Application of Integrals Test - 54

$$\displaystyle \sin^{ 2 }\frac { x }{ 4 } +\cos\frac { x }{ 4 } =1-\cos ^{ 2 }{ \frac { x }{ 4 } + } \cos\frac { x }{ 4 }$$

$$\displaystyle =1-(\cos ^{ 2 }{ \frac { x }{ 4 }  } - \cos\frac { x }{ 4 } )$$

$$\displaystyle=\frac { 5 }{ 4 } -\left( \cos\frac { x }{ 4 } -\frac { 1 }{ 2 }  \right) ^{ { 2 } }$$

Now, for $$ -2\le x\le 2$$
$$\Rightarrow \displaystyle -\frac{1}{2}\le \frac{x}{4}\le \frac{1}{2}  $$

$$\Rightarrow \cos^{2}\displaystyle \frac{x}{4}\leq \cos\displaystyle \frac{x}{4}$$ for all $$x\in \left [-2,2  \right ]$$

Inverse of $$y = x^2$$ is $$y = \sqrt x $$ and required area $$= 2 \int_{o}^{1} (x - x^2) dx$$
$$\displaystyle = 2[{\frac {x^2}{2} - \frac {x^3}{3}}]_0^1 = 2[\frac {1}{2} - \frac {1}{3}] = \frac {1}{3}$$ sq. units

$$sin^{4}(\pi(x))$$
$$=(sin^{2}\pi(x))^{2}$$
$$=(\dfrac{1-cos2\pi(x)}{2})^{2}$$

$$=\dfrac{1}{4}[1+cos^{2}2\pi(x)-2cos2\pi(x)]$$

$$=\dfrac{1}{4}[1+\dfrac{1+cos4\pi(x)}{2}-2cos2\pi(x)]$$

$$=\dfrac{1}{4}[\dfrac{3+cos4\pi(x)}{2}-2cos2\pi(x)]$$
Hence the required area
$$=\int_{0} ^{1} \dfrac{1}{4}[\dfrac{3+cos4\pi(x)}{2}-2cos2\pi(x)]-ln(x).dx$$

$$=[\dfrac{3x}{8}+\dfrac{sin4\pi.x}{32\pi}-\dfrac{sin2\pi(x)}{4\pi}-x(ln(x)-1)]_{0} ^{1}$$

$$=\dfrac{3}{8}-(-1)$$

$$=\dfrac{11}{8}$$ sq units.

The two curves are $$y=\sqrt { 1-{ x }^{ 2 } } $$    ....(1)

The circles intersect at $$x=0.5$$
Hence the required area will be 
$$=\displaystyle 2\int_{\frac{1}{2}} ^{1} \sqrt{1-x^{2}}-\sqrt{1-(x-1)^{2}}.dx$$

$$=2\times\dfrac{1}{2}\left[\sin^{-1}(x)+x\sqrt{1-x^{2}}-\sin^{-1}(x-1)-(x-1)\sqrt{1-(x-1)^{2}}\right]_{\frac{1}{2}} ^{1}$$

Let $$f(x)=x^{2}+1$$
Hence $$f(0)=1$$ and 
$$f(x+1)$$
$$=(x+1)^{2}+1$$
$$=x^{2}+2x+2$$
$$=(x^{2}+1)+2x+1$$
$$=f(x)+2x+1$$.
Therefore $$f(x)=x^{2}+1$$ satisfies all the conditions required.
Let the tangents be $$y=mx$$
Hence
$$f'(x)_{x=h}=m$$
Hence
$$2x=m$$
$$x=\dfrac{m}{2}$$ ...(i)
Now
$$mx=x^{2}+1$$

$$x^{2}-mx+1=0$$

$$x=\dfrac{m\pm\sqrt{m^{2}-4}}{2}$$

$$=\dfrac{m}{2}$$ ... from i.
Hence
$$\sqrt{m^{2}-4}=0$$
Or 
$$m=\pm2$$.
Hence the equations of the tangents are $$y=\pm2x$$
The required area will be 
$$=|2\int_{0}^{1} x^{2}+1-2x.dx|$$ or $$|2\int _{0}^{-1} x^{2}+1+2x.dx|$$
Hence
$$=|2\int_{0}^{1} x^{2}+1-2x.dx|$$
$$=2\int_{0} ^{1}(x-1)^{2}.dx$$

$$=2.|[\dfrac{(x-1)^{3}}{3}]|_{0} ^{1}$$

$$=\dfrac{2}{3}$$sq units.

