Application of Integration

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.

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

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$

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

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

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.

$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.

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} $