Application of Integration

$ \text { Area }=\underset{\substack{\downarrow \\ \text { Trapezium APBC }}}{\frac{1}{2}\left(1+\frac{9}{4}\right)}+\underset{\substack{\downarrow \\ \text { Trapezium PDCE }}}{\frac{1}{4}\left(\frac{9}{4}+\frac{1}{4}\right)} \cdot 2-\int_1^4 \frac{\mathrm{dx}}{\mathrm{x}} $

$ \text { Area }=\frac{13}{8}+\frac{20}{8}-\log _e 4=\left(\frac{33}{8}-\log _e 4\right) $

Point of intersection of all curves is $(2,2 \sqrt{2})$

$\begin{aligned} & \text { Area }=\mathrm{A}_1+\mathrm{A}_2 \\ & \alpha \sqrt{2}=\int_\limits0^2 2 \sqrt{\mathrm{x}} \mathrm{dx}+\frac{1}{2} \times 3 \times 2 \sqrt{2} \\ & \alpha \sqrt{2}=2\left[\frac{\mathrm{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^2+3 \sqrt{2} \\ & \alpha \sqrt{2}=\frac{17 \sqrt{2}}{3} \\ & \alpha=\frac{17}{3} \end{aligned}$

Required area = Shaded region

On solving x + y = 2 and x = 3y, we get

$A \equiv \left( {{3 \over 2},{1 \over 2}} \right)$

On solving y = 0 and x + y = 2, we get

B ≡ (2, 0)

On solving x = ${9 \over 4}$ and x = 3y, we get

D ≡ $\left( {{9 \over 4},{3 \over 4}} \right)$

and C ≡ $\left( {{9 \over 4},0} \right)$

Required area = Area of ∆OCD − Area of ∆OBA

= ${1 \over 2} \times \left( {{9 \over 4} - 0} \right) \times {3 \over 4} - {1 \over 2} \times \left( {2 - 0} \right) \times {1 \over 2}$

= ${{27} \over {32}} - {1 \over 2}$

= ${{11} \over {32}}$

Given, ${{dy} \over {dx}} + \alpha y = x\,.\,{e^{\beta x}}$ which is a linear differential equation.

Integrating factor $(IF) = {e^{\int {\alpha dx} }} = {e^{\alpha x}}$

So, the solution is $y \times {e^{\alpha x}} = \int {x{e^{\beta x}}\,.\,{e^{\alpha x}}dx} $

$ \Rightarrow y \times {e^{\alpha x}} = \int {x{e^{(\alpha + \beta )x}}dx} $ .... (i)

Case (I) If $\alpha$ + $\beta$ = 0

From Eq. (i), we get

$ \Rightarrow y{e^{\alpha x}} = \int {x{e^{0.x}}dx = \int {xdx = {{{x^2}} \over 2} + C} } $ .... (ii)

Given, y(1) = 1 i.e. when x = 1, then y = 1

From Eq. (ii), we get

$1.{e^\alpha } = {1 \over 2} + C \Rightarrow C = {e^\alpha } - {1 \over 2}$

From Eq. (ii), we get

$y{e^{\alpha x}} = {{{x^2} - 1} \over 2} + {e^\alpha }$

For $\alpha$ = 1

$y{e^x} = {{{x^2} - 1} \over 2} + e \Rightarrow y = {{{x^2}} \over 2}{e^{ - x}} + \left( {e - {1 \over 2}} \right){e^{ - x}}$

Option (a) is correct.

Case (II) If $\alpha$ + $\beta$ $\ne$ 0

$ \Rightarrow y{e^{\alpha x}} = x.{{{e^{(\alpha + \beta )x}}} \over {(\alpha + \beta )}} - \int {1 \times {{{e^{(\alpha + \beta )x}}} \over {(\alpha + \beta )}}dx} $

$ \Rightarrow y{e^{\alpha x}} = x.{{{e^{(\alpha + \beta )x}}} \over {(\alpha + \beta )}} - {{{e^{(\alpha + \beta )x}}} \over {{{(\alpha + \beta )}^2}}} + {c_1}$

$ \Rightarrow y = {{x\,.\,{e^{\beta x}}} \over {(\alpha + \beta )}} - {{{e^{\beta x}}} \over {{{(\alpha + \beta )}^2}}} + {c_1}{e^{ - \alpha x}}$ (Cancelling e^$\alpha$x from both sides)

