Application of Integrals Test - 55

First plot the graph of $$y=\sin x$$, $$y=cos x$$ by a dotted curve & find their point of intersections.
Now we find any two consecutive points of intersections. In between these points either $$\sin x> \cos x$$ or $$\cos x> \sin x$$ then in order to get max $$\left ( \sin x, \cos x \right )$$ we take those segments for which one function is greater than the other function.
In the figure, A is the point of intersection of $$\sin x$$ and $$\cos x$$.
$$\therefore $$   $$\displaystyle A\left ( \frac{\pi }{4}, \frac{1}{\sqrt{2}} \right )$$
$$\therefore $$   max $$\left ( \sin x, \cos x \right )$$ $$\forall 0\leq x\leq \pi $$
$$\therefore $$   $$\displaystyle =\begin{cases}
\cos x & \forall  0\leq x\leq \dfrac{\pi }{4} \\ 
\sin x & \forall  \dfrac{\pi }{4}\leq x\leq \pi  
\end{cases}$$
$$\therefore $$   $$\int_{0}^{\pi }max \left ( \sin x\cos x \right )dx=\int_{0}^{\pi /4}\cos xdx+\int_{\pi /4}^{\pi }\sin xdx =\sqrt{2}+1$$

For the point of intersection
$$2\sqrt{3x}-\dfrac{x^{2}}{12}=0$$
Or 
$$\sqrt{x}[24\sqrt{3}-x\sqrt{x}]=0$$

$$x=0$$ and $$x=12$$.
Hence
$$\int _{0} ^{3} (2\sqrt{3x}-\dfrac{x^{2}}{12}).dx$$

$$=[\dfrac{4\sqrt{3}}{3}.x^{\frac{3}{2}}-\dfrac{x^{3}}{36}]_{0} ^{3}$$

$$=\dfrac{45}{4}$$ ...(i)

And
$$\int _{3} ^{12} (2\sqrt{3x}-\dfrac{x^{2}}{12}).dx$$

$$=[\dfrac{4\sqrt{3}}{3}.x^{\frac{3}{2}}-\dfrac{x^{3}}{36}]_{3} ^{12}$$

$$=\dfrac{147}{4}$$...(ii)
Hence
Ratio: $$\dfrac{i}{ii}$$

$$=\dfrac{45}{147}$$

$$=\dfrac{15}{49}$$.

Solving $$y={ x }^{ 2 }$$ and $$y=2x\left( 1-x \right) $$ we get $$\displaystyle x=0,\frac { 2 }{ 3 } $$

And solving $$y=1-{ x }^{ 2 }$$ and $$y=2x\left( 1-x \right) $$, we get $$\displaystyle x=1,\frac { 1 }{ 3 } $$

Therefore 
The required area $$\displaystyle A=\int _{ 0 }^{ 1 }{ f\left( x \right) dx } =\int _{ 0 }^{ \frac { 1 }{ 3 }  }{ { \left( 1-x \right)  }^{ 2 } } dx+\int _{ \frac { 1 }{ 3 }  }^{ \frac { 2 }{ 3 }  }{ 2x } \left( 1-x \right) dx\quad +\int _{ \frac { 2 }{ 3 }  }^{ 1 }{ { x }^{ 2 }dx } $$

$$\displaystyle ={ \left( -\frac { 1 }{ 3 } { \left( 1-x \right)  }^{ 3 } \right)  }_{ 0 }^{ \frac { 1 }{ 3 }  }+{ \left( { x }^{ 2 }-\frac { 2{ x }^{ 3 } }{ 3 }  \right)  }_{ \frac { 2 }{ 3 }  }^{ \frac { 2 }{ 3 }  }+{ \left( \frac { { x }^{ 3 } }{ 3 }  \right)  }_{ \frac { 2 }{ 3 }  }^{ 1 }$$

$$\displaystyle =\frac { 19 }{ 81 } +\frac { 13 }{ 81 } +\frac { 19 }{ 81 } =\frac { 17 }{ 27 } $$

The required area 

Eliminating $$\theta $$, we have
$$\displaystyle { x }^{ \dfrac { 2 }{ 3 }  }+{ y }^{ \dfrac { 2 }{ 3 }  }={ a }^{ \dfrac { 2 }{ 3 }  }\\ x=0\Rightarrow y=\pm a\\ y=0\Rightarrow x=\pm a$$
Symmetric about both axis 
Required area $$\displaystyle =4\int _{ 0 }^{ a }{ y } dx=4\int _{ \dfrac { \pi  }{ 2 }  }^{ 0 }{ y } \dfrac { dx }{ dt } dt$$
$$\displaystyle y=a\sin ^{ 3 }{ t } ,x=a\cot ^{ 3 }{ t } \Rightarrow \frac { dx }{ dt } =-3a\cos ^{ 3 }{ t } \sin { t } $$
Area $$\displaystyle =4\int _{ \dfrac { \pi  }{ 2 }  }^{ 0 }{ a } \sin ^{ 3 }{ t } .\left( -3a\cos ^{ 2 }{ t } \sin { t }  \right) dt$$
$$\displaystyle =12{ a }^{ 2 }\int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \sin ^{ 4 }{ t }  } \cos ^{ 2 }{ t } dt$$
$$\displaystyle =\dfrac { { 12a }^{ 2 }\dfrac { 5 }{ 2 } !\dfrac { 3 }{ 2 } ! }{ 2\dfrac { 8 }{ 2 } ! } =\dfrac { { 6a }^{ 2 }\times \dfrac { 3 }{ 2 } \times \dfrac { 1 }{ 2 } \times \sqrt { \pi  } .\dfrac { 1 }{ 2 } \sqrt { \pi  }  }{ 3\times 2 } $$
$$\displaystyle =\dfrac { 3 }{ 8 } { \pi a }^{ 2 }$$

