Application of Integration
Consider the curve $C_1$ given by
$ y=e^{-x} \quad \text { for } x \in[0,10 \pi], $
and the curve $C_2$ given by
$ y=e^{-x}(\sin x+\cos x) \quad \text { for } x \in[0,10 \pi] . $
Let $n$ be the total number of points of intersection of the curves $C_1$ and $C_2$.
Suppose that $\alpha_1, \alpha_2, \ldots, \alpha_n \in[0,10 \pi]$ are the $x$-coordinates of the points of intersection of the curves $C_1$ and $C_2$ such that
$ \alpha_1<\alpha_2<\cdots<\alpha_n . $
Let $\beta$ be the area of the region enclosed between the curves $C_1, C_2$, and the lines $x=\alpha_1$ and $x=\alpha_4$. Then the value of
$ -\frac{1}{\pi} \log _e\left(\beta-2 e^{-\frac{\pi}{2}}\right) $
is $\_\_\_\_$ .
Explanation:
$\alpha_1=0 \& \alpha_4=\frac{5 \pi}{2}$
$\begin{aligned} & \beta=\int_0^{\frac{5 \pi}{2}}\left|\left(e^{-x}(\sin x+\cos x)-e^{-x}\right)\right| d x \\ & \beta=\int_0^{\frac{\pi}{2}}\left(e^{-x}(\sin x+\cos x)-e^{-x}\right) d x+\int_{\frac{\pi}{2}}^{2 \pi}\left(e^{-x}-e^{-x}(\sin x+\cos x)\right) d x+\int_{2 \pi}^{\frac{5 \pi}{2}}\left(e^{-x}(\sin x+\cos x)-e^{-x}\right) d x \\ & \beta=2 e^{\frac{-\pi}{2}}+e^{\frac{-5 \pi}{2}} \\ & \text { so } \frac{-1}{\pi} \log \left(\beta-2 e^{\frac{-\pi}{2}}\right)=\frac{5}{2}=2.50\end{aligned}$
Consider the ellipses given by
$ x^2+4 y^2=1 \quad \text { and } \quad 4 x^2+y^2=1 . $
If $\alpha$ is the area of the common region that lies inside both the given ellipses, then the value of $\cot \alpha$ is $\_\_\_\_$ .
Explanation:
$ A=\int_0^{\frac{1}{\sqrt{5}}}\left(\frac{\sqrt{1-x^2}}{2}-x\right) d x $
Total required area
$ \begin{aligned} & \alpha=8 \mathrm{~A}=8\left[\int_0^{\frac{1}{\sqrt{5}}}\left(\frac{\sqrt{1-\mathrm{x}^2}}{2}-\mathrm{x}\right) \mathrm{dx}\right] \\ & \alpha=8\left[\frac{\mathrm{x}}{4} \sqrt{1-\mathrm{x}^2}+\frac{1}{4} \sin ^{-1} \mathrm{x}-\frac{\mathrm{x}^2}{2}\right]_0^{\frac{1}{\sqrt{5}}} \\ & \alpha=8\left[\frac{1}{4 \sqrt{5}} \sqrt{\frac{4}{5}}+\frac{1}{4} \sin ^{-1} \frac{1}{\sqrt{5}}-\frac{1}{10}\right] \end{aligned} $
$\begin{aligned} & \alpha=2 \sin ^{-1} \frac{1}{\sqrt{5}} \\ & \cot \alpha=\cot \left(2 \sin ^{-1} \frac{1}{\sqrt{5}}\right) \\ & =\cot \left(2 \tan ^{-1} \frac{1}{2}\right) \\ & =\cot \left(\tan ^{-1}\left(\frac{2 \times \frac{1}{2}}{1-\frac{1}{4}}\right)\right) \\ & =\cot \left(\tan ^{-1}\left(\frac{4}{3}\right)\right) \\ & \cot \alpha=\frac{3}{4}=0.75\end{aligned}$
Let the function $f:[1, \infty) \rightarrow \mathbb{R}$ be defined by
$ f(t)=\left\{\begin{array}{cc} (-1)^{n+1} 2, & \text { if } t=2 n-1, n \in \mathbb{N}, \\ \frac{(2 n+1-t)}{2} f(2 n-1)+\frac{(t-(2 n-1))}{2} f(2 n+1), & \text { if } 2 n-1 < t < 2 n+1, n \in \mathbb{N} . \end{array}\right. $
Define $g(x)=\int_1^x f(t) d t, x \in(1, \infty)$. Let $\alpha$ denote the number of solutions of the equation $g(x)=0$ in the interval $(1,8]$ and $\beta=\lim \limits_{x \rightarrow l+} \frac{g(x)}{x-1}$.