$ \Rightarrow y = {{{e^{\beta x}}} \over {\alpha + \beta }}\left( {x - {1 \over {\alpha + \beta }}} \right) + {c_1}{e^{ - \alpha x}}$ .... (iii)

Putting $\alpha$ = $\beta$ = 1 in Eq. (iii), we get

$y = {{{e^x}} \over 2}\left( {x - {1 \over 2}} \right) + {c_1}{e^{ - x}}$

Given, y(1) = 1

$\therefore$ $1 = {e \over 2} \times {1 \over 2} + {{{c_1}} \over e} \Rightarrow {c_1} = e - {{{e^2}} \over 4}$

So, $y = {{{e^x}} \over 2}\left( {x - {1 \over 2}} \right) + \left( {e - {{{e^2}} \over 4}} \right){e^{ - x}}$ $\to$ option (c) is correct.

The given functions f : R $ \to $ R and g : R $ \to $ R be defined by

$f(x) = {e^{x - 1}} - {e^{ - |x - 1|}}$

$ = \left| \matrix{ {e^{x - 1}} - {e^{1 - x}},\,x\, \ge \,1 \hfill \cr 0,\,x < 1 \hfill \cr} \right.$

and $g(x) = {1 \over 2}({e^{x - 1}} + {e^{1 - x}})$

For point of intersection of curves f(x) and g(x) put f(x) = g(x)

for $x\, \ge \,1,\,{e^{x - 1}} - {e^{1 - x}} = {1 \over 2}({e^{x - 1}} + {e^{1 - x}})$

$ \Rightarrow {e^{x - 1}} = 3{e^{1 - x}} \Rightarrow {e^{2x}} = 3{e^2}$

$ \Rightarrow x = {1 \over 2}\log _e^3 + 1$

So, required area is

$\int_0^{1/2\log _e^3 + 1} {(g(x) - f(x))dx} $

$ = \int_0^{1/2\log _e^3 + 1} {g(x)dx - \int_1^{1/2\log _e^3 + 1} {f(x)dx} } $

$ = {1 \over 2}\int_0^{1/2\log _e^3 + 1} {({e^{x - 1}} + {e^{1 - x}})dx - } \int_1^{1/2\log _e^3 + 1} {({e^{x - 1}} - {e^{1 - x}})dx} $



$ = {1 \over 2}[{e^{x - 1}} - {e^{1 - x}}]_0^{{1 \over 2}\log _e^3 + 1} - [{e^{x - 1}} + {e^{1 - x}}]_1^{{1 \over 2}\log _e^3 + 1}$

$ = {1 \over 2}\left[ {\sqrt 3 - {1 \over {\sqrt 3 }} - {e^{ - 1}} + e} \right] - \left[ {\sqrt 3 + {1 \over {\sqrt 3 }} - 1 - 1} \right]$

$ = {1 \over {\sqrt 3 }} + {1 \over 2}(e - {e^{ - 1}}) - {4 \over {\sqrt 3 }} + 2$

$ = (2 - \sqrt 3 ) + {1 \over 2}(e - {e^{ - 1}})$

The given region

{(x, y) : xy $ \le $ 8, 1 $ \le $ y $ \le $ x²}.

From the figure, region A and B satisfy the given region, but only A is bounded region, so area of bounded region



$A = \int_1^2 {({x^2} - 1)} dx + \int_2^8 {\left( {{8 \over x} - 1} \right)} dx$

[$ \therefore $ Points P(1, 1), Q(2, 4) and R(8, 1)]

$ = \left[ {{{{x^3}} \over 3} - x} \right]_1^2 + [8\log |x| - x]_2^8$

$ = \left( {{8 \over 3} - 2 - {1 \over 3} + 1} \right) + 8\log 8 - 8 - 8\log 2 + 2$

$ = - {{14} \over 3} + 16\log 2$

$ = 16\log 2 - {{14} \over 3}$

We have, $f(x) = 1 - 2x + \int_0^x {{e^{x - t}}f(t)dt} $

On multiplying e^$-$x both sides, we get

${e^{ - x}}f(x) = {e^{ - x}} - 2x{e^{ - x}} + \int_0^x {{e^{x - t}}f(t)dt} $

On differentiating both side w.r.t. x, we get

${e^{ - x}}f'(x) - {e^{ - x}}f(x) = - {e^{ - x}} - 2{e^{ - x}} + 2{e^{ - x}} + {e^{ - x}}f(x)$