Both the curves are defined for $$x>0$$
Both are positive when $$x>1$$ and negative when $$0<x<1$$
We know $$\displaystyle\lim _{ x\rightarrow { 0 }^{ + } }{ \left( \log { x }  \right) \rightarrow -\infty  } $$
Hence, $$\displaystyle\lim _{ x\rightarrow { 0 }^{ + } }{ \left( \log { x }  \right) \rightarrow -\infty  } $$ .Thus y-axis is asymptote of second curve.
And $$\displaystyle\lim _{ x\rightarrow { 0 }^{ + } }{ ex\log { x }  } \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \left[ \left( 0 \right) \times \infty form \right] $$
$$\displaystyle =\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { e\log { x }  }{ \frac { 1 }{ x }  }  } \left( -\frac { \infty  }{ \infty  } form \right) $$
$$\displaystyle =\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { e\left( \frac { 1 }{ x }  \right)  }{ \left( -\frac { 1 }{ { x }^{ 2 } }  \right)  }  } =0$$ (using L'hopital rule)
Thus, the first curve starts from $$(0,0)$$ but does not include $$(0,0)$$
Now, the given curves intersect, therefore
$$\displaystyle ex\log { x } =\frac { \log { x }  }{ ex } \Rightarrow \left( { e }^{ 2 }{ x }^{ 2 }-1 \right) \log { x } =0$$
$$\displaystyle \Rightarrow x=1,\frac { 1 }{ e } \quad \quad \quad \quad \left( since\quad x>0 \right) $$
Therefore required area 
$$\displaystyle =\int _{ \frac { 1 }{ e }  }^{ 1 }{ \left( \frac { \left( \log { x }  \right)  }{ ex } -ex\log { x }  \right)  } dx$$$$\displaystyle=\frac { 1 }{ e } { \left[ \frac { { \left( \log { x }  \right)  }^{ 2 } }{ 2 }  \right]  }_{ \frac { 1 }{ e }  }^{ 1 }-e{ \left[ \frac { { x }^{ 2 } }{ 4 } \left( 2\log { x } -1 \right)  \right]  }_{ \frac { 1 }{ e }  }^{ 1 }$$
$$\displaystyle =\frac { { e }^{ 2 }-5 }{ 4e } $$

The graphs $$\displaystyle \left| x \right| +\left| y \right| =a$$ and $$\displaystyle { \left| x \right|  }^{ 2 }+{ \left| y \right|  }^{ 2 }={ a }^{ 2 }$$ are as shown in the figure 

$$y=x^{2}$$ and $$y=\dfrac{2}{1+x^{2}}$$
Therefore 
$$x^{2}=\dfrac{2}{1+x^{2}}$$

$$x^{4}+x^{2}-2=0$$

$$(x^{2}+2)(x^{2}-1)=0$$
Now x is real.
Therefore $$x^{2}+2=0$$ is not possible.
Hence
$$x=\pm1$$.
Hence the required area is 
$$=\int_{-1} ^{1} x^{2}-\dfrac{2}{1+x^{2}}$$

$$=2\int _{0} ^{1} x^{2}-\dfrac{2}{1+x^{2}}$$

$$=2[\dfrac{x^{3}}{3}-2tan^{-1}(x)]_{0} ^{1}$$

$$=2[\dfrac{1}{3}-2.\dfrac{\pi}{4}]$$

$$=|\dfrac{2}{3}-\pi|$$

$$=\pi-\dfrac{2}{3}$$ sq.units.

For the point of intersection.
$$\dfrac{x^{4}}{16}=4x$$
$$x=0$$ and $$x=4$$...(i)
Now the area bounded by the curve between x=0 and x=1
$$=\int_{0} ^{1} 2\sqrt{x}-\dfrac{x^{2}}{4}$$

$$=[\dfrac{4x\sqrt{x}}{3}-\dfrac{x^{3}}{12}]_{0}^{1}$$

$$=\dfrac{4}{3}-\dfrac{1}{12}$$

$$=\dfrac{15}{12}$$ ...(i)

The area bounded by the curve between x=1 and x=4 is 
$$=\int_{1} ^{4}2\sqrt{x}-\dfrac{x^{2}}{4}$$

$$=[\dfrac{4x\sqrt{x}}{3}-\dfrac{x^{3}}{12}]_{1}^{4}$$

$$=\dfrac{32}{3}-\dfrac{64}{12}-\dfrac{15}{12}$$

$$=\dfrac{128-64-15}{12}$$

$$=\dfrac{128-79}{12}$$

$$=\dfrac{49}{12}$$ ...(ii)
Hence the required ratio
$$=\dfrac{i}{ii}$$

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