Then the value of $\alpha+\beta$ is equal to _______.
Explanation:
$\mathrm{f}(\mathrm{t})=\left\{\begin{array}{ccc} 2 & ; & \mathrm{t}=1 \\ 4-2 \mathrm{t} & ; & 1<\mathrm{t}<3 \\ -2 & ; & \mathrm{t}=3 \\ -8-2 \mathrm{t} & ; & 3<\mathrm{t}<5 \\ 2 & ; & \mathrm{t}=5 \\ 12-2 \mathrm{t} & ; & 5<\mathrm{t}<7 \\ -2 & ; & \mathrm{t}=7 \\ -16+2 \mathrm{t} & ; & 7<\mathrm{t}<9 \end{array}\right.$
$\begin{aligned} & g(x)=\int_\limits1^x f(t) d t ; g^{\prime}(x)=f(x) \\ & \text { for } x \in(1,8] \\ & g(x)=0 \Rightarrow x=3,5,7 \therefore \alpha=3 \\ & \beta=\lim _{x \rightarrow 1^{+}} \frac{g(x)}{x-1} \end{aligned}$
Apply L'pital
$\begin{aligned} & =\frac{g^{\prime}\left(1^{+}\right)}{1}=f\left(1^{+}\right) \\ & \beta=2 \\ & \therefore \alpha+\beta=5 \end{aligned}$
$ f(x)= \begin{cases}n(1-2 n x) & \text { if } 0 \leq x \leq \frac{1}{2 n} \\\\ 2 n(2 n x-1) & \text { if } \frac{1}{2 n} \leq x \leq \frac{3}{4 n} \\\\ 4 n(1-n x) & \text { if } \frac{3}{4 n} \leq x \leq \frac{1}{n} \\\\ \frac{n}{n-1}(n x-1) & \text { if } \frac{1}{n} \leq x \leq 1\end{cases} $
If $n$ is such that the area of the region bounded by the curves $x=0, x=1, y=0$ and $y=f(x)$ is 4 , then the maximum value of the function $f$ is :
Explanation:
increasing in $\left[\frac{1}{2 n}, \frac{3}{4 n}\right]$
decreasing in $\left[\frac{3}{4 n}, \frac{1}{n}\right]$
increasing in $\left[\frac{1}{n}, 1\right]$
$\begin{aligned} \text { Area } & =\frac{1}{2} \times \frac{1}{2 n} \times n+\frac{1}{2} \times \frac{1}{2 n} \times n+\frac{1}{2} \times\left(1-\frac{1}{n}\right) \times n \\\\ 4 & =\frac{1}{4}+\frac{1}{4}+\frac{n-1}{2} \\\\ 4 & =\frac{1}{2}+\frac{n-1}{2} \\\\ 4 & =\frac{n}{2} \\\\ n & =8\end{aligned}$
$ f(x)=x^{2}+\frac{5}{12} \quad \text { and } \quad g(x)= \begin{cases}2\left(1-\frac{4|x|}{3}\right), & |x| \leq \frac{3}{4} \\ 0, & |x|>\frac{3}{4}\end{cases} $
If $\alpha$ is the area of the region
$ \left\{(x, y) \in \mathbb{R} \times \mathbb{R}:|x| \leq \frac{3}{4}, 0 \leq y \leq \min \{f(x), g(x)\}\right\}, $
then the value of $9 \alpha$ is
Explanation:
$f(x) = {x^2} + {5 \over {12}}$
This represent upward parabola.
