Engineering Mathematics Test 3

\(f\left( x \right) + 2f\left( {\frac{1}{x}} \right) = {x^2}\) 

\(f\left( {\frac{1}{x}} \right) = 2f\left( x \right) = \frac{1}{{{x^2}}}\) 

By solving the above two equations,

\(2f\left( {\frac{1}{x}} \right) + 4\;f\left( x \right) - f\left( x \right) - 2f\left( {\frac{1}{x}} \right) = \frac{2}{{{x^2}}} - {x^2}\) 

\(\Rightarrow 3f\left( x \right) = \frac{2}{{{x^2}}} - {x^2}\) 

\(\Rightarrow {x^2}f\left( x \right) = \frac{1}{3}\left( {1 - {x^4}} \right)\) 

\(\mathop \smallint \limits_1^2 {x^2}f\left( x \right)dx = \mathop \smallint \limits_1^2 \frac{1}{3}\left( {2 - {x^4}} \right)dx\) 

\( = \frac{1}{3}\left[ {2x - \frac{{{x^5}}}{5}} \right]_1^2\) 

\(= \frac{1}{3}\left[ {\left( {2 - 1} \right) - \frac{1}{5}\left( {32 - 1} \right)} \right] = \frac{{ - 7}}{5}\) 

The given region is, R: 0 ≤ x ≤ y ≤ L

The limits of x: 0 ≤ x ≤ L

The limits of y: x ≤ y ≤ L

\(\mathop \int\!\!\!\int \limits_R \left( {{x^2} + {y^2}} \right)dx\;dy\) 

\( = \mathop \smallint \limits_{x = 0}^{x = L} \mathop \smallint \limits_{y = x}^{y = L} \left( {{x^2} + {y^2}} \right)dx\;dy\) 

\(= \mathop \smallint \limits_{x = 0}^L \left[ {{x^2}y + \frac{{{y^3}}}{3}} \right]_x^2dx\) 

\( = \mathop \smallint \limits_{x = 0}^L \left[ {{x^2}L + \frac{{{L^3}}}{3} - \frac{{{x^3}}}{3} - {x^3}} \right]dx\;\) 

\(= \mathop \smallint \limits_{x = 0}^L \left[ {{x^2}L + \frac{{{L^3}}}{3} - \frac{{4{x^3}}}{3}} \right]dx\) 

\(= \left[ {L\frac{{{x^3}}}{3} + \frac{{{L^3}}}{3}x - \frac{{{x^4}}}{3}} \right]_0^L\) 

\(= \frac{{{L^4}}}{3} + \frac{{{L^4}}}{3} - \frac{{{L^4}}}{3} = \frac{{{L^4}}}{3}\) 

The volume of solid generated \( = \smallint \pi {y^2}dx\)

\( = \mathop \smallint \limits_{ - 4}^4 \pi \frac{{\left( {16 - {x^2}} \right)}}{2}dx\)

\( = \frac{\pi }{2}\left[ {16x - \frac{{{x^3}}}{3}} \right]_{ - 4}^4 = \frac{{128\pi }}{3}\)

Let \(I = \mathop \smallint \limits_0^3 \frac{{2x}}{{{{\left( {1 - {x^2}} \right)}^{2/3}}}}dx\)

The given integral is undefined at x = 1. So, we can split the above integral as follows.

\(= \mathop \smallint \limits_0^1 \frac{{2x}}{{{{\left( {1 - {x^2}} \right)}^{2/3}}}}dx + \mathop \smallint \limits_1^2 \frac{{2x}}{{{{\left( {1 - {x^2}} \right)}^{2/3}}}}dx\)

\(= \left[ { - 3{{\left( {1 - {x^2}} \right)}^{1/3}}} \right]_0^1 + \left[ { - 3{{\left( {1 - {x^2}} \right)}^{1/3}}} \right]_1^3\)

Length of curve is given by

\(L = \mathop \smallint \limits_{x = a}^{x = b} \sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx\)

\(y = \frac{2}{3}{x^{\frac{3}{2}}}\)

\(\frac{{dy}}{{dx}} = \sqrt x \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)^2} = x\) ---(1)

