Application of Integration

$\mathrm{f}(\mathrm{t})=\left\{\begin{array}{ccc} 2 & ; & \mathrm{t}=1 \\ 4-2 \mathrm{t} & ; & 1<\mathrm{t}<3 \\ -2 & ; & \mathrm{t}=3 \\ -8-2 \mathrm{t} & ; & 3<\mathrm{t}<5 \\ 2 & ; & \mathrm{t}=5 \\ 12-2 \mathrm{t} & ; & 5<\mathrm{t}<7 \\ -2 & ; & \mathrm{t}=7 \\ -16+2 \mathrm{t} & ; & 7<\mathrm{t}<9 \end{array}\right.$

$\begin{aligned} & g(x)=\int_\limits1^x f(t) d t ; g^{\prime}(x)=f(x) \\ & \text { for } x \in(1,8] \\ & g(x)=0 \Rightarrow x=3,5,7 \therefore \alpha=3 \\ & \beta=\lim _{x \rightarrow 1^{+}} \frac{g(x)}{x-1} \end{aligned}$

Apply L'pital

$\begin{aligned} & =\frac{g^{\prime}\left(1^{+}\right)}{1}=f\left(1^{+}\right) \\ & \beta=2 \\ & \therefore \alpha+\beta=5 \end{aligned}$

$f(x)$ is decreasing in $\left[0, \frac{1}{2 n}\right]$

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 $\left[\frac{1}{n}, 1\right]$



$\begin{aligned} \text { Area } & =\frac{1}{2} \times \frac{1}{2 n} \times n+\frac{1}{2} \times \frac{1}{2 n} \times n+\frac{1}{2} \times\left(1-\frac{1}{n}\right) \times n \\\\ 4 & =\frac{1}{4}+\frac{1}{4}+\frac{n-1}{2} \\\\ 4 & =\frac{1}{2}+\frac{n-1}{2} \\\\ 4 & =\frac{n}{2} \\\\ n & =8\end{aligned}$

$f(x) = {x^2} + {5 \over {12}}$

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$

${f_1}(x) = \int_0^x {{{(t - 1)}^1}{{(t - 2)}^2}{{(t - 3)}^3}{{(t - 4)}^4}....{{(t - 21)}^{21}}dt} $

$ \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, $2{m_1} + 3{n_1} + {m_1}{n_1} = 2 \times 6 + 3 \times 5 + 6 \times 5 = 57$

${f_1}(x) = \int_0^x {{{(t - 1)}^1}{{(t - 2)}^2}{{(t - 3)}^3}{{(t - 4)}^4}....{{(t - 21)}^{21}}dt} $

$ \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, $6{m_2} + 4{n_2} + 8{m_2}{n_2} = 6 + 0 + 0 = 6$

We have,

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$

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, $\mathop {\lim }\limits_{x \to 1} {{F(x)} \over {G(x)}} = {1 \over {14}}$

$\therefore$ ${{{1 \over 2}} \over {\left| {f\left( {{1 \over 2}} \right)} \right|}} = {1 \over {14}} \Rightarrow \left| {f\left( {{1 \over 2}} \right)} \right| = 7$

$\text { Given, } f(x)=\int_\limits x^{x^2+\frac{\pi}{6}} 2 \cos ^2 t d t \forall x \in \mathrm{R}$

$\text { As we know, if } \mathrm{I}(x)=\int_\limits{g(x)}^{h(x)} \phi(t) d t \text {, then }$

$\mathrm{I}^{\prime}(x)=\phi\{h(x)\} h^{\prime}(x)-\phi\{g(x)\} g^{\prime}(x)$

$\Rightarrow f(x)=2\left\{\cos \left(x^2+\frac{\pi}{6}\right)\right\}^2 \cdot \frac{d}{d x}\left(x^2+\frac{\pi}{6}\right) -2 \cos ^2 x \cdot \frac{d x}{d x}$

$\Rightarrow f(x)=4 x\left\{\cos \left(x^2+\frac{\pi}{6}\right)\right\}^2-2 \cos ^2 x$

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.