$g(x) = 2\left( {1 - {{4|x|} \over 3}} \right)$
At $x = 0$
$g(0) = 2$
At $x = {3 \over 4}$
$g\left( {{3 \over 4}} \right) = 0$
$\therefore$ Graph is
Intersection point of f(x) and g(x) at first quadrant,
$f(x) = g(x)$
$ \Rightarrow {x^2} + {5 \over {12}} = 2\left( {1 - {{4x} \over 3}} \right)$ [ $|x| = x$ as in first quadrant $x > 0$ ]
$ \Rightarrow 12{x^2} + 5 = 24 - 32x$
$ \Rightarrow 12{x^2} + 32x - 19 = 0$
$ \Rightarrow 12{x^2} + 38x - 6x - 19 = 0$
$ \Rightarrow 2x(6x + 19) - 1(6x + 19) = 0$
$ \Rightarrow (2x - 1)(6x + 19) = 0$
$\therefore$ $x = {1 \over 2},\, - {{19} \over 6}$
In first quadrant $x = {1 \over 2}$
When $x = {1 \over 2}$,
$y = {\left( {{1 \over 2}} \right)^2} + {5 \over {12}} = {2 \over 3}$
$\therefore$ Area $A = 2$ ( $\int\limits_0^{{1 \over 2}} {\left( {{x^2} + {5 \over {12}}} \right)dx + } $ Area of $\Delta ABC$ )
$ = 2\left[ {\int\limits_0^{{1 \over 2}} {\left( {{x^2} + {5 \over {12}}} \right)dx + {1 \over 2} \times \left( {{3 \over 4} - {1 \over 2}} \right) \times {2 \over 3}} } \right]$
$ = 2\left[ {\left[ {{{{x^3}} \over 3} + {{5x} \over {12}}} \right]_0^{{1 \over 2}} + {1 \over 2} \times {1 \over 4} \times {2 \over 3}} \right]$
$ = 2\left[ {\left( {{1 \over {24}} + {5 \over {12}} \times {1 \over 2}} \right) + {1 \over {12}}} \right]$
$ = 2\left[ {{1 \over {24}} + {5 \over {24}} + {2 \over {24}}} \right]$
$ = 2\left[ {{{1 + 5 + 2} \over {24}}} \right]$
$ = 2 \times {8 \over {24}}$
$ = {2 \over 3} = \alpha $
$\therefore$ $ = 9\alpha = {2 \over 3} \times 9 = 6$
The value of $2{m_1} + 3{n_1} + {m_1}{n_1}$ is ___________.
Explanation:
$ \Rightarrow {f_1}'(x) = (x - 1){(x - 2)^2}{(x - 3)^3}{(x - 4)^4}....{(x - 21)^{21}}$
Sign Scheme for f1'(x)
From sign scheme of f1'(x), we observe that f(x) has local minima at x = 4k + 1, k$\in$W i.e. f1'(x) changes sign from $-$ve to + ve which are x = 1, 5, 9, 13, 17, 21 and f(x) has local maxima at x = 4k + 3, k$\in$W i.e. f1'(x) changes sign from + ve to $-$ ve, which are x = 3, 7, 11, 15, 19.
So, m1 = number of local minima points = 6
and n1 = number of local maxima points = 5
Hence, $2{m_1} + 3{n_1} + {m_1}{n_1} = 2 \times 6 + 3 \times 5 + 6 \times 5 = 57$
The value of $6{m_2} + 4{n_2} + 8{m_2}{n_2}$ is ___________.
Explanation:
$ \Rightarrow {f_1}'(x) = (x - 1){(x - 2)^2}{(x - 3)^3}{(x - 4)^4}....{(x - 21)^{21}}$
Sign Scheme for f1'(x)
From sign scheme of f1'(x), we observe that f(x) has local minima at x = 4k + 1, k$\in$W i.e. f1'(x) changes sign from $-$ve to + ve which are x = 1, 5, 9, 13, 17, 21 and f(x) has local maxima at x = 4k + 3, k$\in$W i.e. f1'(x) changes sign from + ve to $-$ ve, which are x = 3, 7, 11, 15, 19.
