Application of Integrals Test - 56

Total area $$=4\times 4=16$$ sq. units
Area of $$\displaystyle S_{3}=\int_{0}^{4}\frac{x^{2}}{4}dx=\frac{16}{3}=S_{1}$$
$$\therefore $$   $$\displaystyle S_{2}=16-\frac{16}{3}\times 2=\frac{16}{3}$$
$$\therefore $$   $$S_{1}:S_{2}:S_{3}$$ is $$1:1:1$$

Given curve is $$y=x^{2}+2x+1=\left ( x+1 \right )^{2}$$          (i)
is upward parabola with vertex at (-1, 0) meet y-axis at (0, 1) and the curve $$y=x^{2}-2x+1=\left ( x-1 \right )^{2}$$          (ii)
is also upward parabola with vertex at (1, 0) meet, y-axis at (0, 1)
Also $$\displaystyle y=\frac{1}{2}$$          (iii)
A line parallel to x-axis meeting(i) at $$\displaystyle \left ( -\frac{1}{2}, \frac{1}{4} \right )$$, $$\displaystyle \left ( -\frac{3}{2}, \frac{1}{4} \right )$$ & meeting (ii) at $$\displaystyle \left ( \frac{3}{2}, \frac{1}{4} \right )$$, $$\displaystyle \left ( \frac{1}{2}, \frac{1}{4} \right )$$
Required area is the shaded region given by 2 Area (LTNL) (by symmetry)
     $$\displaystyle =2\int_{0}^{1/2}\left \{ \left ( x-1 \right )^{2}-\frac{1}{4} \right \}dx=2\left [ \frac{\left ( x-1 \right )^{3}}{3}-\frac{x}{4} \right ]_{0}^{1/2}$$
     $$\displaystyle =2\left [ \left ( -\frac{1}{24}-\frac{1}{8} \right )+\left ( \frac{1}{3} \right ) \right ]=2\left [ -\frac{1}{6}+\frac{1}{3} \right ]=\frac{1}{3}$$ square unit

min $$\left \{ \sin x, \cos x \right \}$$ $$\displaystyle \forall 0\leq x\leq \frac{\pi }{2}$$          (from the graph of $$y=\sin x$$,

$$y=\cos x$$ $$\displaystyle \forall 0\leq x\leq \frac{\pi }{2}$$)
$$\displaystyle =\begin{cases}
\sin x &  0\leq x\leq \frac{\pi }{4} \\
\cos x & \frac{\pi }{4}\leq x\leq \frac{\pi }{2}
\end{cases}$$
$$\therefore $$   Required area $$\displaystyle =\int_{0}^{\pi /4}\sin xdx+\int_{\pi /4}^{\pi /2}\cos xdx$$
    

$$\displaystyle =\left ( -\cos x \right )_{0}^{\pi /4}+\left ( \sin x

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

1-\frac{1}{\sqrt{2}} \right )$$
     $$=2-\sqrt{2}=\sqrt{2}\left ( \sqrt{2}-1 \right )$$

The curve $${ y }^{ 2 }+8x=16$$ and $${ y }^{ 2 }-24x=48$$ cuts at $$x=-1$$

$$y=\sqrt { x } $$ or $${ y }^{ 2 }=x\left( y\ge 0 \right) $$ and $$y={ x }^{ 3 }.$$

$$y=\cos ^{ -1 }{ x } $$ and $$y=\sin ^{ -1 }{ x } $$ intersect at $$P\equiv\displaystyle \left( \frac { 1 }{ \sqrt { 2 }  } ,\frac { \pi  }{ 4 }  \right) $$
Hence, required area is,
$$\displaystyle \int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 }  }  }{ \left( \cos ^{ -1 }{ x } -\sin ^{ -1 }{ x }  \right) dx } =\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 }  }  }{ \cos ^{ -1 }{ x } dx } -\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 }  }  }{ \sin ^{ -1 }{ x } dx } $$
$$\displaystyle ={ \left[ x\cos ^{ -1 }{ x }  \right]  }_{ 0 }^{ \frac { 1 }{ \sqrt { 2 }  }  }-\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 }  }  }{ x.\frac { -1 }{ \sqrt { 1-{ x }^{ 2 } }  } dx } -{ \left[ x\sin ^{ -1 }{ x }  \right]  }_{ 0 }^{ \frac { 1 }{ \sqrt { 2 }  }  }+\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 }  }  }{ x.\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } }  } dx }$$ 

Let the x point of intersection of the two curves be (k)
$$D1 =  \int _1 ^k (\cfrac{1}{x} - \ln \ \ x)dx$$
$$D2 = \int _k ^a (- \cfrac{1}{x} \ln \ \ x)dx$$
$$ D1 +D2 =0$$
$$\int _1 ^a (\cfrac{1}{x}dx = \int _1 ^a (\ln \ \ x)dx$$
$$ \ln \ \ a =  a \ \ \ln \ \ a - a +1$$
$$ (\ln \ \ a -1)(1-a) = 0$$
$$ a =e$$

$$f(x) > g(x)$$ for $$0 < x < 2$$ and $$f(x) < g(x)$$ for $$2 < x < 4 $$.
 if $$\displaystyle \int_{0}^{4}\left [ f(x)-g(x) \right ]dx=10$$ and $$\displaystyle \int_{2}^{4}\left [ g(x)-f(x)) \right ]dx=5$$,
 $$\displaystyle \int_{0}^{4}\left [ f(x)-g(x) \right ]dx=10$$
 $$\displaystyle \int_{0}^{2} f(x)-g(x) dx + \int_{2}^{4}  f(x)-g(x) dx  =10$$
$$\displaystyle \int_{0}^{2}  ( f(x)-g(x)) dx + \int_{2}^{4}  ( g(x)-f(x))  dx  =10$$
$$\displaystyle \int_{0}^{2} ( f(x)-g(x))  dx  =15$$

Required area is,
$$\displaystyle A=2\int_{0}^{1}\left [ y\sqrt{1-y^{2}}-\left ( y^{2}-1 \right ) \right ]dy$$
$$\displaystyle =\left(\frac{-2}{3}\left ( 1-y^{2} \right )^{3/2}\right)_{0}^{1}-\left ( \frac{2y^{3}}{3}-2y \right )^{1}_{0}=2$$

Therefore the required area
$$\displaystyle =\int _{ \dfrac { 1 }{ 2 }  }^{ 2 }{ \left( { 2 }^{ x }-\log { x }  \right) dx } $$

$$\displaystyle ={ \left[ \dfrac { { 2 }^{ x } }{ \log { 2 }  } -\left( x\log { x } -x \right)  \right]  }_{ \dfrac { 1 }{ 2 }  }^{ 2 }$$

$$\displaystyle =\frac { 4-\sqrt { 2 }  }{ \log  2 } -\frac { 5 }{ 2 } \log  2+\frac { 3 }{ 2 } $$