$ \Rightarrow $ f'(x) $-$ 2f (x) = 2x $-$ 3

[dividing both sides by e^$-$x]

Let f(x) = y

$ \Rightarrow $ $f'(x) = {{dy} \over {dx}}$

$ \therefore $ ${{dy} \over {dx}} - 2y = 2x - 3$

which is linear differential equation of the form ${{dy} \over {dx}} + Py = Q$. Here, P = $-$2 and Q = 2x $-$ 3

Now, $IF = {e^{\int {P\,dx} }} = {e^{\int { - 2\,dx} }} = {e^{ - 2x}}$

$ \therefore $ Solution of the given differential equation is

$y.{e^{ - 2x}} = \int {(2x - 3){e^{ - 2x}}} dx + C$

$y.{e^{ - 2x}} = {{ - (2x - 3).{e^{ - 2x}}} \over 2} + 2\int {{{{e^{ - 2x}}} \over 2}dx + C} $

[by using integration by parts]

$ \Rightarrow $ $y.{e^{ - 2x}} = {{ - (2x - 3)\,{e^{ - 2x}}} \over 2} - {{{e^{ - 2x}}} \over 2} + C$

$ \Rightarrow $ y = (1 $-$ x) + Ce^2x

On putting x = 0 and y = 1, we get

1 = 1 + C $ \Rightarrow $ C = 0

$ \therefore $ y = 1 $-$ x

y = 1 $-$ x passes through (2, $-$ 1)

Now, area of region bounded by curve y = $\sqrt {1 - {x^2}} $ and y = 1 $-$ x is shows as



$ \therefore $ Area of shaded region

= Area of 1st quadrant of a circle $-$ Area of $\Delta $OAB

= ${\pi \over 4}{(1)^2} - {1 \over 2} \times 1 \times 1$

$ = {\pi \over 4} - {1 \over 2} = {{\pi - 2} \over 4}$

Hence, options b and c are correct.

$\int_0^1 {(x - {x^3})dx = 2\int_0^\alpha {x - {x^3})dx} } $

${1 \over 4} = 2\left( {{{{\alpha ^2}} \over 2} - {{{\alpha ^2}} \over 4}} \right)$

$2{\alpha ^2} - 4{\alpha ^2} + 1 = 0$

${\alpha ^2} = {{4 - \sqrt {16 - 8} } \over 4}$ ($ \because $ $\alpha $$ \in $(0, 1))

$ \because $ ${\alpha ^2} = 1 - {1 \over {\sqrt 2 }}$

Here, $\{ (x,y) \in {R^2}:y \ge \sqrt {\left| {x + 3} \right|} ,\,5y \le (x + 9) \le 15\} $

$\therefore$ $y \ge \sqrt {x + 3} $

$ \Rightarrow y \ge \left\{ \matrix{ \sqrt {x + 3} ,\,when\,x \ge - 3 \hfill \cr \sqrt { - x - 3} ,\,when\,x \le - 3 \hfill \cr} \right.$

or, ${y^2} \ge \left\{ \matrix{ x + 3,\,when\,x \ge - 3 \hfill \cr - 3 - x,\,when\,x \le - 3 \hfill \cr} \right.$

Shown as

Also, $5y \le (x + 9) \le 15$

$ \Rightarrow (x + 9) \ge 5y$ and $x \le 6$

Shown as

$\therefore$ $\{ (x,y) \in {R^2}:y \ge \sqrt {\left| {x + 3} \right|} ,\,5y \le (x + 9) \le 15\} $

$\therefore$ Required area = Area of trapezium ABCD $-$ Area of ABE under parabola $-$ Area of CDE under parabola

$ = {1 \over 2}(1 + 2)(5) - \int_{ - 4}^{ - 3} {\sqrt { - (x + 3)} dx - \int_{ - 3}^1 {\sqrt {(x + 3)} dx} } $

$ = {{15} \over 2} - \left[ {{{{{( - 3 - x)}^{3/2}}} \over { - {3 \over 2}}}} \right]_{ - 4}^{ - 3} - \left[ {{{{{(x + 3)}^{3/2}}} \over {{3 \over 2}}}} \right]_{ - 3}^1$

