2$ is equal to __________.
Explanation:
$y=f(x)=\min \{\sin x, \cos x\}$
$\begin{aligned} & A=-\int_\limits{-\pi}^{\frac{-3 \pi}{4}} \cos x d x-\int_\limits{\frac{-3 \pi}{4}}^0 \sin x d x+\int_\limits0^{\frac{\pi}{4}} \sin x d x+\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x d x \\ & -\int_\limits{\frac{\pi}{2}}^\pi \cos x d x \\ & A=4 \\ & A^2=16 \end{aligned}$
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:
$\begin{aligned} & A=\int_\limits0^1\left[\left(8-4 y^2\right)-\left(-2 y^2\right)\right] d y+ \\ & \int_\limits1^{3 / 2}\left[\left(8-4 y^2\right)-(2 y-4)\right] d y \\ & =\left[8 y-\frac{2 y^3}{3}\right]_0^1+\left[12 y-y^2-\frac{4 y^3}{3}\right]_1^{3 / 2}=\frac{107}{12}=\frac{m}{n} \\ & \therefore m+n=119 \end{aligned}$
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
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, $\frac{R_{1}+R_{2}}{R_{1}}=\frac{5 / 12}{1 / 48} \Rightarrow 1+\frac{R_{2}}{R_{1}}=20$
$\Rightarrow \frac{R_{2}}{R_{1}}=19$
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$