For the point of intersection
$$tan(x)=\dfrac{2}{3}cos(x)$$
Or 
$$3tan(x)=2cos(x)$$
$$3sin(x)=2cos^{2}(x)$$
Or 
$$3sin(x)=2-2sin^{2}(x)$$
Or 
$$2sin^{2}(x)+3sin(x)-2=0$$
Hence
$$sin(x)=\dfrac{-3\pm\sqrt{9+16}}{4}$$

$$=\dfrac{-3\pm5}{4}$$
Hence
$$sin(x)=-2$$ ...(not possible) and $$sin(x)=\dfrac{1}{2}$$
Hence
$$x=\dfrac{\pi}{6}$$ , $$x=\dfrac{5\pi}{6}$$.
Hence the required area will be 
$$=|\int_{0} ^{\frac{\pi}{6}} tan(x)-\dfrac{2}{3}cos(x)|$$

$$=ln|sec(x)|-\dfrac{2}{3}sin(x)|_{0} ^{\frac{\pi}{6}}$$

$$=|ln(\dfrac{2}{\sqrt{3}})-\dfrac{1}{3}|$$

$$=\dfrac{1}{3}-ln(\dfrac{2}{\sqrt{3}})$$

$$=\dfrac{1}{3}+ln|\dfrac{\sqrt{3}}{2}|$$ sq units.

min $$\left \{ 1, 1-\cos x, 2\sin x\forall 0\leq x\leq \pi  \right \}$$
   $$\displaystyle =\begin{cases}
1-\cos x &  0\leq x\leq \dfrac{\pi }{2} \\
1 &  \dfrac{\pi }{2}\leq x\leq \dfrac{5\pi }{6} \\
2\sin x &  \dfrac{5\pi }{6}\leq x\leq \pi 
\end{cases}$$
$$\therefore

$$   $$\displaystyle \int_{0}^{\pi }f\left ( x \right )dx=\int_{0}^{\pi /2}\left (

1-\cos x \right )dx+\int_{\pi /2}^{5\pi /6}1dx+\int_{5\pi /6}^{\pi }2\sin

xdx$$
     $$=\left [ x-\sin x \right ]_{0}^{\pi /2}+\left [ x \right ]_{\pi /2}^{5\pi /6}-2\left [ \cos x \right ]_{5\pi /6}^{\pi }$$
    

$$\displaystyle =\frac{\pi }{2}-1+\frac{5\pi }{6}-\frac{\pi }{2}-2\left

[ -1+\frac{\sqrt{3}}{2} \right ]=\frac{5\pi }{6}+1-\sqrt{3}$$

Solving the equation $$x+2y^{2}=0$$ & $$3y^{2}+x=1$$
we get $$A(-2, 1)$$ & $$B(-2, -1)$$
Required Area $$\displaystyle =2\left [ \int_{0}^{1}xdy-\int_{0}^{1}xdy \right ]$$
   $$\displaystyle =2\int_{0}^{1}\left ( 1-3y^{2}+2y^{2} \right )dy$$
   $$\displaystyle =2\int_{0}^{1}\left ( 1-y^{2} \right )dy$$
   $$\displaystyle =2\left [ y-\frac{y^{3}}{3} \right ]_{0}^{1}=2\left ( 1-\frac{1}{3} \right )=\frac{4}{3}$$ sq.units

The two curves represent parabolas with vertices at $$\left( 0,0 \right) $$ and $$\left( 3,0 \right) $$
They intersect at $$\left( 1,1 \right) $$ and $$\left( 1,-1 \right) $$
So the required area is 
area of OPMQO $$= 2$$( area of OPMO)
$$\displaystyle =2\left( \int _{ 0 }^{ 1 }{ \sqrt { x }  } dx+\int _{ 1 }^{ 3 }{ \sqrt { \frac { 3-x }{ 2 }  } dx }  \right) $$

$$\displaystyle =2\left( \frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 }  } \right) _{ 0 }^{ 1 }{ - }\left[ 2\frac { 1 }{ \sqrt { 2 }  } .\frac { 2 }{ 3 } { \left( 3-x \right)  }^{ \frac { 3 }{ 2 }  } \right] _{ 1 }^{ 3 }$$

$$\displaystyle =2\left[ \frac { 2 }{ 3 } -\left( 0-\frac { 1 }{ \sqrt { 2 }  } \frac { 2 }{ 3 } { 2 }^{ \frac { 3 }{ 2 }  } \right)  \right] =2\left( \frac { 2 }{ 3 } +\frac { 4 }{ 3 }  \right) =4$$

Required Area $$=2\int_{0}^{1}\left | \log \left | x \right | \right |dx$$
     $$\displaystyle =2\left [ \left ( x\left | \log \left | x \right | \right | \right ) \right ]_{0}^{1}-\int_{0}^{1}\left ( -\frac{1}{x} \right )xdx$$
     $$=2\left [ \left ( 1-0 \right )+\left ( x \right )_{0}^{1} \right ]=4$$ units