$ = {{15} \over 2} + {2 \over 3}[0 - 1] - {2 \over 3}[8 - 0]$

$ = {{15} \over 2} - {2 \over 3} - {{16} \over 3} = {{15} \over 2} - {{18} \over 3} = {3 \over 2}$

Given, $\int_1^3 {{x^2}F'(x)dx = - 12} $

$ \Rightarrow [{x^2}F(x)]_1^3 - \int_1^3 {2x\,.\,F(x)dx = - 12} $

$ \Rightarrow 9F(3) - F(1) - 2\int_1^3 {f(x)dx = - 12} $

[$\because$ $xF(x) = f(x)$ given]

$ \Rightarrow - 36 - 0 - 2\int_1^3 {f(x)dx = - 12} $

$\therefore$ $\int_1^3 {f(x)dx = - 12} $

and $\int_1^3 {{x^3}F''(x)dx = 40} $

$ \Rightarrow [{x^3}F'(x)]_1^3 - \int_1^3 {3{x^2}F'(x)dx = 40} $

$ \Rightarrow [{x^2}(xF'(x)]_1^3 - 3 \times ( - 12) = 40$

$ \Rightarrow \{ {x^2}\,.\,[f'(x) - F(x)]\} _1^3 = 4$

$ \Rightarrow 9[f'(3) - F(3)] - [f'(1) - F(1)] = 4$

$ \Rightarrow 9[f'(3) + 4] - [f'(1) - 0] = 4$

$ \Rightarrow 9f'(3) - f'(1) = - 32$

Draw the graph of $y=\sin x+\cos x$ and $y=|\cos x-\sin x|$

$\text { For } x \in\left[0, \frac{\pi}{2}\right]$

Let A be the area bounded by curves $y=\sin x +\cos x$ and $y=|\cos x+\sin x|$ for $x \in\left[0, \frac{\pi}{2}\right]$

$\begin{aligned} \mathrm{A} & =2 \int_\limits0^{\frac{\pi}{4}}((\sin x+\cos x)-(\cos x-\sin x)) d x \\ \mathrm{~A} & =4 \int_\limits0^{\frac{\pi}{4}} \sin x d x \\ \Rightarrow \mathrm{A} & =4[-\cos x]_0^{\frac{\pi}{4}} \\ \Rightarrow \mathrm{A} & =4\left[\frac{-1}{\sqrt{2}}+1\right] \\ \Rightarrow \mathrm{A} & =2 \sqrt{2}(\sqrt{2}-1) \text { sq. units } \end{aligned}$

Hints:

(i) the area bounded by curves $y=f(x)$ and $y=g(x)$ and the lines $x=a$ and $x=b(b>a)$ is $\int_\limits a^b|f(x)-g(x)| d x$

(ii) Recall the graph of $y=\sin x+\cos x$ and $y=\cos x-\sin x$

(iii) Recall the graphical transformation $y=f (x)$ in to $y=|f(x)|$.

$\begin{array}{lc} & \mathrm{S}=\int_0^1 e^{-x^2} d x \\ \text { Now, } & -x^2 \leq 0 \\ \Rightarrow & e^{-x^2} \leq 1 \\ \Rightarrow & \int_0^1 e^{-x^2} d x \leq 1 \\ \text { Now, } & x<1 \\ \Rightarrow & x^2 \leq x \\ \Rightarrow & -x^2 \geq-x \\ \Rightarrow & e^{-x^2} \geq e^{-x} \\ \Rightarrow & \mathrm{S} \geq \int_0^1 e^{-x} d x \end{array}$

$\begin{array}{ll} \Rightarrow & S \geq-\left(e^{-x}\right)_0^1 \\ \Rightarrow & S \geq-\left(\frac{1}{e}-1\right) \\ \Rightarrow & S \geq 1-\frac{1}{e} \Rightarrow(B) \text { is correct. } \\ \text { Since, } & S \geq 1-\frac{1}{e} \\ \Rightarrow & S>\frac{1}{e} \Rightarrow(\mathrm{A}) \text { is correct. } \end{array}$

Now, Area of rectangle OAPQ + Area of rectangle QBRS > S

$\Rightarrow \mathrm{S}<\frac{1}{\sqrt{2}}(1)+\left(1-\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{e}}\right) \Rightarrow(\mathrm{D})$ is correct.