So, m1 = number of local minima points = 6
and n1 = number of local maxima points = 5
Hence, $2{m_1} + 3{n_1} + {m_1}{n_1} = 2 \times 6 + 3 \times 5 + 6 \times 5 = 57$
Also, ${f_2}(x) = 98{(x - 1)^{50}} - 600{(x - 1)^{49}} + 2450$
$ \Rightarrow {f_2}'(x) = 98 \times 50{(x - 1)^{49}} - 600 \times 49{(x - 1)^{48}}$
$ \Rightarrow {f_2}'(x) = 98 \times 50{(x - 1)^{48}}(x - 1 - 6)$
$ \Rightarrow {f_2}'(x) = 98 \times 50{(x - 1)^{48}}(x - 7)$
Clearly, m2 = 1 and n2 = 0
So, $6{m_2} + 4{n_2} + 8{m_2}{n_2} = 6 + 0 + 0 = 6$
Explanation:
y = xn, n > 1
$ \because $ P(0, 0) Q(1, 1) and R(2, 0) are vertices of $\Delta $PQR.
$ \therefore $ Area of shaded region = 30% of area of $\Delta $PQR
$ \Rightarrow \int_0^1 {(x - {x^n})dx = {{30} \over {100}} \times {1 \over 2} \times 2 \times 1} $
$ \Rightarrow \left[ {{{{x^2}} \over 2} - {{{x^{n + 1}}} \over {n + 1}}} \right]_0^1 = {3 \over {10}}$
$ \Rightarrow \left( {{1 \over 2} - {1 \over {n + 1}}} \right) = {3 \over {10}}$
$ \Rightarrow {1 \over {n + 1}} = {1 \over 2} - {3 \over {10}} = {2 \over {10}} = {1 \over 5}$
$ \Rightarrow n + 1 = 5 \Rightarrow n = 4$
Explanation:
Here, $\mathop {\lim }\limits_{x \to 1} {{F(x)} \over {G(x)}} = {1 \over {14}}$
$ \Rightarrow \mathop {\lim }\limits_{x \to 1} {{F'(x)} \over {G'(x)}} = {1 \over {14}}$ [using L' Hospital's rule] ....... (i)
As $F(x) = \int_{ - 1}^x {f(t)dt} $
$ \Rightarrow F'(x) = f(x)$ ...... (ii)
and $G(x) = \int_{ - 1}^x {t|f\{ f(t)\} |dt} $
$ \Rightarrow G'(x) = x|f\{ f(x)\} |$ ...... (iii)
$\therefore$ $\mathop {\lim }\limits_{x \to 1} {{F(x)} \over {G(x)}} = \mathop {\lim }\limits_{x \to 1} {{F'(x)} \over {G'(x)}} = \mathop {\lim }\limits_{x \to 1} {{f(x)} \over {x|f\{ f(x)\} |}}$
$ = {{f(1)} \over {1|f\{ f(1)\} |}} = {{1/2} \over {|f(1/2)|}}$ ....... (iv)
Given, $\mathop {\lim }\limits_{x \to 1} {{F(x)} \over {G(x)}} = {1 \over {14}}$
$\therefore$ ${{{1 \over 2}} \over {\left| {f\left( {{1 \over 2}} \right)} \right|}} = {1 \over {14}} \Rightarrow \left| {f\left( {{1 \over 2}} \right)} \right| = 7$
Explanation:
$\text { Given, } f(x)=\int_\limits x^{x^2+\frac{\pi}{6}} 2 \cos ^2 t d t \forall x \in \mathrm{R}$
$\text { As we know, if } \mathrm{I}(x)=\int_\limits{g(x)}^{h(x)} \phi(t) d t \text {, then }$
$\mathrm{I}^{\prime}(x)=\phi\{h(x)\} h^{\prime}(x)-\phi\{g(x)\} g^{\prime}(x)$
$\Rightarrow f(x)=2\left\{\cos \left(x^2+\frac{\pi}{6}\right)\right\}^2 \cdot \frac{d}{d x}\left(x^2+\frac{\pi}{6}\right) -2 \cos ^2 x \cdot \frac{d x}{d x}$
$\Rightarrow f(x)=4 x\left\{\cos \left(x^2+\frac{\pi}{6}\right)\right\}^2-2 \cos ^2 x$
Putting $x=a$ in the above equation, we get
$f(a)=4 a\left\{\cos \left(a^2+\frac{\pi}{6}\right)\right\}^2-2 \cos ^2 a$
Also, the area of the region bounded by $x=0$,
