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- Feb 14, 2012

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- Thread starter anemone
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- Feb 14, 2012

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- Feb 7, 2012

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Prove that if $[a,\,b]\subset \left(0,\,\dfrac{\pi}{2}\right)$, $\displaystyle \int_a^b \sin x\,dx>\sqrt{b^2+1}-\sqrt{{\color{red}a}^2+1}$.

[scale=3]

\clip (-0.25,-0.25) rectangle (3.5,3.5) ;

\draw (0,0) circle (3cm) ;

\draw [help lines, ->] (-0.5,0) -- (3.25,0) ;

\draw [help lines, ->] (0,-0.5) -- (0,3.25) ;

\coordinate [label=below left:$0$] (O) at (0,0) ;

\coordinate [label=below left:$1$] (I) at (3,0) ;

\coordinate [label=above right:$A$] (A) at (45:3) ;

\coordinate [label=above:$B$] (B) at (3,2.36) ;

\draw [thick, blue] (I) -- node[ right ]{$x$} (B) ;

\draw [thick, red] (3,0) arc (0:45:3cm) ;

\draw (I) -- (O) -- (A) ;

\draw (O) -- (B) ;

\draw [blue] (0.3,0.1) node{$\theta$} ;

\draw [red] (0.55,0.15) node{$x$} ;

\draw [blue] (0.45,0) arc (0:38:0.45cm) ;

\draw [red](0.7,0) arc (0:45:0.7cm) ;

\end{tikzpicture}

If $0 < x < \frac\pi2$ then $x < \tan x$ and so $\arctan x < x$.

In the above diagram, the red arc of the unit circle and the blue vertical line both have length $x$. So the line $0A$ makes an angle $x$ (radians) with the horizontal, and $\theta = \arctan x < x$. Therefore $\dfrac x{\sqrt{x^2+1}} = \sin\theta < \sin x$.

Now integrate from $a$ to $b$, to get \(\displaystyle \int_a^b\!\! \sin x\,dx > \int_a^b\!\!\frac x{\sqrt{x^2+1}\,}dx = \left[\sqrt{x^2+1}\right]_a^b = \sqrt{b^2+1}-\sqrt{a^2+1}.\)