Since, $\quad \frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)<1-\frac{1}{e}$

Hence, $(\mathrm{C})$ is incorrect.

We can write the integral

$\int\limits_0^b {{{(1 - x)}^2}dx - \int\limits_0^1 {{{(1 - x)}^2}dx = {1 \over 4}} } $

$ \Rightarrow \left. {{{{{(x - 1)}^3}} \over 3}} \right|_0^b - \left. {{{{{(x - 1)}^3}} \over 3}} \right|_b^1 = {1 \over 4}$

$ \Rightarrow {{{{(b - 1)}^3}} \over 3} + {1 \over 3} - \left( {0 - {{{{(b - 1)}^3}} \over 3}} \right) = {1 \over 4}$

$ \Rightarrow {{2{{(b - 1)}^3}} \over 3} = - {1 \over {12}} \Rightarrow {(b - 1)^3} = - {1 \over 8} \Rightarrow b = {1 \over 2}$

${R_1} = \int\limits_{ - 1}^2 {xf(x)dx = \int\limits_{ - 1}^2 {(2 - 1 - x)f(2 - 1 - x)dx} } $

$ = \int\limits_{ - 1}^2 {(1 - x)f(1 - x)dx = \int\limits_{ - 1}^2 {(1 - x)f(x)dx} } $

Hence, $2{R_1} = \int\limits_{ - 1}^2 {f(x)dx = {R_2}} $.

$\int\limits_0^{1/2} {f(x)dx < \int\limits_0^t {f(x)dx < \int\limits_0^{3/4} {f(x)dx} } } $

Now, $\int\limits_{}^{} {f(x)dx} $

$ = \int\limits_{}^{} {(1 + 2x + 3{x^2} + 4{x^3})dx} $

$ = x + {x^2} + {x^3} + {x^4}$

$ \Rightarrow \int\limits_0^{1/2} {f(x)dx = {{15} \over {16}} > {3 \over 4}} $

$\int\limits_0^{3/4} {f(x)dx = {{530} \over {256}} < 3} $

$f(x) = \ln x + \int\limits_0^x {\sqrt {1 + \sin t} \,dt} $

$f'(x) = {1 \over x} + \sqrt {1 + \sin x} $

f' is not differentiable at sin x = $-$1

i.e. $x = 2n\pi - {\pi \over 2},n \in N$ in the interval (0, $\infty$)

$f''(x) = - {1 \over {{x^2}}} + {{\cos x} \over {2\sqrt {1 + \sin x} }}$

f'' does not exist for all x $\in$ (0, $\infty$)

f' exist for x > 0

we have ${1 \over x} + \sqrt {1 + \sin x} < \ln x + \int\limits_0^x {\sqrt {1 + \sin x} dx} $

because L.H.S. is bounded and R.H.S. is not bounded so $\exists $ some $\alpha$ beyond which R.H.S. is greater than L.H.S.

i.e. $|f'(x)| < |f(x)|$ for all x $\in$ ($\alpha$, $\infty$)

$|f| + |f'| \le \beta $ is wrong as f is unbounded while f' is bounded.

We have, $f' = \pm \sqrt {1 - {f^2}} $

$ \Rightarrow f(x) = \sin x$ or $f(x) = - \sin x$ (not possible)

$ \Rightarrow f(x) = \sin x$

Also, $x > \sin x\forall x > 0$.

The required area is obtained as follows:

$\int\limits_1^e {\ln y\,dy = (y\ln y - y)_1^e = (e - e) - \{ - 1\} = 1} $.

Also,

$\int\limits_1^e {\ln y\,dy = \int\limits_1^e {\ln (e + 1 - y)dy} } $

Further, the area bounded by the region is

$ = e \times 1 - \int\limits_0^e {{e^x}dx} $

Let us take the required equation

$y = \sqrt {{{1 + \sin x} \over {\cos x}}} $ and $y = \sqrt {{{1 - \sin x} \over {\cos x}}} $

$\int\limits_0^{{\pi \over 4}} {\left( {\sqrt {{{1 + \sin x} \over {\cos x}}} - \sqrt {{{1 - \sin x} \over {\cos x}}} } \right)dx} $