$\begin{aligned} & y=0, y=f(x) \text { and } x=a \text { is } \int_0^a f(x) d x \\ & \Rightarrow f(a)+2=\int_\limits0^a f(x) d x \\ & \Rightarrow 4 a\left\{\cos \left(a^2+\frac{\pi}{6}\right)\right\}^2-2 \cos ^2 a+2=\int_\limits0^a f(x) d x \end{aligned}$
Differentiating above equation w.r.t. a, we get
$\begin{aligned} \Rightarrow & -4 a \cdot 2 \cos \left(a^2+\frac{\pi}{6}\right) \cdot \sin \left(a^2+\frac{\pi}{6}\right) \\ & 2 a+4\left\{\cos \left(a^2+\frac{\pi}{6}\right)\right\}^2 \\ & -4 \cos a(-\sin a)=f(a) \\ \Rightarrow & -8 a^2 \sin \left(2 a^2+\frac{\pi}{3}\right)+4\left\{\cos \left(a^2+\frac{\pi}{6}\right)\right\}^2 \\ & +2 \sin 2 a=f(a)\{\because 2 \sin x \cos x=\sin 2 x\} \end{aligned}$
Putting $a=0$ in the above equation, we get.
$\begin{aligned} & 0+4 \cos ^2\left(\frac{\pi}{6}\right)+2 \sin (0)=f(0) \\ \Rightarrow & f(0)=4\left(\frac{\sqrt{3}}{2}\right)^2 \quad\left\{\because \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}\right\} . \\ \Rightarrow & f(0)=3 \end{aligned}$
(i) Use if $I(x)=\int_\limits{g(x)}^{h(x)} \phi(t) d t$, then $\mathrm{I}^{\prime}(x)=\phi\{h(x)\} h^{\prime}(x)-\phi\left\{g(x) g^{\prime}(x)\right.$
(ii) Use the area of the region bounded by $x=0, y=0, y=g(x)$ and $x=k$ is $\int_\limits0^k g(x) d x$
(iii) Use the product rule of differentiation for further simplification.
function and its maximum value occurs at a point $V$. $A$ is a point of intersection of $y=f(x)$ with $x$-axis and point $B$ is such that chord $AB$ subtends a right angle at $V$. Find the area enclosed by $f(x)$ and chord $AB$.
the region bounded by the $y$-axis and the curve $x{e^{ay}} = \sin $ by,
${{jr} \over b} \le y \le {{\left( {j + 1} \right)\pi } \over b}.$ Show that ${S_0},{S_1},{S_2},\,....,\,{S_n}$ are in
geometric progression. Also, find their sum for $a=-1$ and $b = \pi .$
Find the area of the region in the third quadrant bounded by the curves $x = - 2{y^2}$ and $y=f(x)$ lying on the left of the line $8x+1=0.$
Determine the area of the region bounded by the curves
$y = f\left( x \right),$ $x$-axes, $x=0$ and $x=1.$
lines $x=0,$ $y=0,$ and $x = {\pi \over 4}.$ Prove that for $n > 2,$
${A_n} + {A_{n - 2}} = {1 \over {n - 1}}$ and deduce ${1 \over {2n + 2}} < {A_n} < {1 \over {2n - 2}}.$
bounded by the parabolas $y = 4x - {x^2}$ and $y = {x^2} - x?$
$y = {2 \over {1 + {x^2}}}.$ Find the area.
intersects the curve ${y^2} + \int\limits_0^x {f\left( t \right)dt = 2!} $
$x = {1 \over 2},x = 2,y = \ln \,x$ and $y = {2^x}.$ Find the area of this region.
$\tan x,$ tangent drawn to $C$ at $x = {\pi \over 4}$ and the $x$-axis.
the ordinates at $x=2$ and $x=4$. If the ordinate at $x=a$ divides the area into two equal parts, find $a$.
hyperbola ${x^2} - {y^2} = 1$. Show that the area bounded by this hyperbola and the lines joining its centre to the points corresponding to ${t_1}$ and $-{t_1}$ is ${t_1}$.