\(L = \mathop \smallint \limits_{x = 0}^{x = 3} \sqrt {1 + x} \;dx\)

\(\left[ {\frac{{{{\left( {x + 1} \right)}^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^3\)

\(= \frac{2}{3}\left[ {{{\left( {x + 1} \right)}^{\frac{3}{2}}}} \right]_0^3\)

\(= \frac{2}{3}\left[ {{4^{\frac{3}{2}}} - 1} \right]\)

We have,  \({\rm{I}} = \mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{6}} {\cos ^4}3{\rm{\theta }}{\sin ^3}6{\rm{\theta \;d\theta }}\) 

Let \(3{\rm{\theta }} = {\rm{t}},{\rm{\;d\theta }} = \frac{{{\rm{dt}}}}{3}\)

At \({\rm{\theta }} = 0,{\rm{t}} = 0;{\rm{\theta }} = \frac{{\rm{\pi }}}{6},{\rm{t}} = \frac{{\rm{\pi }}}{2}\)

\(\begin{array}{l} {\rm{I}} = \mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} {\cos ^4}{\rm{t}}{\sin ^3}2{\rm{t}}\frac{{{\rm{dt}}}}{3}\\ = \mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} {\rm{co}}{{\rm{s}}^4}{\rm{t}}{\left( {2\sin {\rm{t}}\cos {\rm{t}}} \right)^3}\frac{{{\rm{dt}}}}{3}\\ {\rm{I}} = \frac{8}{3}\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} {\cos ^7}{\rm{t}}{\sin ^3}{\rm{t\;dt}} \end{array}\)  

Applying beta function definition,

\(\mathop \smallint \limits_0^{\frac{\pi }{2}} {\sin ^{\rm{p}}}x{\cos ^{\rm{q}}}xdx = \frac{1}{2}\beta \left( {\frac{{p + 1}}{2},\frac{{q + 1}}{2}} \right)\)

\(I = \frac{8}{3} \times \frac{1}{2}\beta \left( {4,2} \right)\)

\({\rm{I}} = \frac{8}{3}.\frac{1}{2} \times \frac{{3! \times 1!}}{{5!}} = \frac{1}{{15}}\)

\(\left[ {{x^2}} \right] = 0\;when\;0 \le x < 1\)

\(\left[ {{x^2}} \right] = 1\;when\;\begin{array}{*{20}{c}} {1 \le x < \sqrt 2 }\\ {1 \le x < 1.414} \end{array}\)

\(\left[ {{x^2}} \right] = 2\;\begin{array}{*{20}{c}} {\sqrt 2 \le x < \sqrt 3 }\\ {\;1.414 \le x < 1.732} \end{array}\)

\(\left[ {{x^2}} \right] = 3,\;1.732 \le x < 2\)

\(I = \mathop \smallint \limits_0^1 {x^0}dx + \mathop \smallint \limits_1^{\sqrt 2 } x'dx + \mathop \smallint \limits_{\sqrt 2 }^{\sqrt 3 } {x^2}dx + \mathop \smallint \limits_{\sqrt 3 }^2 {x^3}dx\)

\(= \left( {1 - 0} \right) + \left( {\frac{{{x^2}}}{2}} \right)_1^{\sqrt 2 } + \left( {\frac{{{x^3}}}{3}} \right)_{\sqrt 2 }^{\sqrt 3 } + \left( {\frac{{{x^4}}}{4}} \right)_{\sqrt 3 }^2\)

\(= 1 + \frac{1}{2}\left( {2 - 1} \right) + \frac{1}{3}\left( {3\sqrt 3 - 2\sqrt 2 } \right) + \frac{1}{4}\left( {{2^4} - {{\sqrt 3 }^4}} \right)\)

\(= 1 + \frac{1}{2} + \sqrt 3 - \frac{2}{3}\sqrt 2 + 4 - \frac{9}{4}\)

\(= \frac{3}{2} + \sqrt {3} \frac{{ - 2}}{3}\sqrt 2 + 4 - \frac{9}{4}\)

\(\Rightarrow 3.25 + \sqrt 3 - \frac{2}{3}\sqrt 2\)

= 4.039