$\because$ $\left( {{{1 + \sin x} \over {\cos x}} > {{1 - \sin x} \over {\cos x}} > 0} \right)$

$ = \int\limits_0^{{\pi \over 4}} {\left( {\sqrt {{{1 + {{2\tan {x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}} \over {{{1 - {{\tan }^2}{x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}}}} - \sqrt {{{1 + {{2\tan {x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}} \over {{{{{(1 - \tan {x \over 2})}^2}} \over {1 - \tan {x \over 2}}}}}} } \right)dx} $

$ = \int\limits_0^{{\pi \over 4}} {\left( {\sqrt {{{{{\left( {1 + {{\tan }^2}{x \over 2}} \right)}^2}} \over {1 - {{\tan }^2}{x \over 2}}}} - \sqrt {{{{{\left( {1 - \tan {x \over 2}} \right)}^2}} \over {1 - {{\tan }^2}{x \over 2}}}} } \right)dx} $

$\int\limits_0^{{\pi \over 4}} {{{1 + \tan {x \over 2} - 1 + \tan {x \over 2}} \over {\sqrt {1 - {{\tan }^2}{x \over 2}} }}dx} $

$ = \int\limits_0^{{\pi \over 4}} {{{2\tan {x \over 2}} \over {\sqrt {1 - {{\tan }^2}{x \over 2}} }}dx} $

Put $\tan {x \over 2} = t$

$dx = {{2\,dt} \over {1 + {t^2}}}$

$A = \int\limits_0^{\sqrt 2 - 1} {{{4t} \over {(1 + {t^2})\sqrt {1 - {t^2}} }}dt} $

Required area $\int\limits_a^b {ydx = \int\limits_a^b {f(x)dx} } $

$ = [f(x).x]_a^b - \int\limits_a^b {f'(x)xdx} $

$ = bf(b) - af(a) - \int\limits_a^b {f'(x)xdx} $

$ = bf(b) - af(a) + \int\limits_a^b {{{xdx} \over {3[{{\{ f(x)\} }^2} - 1]}}} $

$\because$ [$f'(x) = {{dy} \over {dx}} = {{ - 1} \over {3({y^2} - 1)}} = {{ - 1} \over {3[{{\{ f(x)\} }^2} - 1]}}$]

Apply given formula.

$\begin{aligned} & a=0, b=\frac{\pi}{2} \text { and } c=\frac{0+\frac{\pi}{2}}{2} =\frac{\pi}{4} \\ & \int_{0}^{\frac{\pi}{2}} \sin d x=\left(\frac{\frac{\pi}{2}-0}{4}\right) \\ &\left[\sin (0)+\sin \left(\frac{\pi}{2}\right)+2 \sin \left(\frac{0+\frac{\pi}{2}}{2}\right)\right] \end{aligned}$

$=\frac{\pi}{8}\left(1+\frac{2 \times 1}{\sqrt{2}}\right)=\frac{\pi}{8}(1+\sqrt{2})$

$\lim_\limits{t \rightarrow a} \frac{\int_\limits{a}^{t} f(x) d x-\left(\frac{t-a}{2}\right)\{f(t)+f(a)\}}{(t-a)^{3}}=0$

Let us assume $t=a+h$

Using L'Hospital's rule

$\lim_\limits{h \rightarrow 0} \frac{\int_\limits{a}^{a+h} f(x) d x-\frac{h}{2}\{f(a+h)+f(a)\}}{h^{3}}=0$

$\lim_\limits{h \rightarrow 0} \frac{f(a+h)-\frac{1}{2}\{f(a+h)+f(a)\}-\frac{h}{2}\left\{f^{\prime}(A+h)\right\}}{3 h^{3}}=0$

Again using L’Hospital’s rule

$\lim_\limits{h \rightarrow 0} \frac{f^{\prime}(a+h)-\frac{1}{2} f^{\prime}(a+h)-\frac{1}{2} f^{\prime}(a+h)-\frac{h}{2} f^{\prime \prime}(a+h)}{6 h}=0$

$\Rightarrow \lim_\limits{h \rightarrow 0} \frac{-\frac{h}{2} f^{\prime \prime}(a+h)}{6 h}=0$

$\Rightarrow f^{\prime \prime}(a)=0 \quad \forall a \in R$

$\Rightarrow f(x)$ must have degree 1

$\begin{aligned} & f(c)=\frac{c-a}{2}(f(a)+f(c))+\left(\frac{b-c}{2}\right)(f(b)+f(c)) \\ & f^{\prime}(c)=\left(\frac{f(a)+f(c)}{2}\right)+\left(\frac{c-a}{2}\right) \\ & f^{\prime}(c)-\left(\frac{f(b)+f(c)}{2}\right)+\frac{(b-c)}{2} f^{\prime}(c) \\ & =\frac{f(a)-f(b)}{2}+f^{\prime}(c)\left(\frac{b-a}{2}\right) \\ & \Rightarrow f^{\prime}(c)=0 \\ & \Rightarrow \frac{f(a)-f(b)}{2}+f^{\prime}(c)\left(\frac{b-a}{2}\right)=0 \\ & \Rightarrow f^{\prime}(c)=\frac{f(b)-f(a)}{(b-a)} \end{aligned}$

$ \Rightarrow f''(c) = f''(c)\left( {{{b - a} \over 2}} \right) < 0$

$\because$ $f''(x) < 0$

$\forall x \in (a,b)$

$\Rightarrow$ c is point of maximum.

$ \text { } \begin{aligned} &(i) \int_0^{\frac{\pi}{2}}(\sin x)^{\cos x}\left(\cos x \cot x-\log (\sin x)^{\sin x}\right) d x \\ & \text { Put } \sin x^{\cos x}=t \\ \Rightarrow & \cos x \cdot \log \sin x=\log t \\ \Rightarrow & {\left[\cos x \frac{1}{\sin x} \cdot \cos x+\log (\sin x)(-\sin x)\right] } \\ & d x=\frac{d x}{t} \\ \Rightarrow & {\left[\cos x \cdot \cot x-\log (\sin x)^{\sin x}\right] d x=\frac{d t}{t} } \\ & \int_0^1 t \frac{d t}{t}=\int_0^1 d t=1 \end{aligned} $

(ii) Area bounded by $-4 y^2=x$ and $x-1=-5 y^2$ Solving two to obtain the point of intersection of curve.

$ \begin{aligned} -4 y^2-1 & =-5 y^2 \\ -y^2 & =-1 \\ \Rightarrow \quad y^2 & =1 \quad \Rightarrow y= \pm 1 \end{aligned} $

$ \therefore \,\,\,\,\,\,\,\,\,\quad x=-4 $

i.e, $(-4,1)$ and $(-4,-1)$

Required area

$ \begin{aligned} & =2\left|\left[\int_0^1\left(1-5 y^2\right) d y-\int_0^1\left(-4 y^2\right) d y\right]\right| \\ & \left.=2\left[\left\lvert\, y-\frac{5 y^3}{3}\right.\right]_0^1+4\left[\frac{y^3}{3}\right]_0^1 \right\rvert\, \\ & =2\left[\left(1-\frac{5}{3}\right)+\frac{4}{3}\right] \\ & =2 \times \frac{2}{3}=\frac{4}{3} \end{aligned} $

(iii) $\quad y=3^{x-1} \log x$ and $y=x^x-1$

$ \begin{aligned} \frac{d y}{d x} & =\frac{3^{x-1}}{x}+3^{x-1} \log 3 \log x \\ y & =x^x-1 \\ \left.\frac{d y}{d x}\right|_{1,0} & =1 \end{aligned} $

$ \begin{aligned} \log y & =x \log x \\ \frac{1}{y} \frac{d y}{d x} & =(\log x+1) \\ \frac{d y}{d x} & =x^x(1+\log x) \end{aligned} $

If $\theta$ is the angle between the curve then

$ \begin{array}{r} \tan \theta=\frac{1-1}{1+1}=0 \\ \theta=0 \end{array} $

Hence, $\quad \cos 0=1$

$ \begin{aligned} & \text { (iv) } \frac{d y}{d x}=\frac{2}{x+y} \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{x}{2}=\frac{y}{2} \\ & \Rightarrow \quad x e^{-y / 2}=\frac{1}{2} / y e^{-y / 2} d y \\ & \Rightarrow \quad x+y+2=k e^{y / 2} \end{aligned} $

Length of tangent $=\left|y \sqrt{1+\left(\frac{d x}{d y}\right)^{2}}\right|=1$

$\Rightarrow y^{2}\left[1+\left(\frac{d x}{d y}\right)^{2}\right]=1$

$\Rightarrow \quad \frac{d y}{d x}= \pm \frac{y}{\sqrt{1-y^{2}}}$

$\Rightarrow \quad \int \frac{\sqrt{1-y^{2}}}{y} d y= \pm \int d x$

Put, $y=\sin \theta \Rightarrow d y=\cos \theta d \theta$

$\therefore \quad \int \frac{\cos \theta}{\sin \theta} \cos \theta \cdot d \theta= \pm x+c$

$\Rightarrow \int \frac{\cos ^{2} \theta}{\sin ^{2} \theta} \sin \theta d \theta= \pm x+c$

Again substitute

$\begin{aligned} & \cos \theta=t \\ & \Rightarrow \quad-\sin \theta d \theta=d t \\ & \therefore \quad-\int \frac{t^{2}}{1-t^{2}} d t= \pm x+c \\ & \Rightarrow \int\left(1-\frac{1}{1-t^{2}}\right) d t= \pm x+c \\ & \Rightarrow t-\frac{1}{2} \log \left|\frac{1+t}{1-t}\right|= \pm x+c \\ & \Rightarrow \sqrt{1-y^{2}}-\frac{1}{2} \log \left|\frac{1+\sqrt{1-y^{2}}}{1-\sqrt{1-y^{2}}}\right|= \pm x+c \end{aligned}$

The curves $x^2=y$ and $x^2=-y$ intersect at origin (0, 0). The curves

$x^2=y$ and $y^2=4x-3$ at (1, 1)

and $x^2=-y$ intersect with $y^2=4x-3$ at (1, $-1$)

Required area

$\begin{aligned} & =2\left[\int_{0}^{1} x^{2} d x-\int_{0.75}^{1} \sqrt{4 x-3} d x\right] \\ & =2\left[\left[\frac{x^{3}}{3}\right]_{0}^{1}-\left[\frac{2}{3} \frac{(4 x-3)^{\frac{3}{2}}}{4}\right]_{0.75}^{1}\right] \\ & =2\left[\frac{1}{3}-\frac{2}{3} \times \frac{1}{4}\right] \\ & =2\left(\frac{1}{3}-\frac{1}{6}\right)=\frac{1}{3} \text { square units }\end{aligned}$

We have

$4{a^2}f( - 1) + 4af(1) + f(2) = 3{a^2} + 3a$ ..... (i)

$4{b^2}f( - 1) + 4bf(1) + f(2) = 3{b^2} + 3b$ ..... (ii)

$4{c^2}f( - 1) + 4cf(1) + f(2) = 3{c^2} + 3c$ .... (iii)

Consider quadratic equation

$4{x^2}f( - 1) + 4xf(1) + f(2) = 3{x^2} + 3x$

$ \Rightarrow [4f( - 1) - 3]{x^2} + [4f( - 1) - 3]x + f(2) = 0$ .... (iv)

Since, equation has three roots.

$x = a,b,c$

It is an identity

$ \Rightarrow f( - 1) = {3 \over 4},f(1) = {3 \over 4}$ and $f(2) = 0$

$ \Rightarrow f(x) = \left( {{{4 - {x^2}} \over 4}} \right)$ ...... (v)

let point A be $(-2,0)$ and B be $(2t, - {t^2} + 1)$

Now, as AB subtends a right angle at the vertex V(0, 1)

${1 \over 2} \times \left( {{{ - {t^2}} \over {2t}}} \right) = - 1$

$ \Rightarrow t = 4$

Let point B be ($8,-15$)

$\therefore$ Area $=\int_\limits{-2}^{8}\left(\frac{4-x^{2}}{4}+\frac{3 x+6}{2}\right) d x$

$=\int_\limits{-2}^{8}\left(1-\frac{x^{2}}{4}+\frac{3 x}{2}+3\right) d x$

$=\left[x-\frac{x^{3}}{12}+\frac{3 x^{2}}{4}+3 x\right]_{-2}^{8}$

$=\left(8-\frac{512}{12}+\frac{192}{4}+24\right)-\left(-2+\frac{8}{12}+3-6\right)$

$=\frac{-128}{3}+80+8-3-\frac{2}{3}$

$=85-\frac{130}{3}$

$=\frac{255-130}{3}=\frac{125}{3} \text { square units. }$