<p>Find $\displaystyle\int_{0}^{n}{\left| \sin x-\cos x \right|}\mathrm{ d}x$, in terms of ${n}$, where ${\frac{\pi}{4} &lt; n &lt; \frac{\pi}{2}}$.</p>

Find $\displaystyle\displaystyle \int_{0}^{n}{\left| \sin x-\cos x \right|}\mathrm{ d}x$, in terms of ${n}$, where ${\frac{\pi}{4} < n < \frac{\pi}{2}}$.

Applications of Intergration - General

Applications of Intergration

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

$y = |\sin x - \cos x|$

$y = \pm(\sin x - \cos x)$

$y = 0$, $\sin x - \cos x = 0$

$\sin x = \cos x$

$\displaystyle\frac{\sin x}{\cos x} = 1$

$\tan x = 1$

$x = \displaystyle\frac{\pi}{4}$

$\displaystyle\int_0^n |\sin x - \cos x| \, \mathrm{d}x$

$\displaystyle = \int_0^{\frac{\pi}{4}} -(\sin x - \cos x) \, \mathrm{d}x + \int_{\frac{\pi}{4}}^n \sin x - \cos x \, \mathrm{d}x$

$\displaystyle = \int_0^{\frac{\pi}{4}} -\sin x + \cos x \, \mathrm{d}x + \int_{\frac{\pi}{4}}^n \sin x - \cos x \, \mathrm{d}x$

$\displaystyle = \Big[\cos x + \sin x\Big]_0^{\frac{\pi}{4}} + \Big[-\sin x - \cos x\Big]_{\frac{\pi}{4}}^n$

$\displaystyle = \left[\left(\cos\frac{\pi}{4} + \sin\frac{\pi}{4}\right) - (\cos 0 + \sin 0)\right] + \left[(-\sin n - \cos n) - \left(-\sin\frac{\pi}{4} - \cos\frac{\pi}{4}\right)\right]$

$\displaystyle = \left[\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - 1\right] + \left[-\sin n - \cos n - \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)\right]$

$\displaystyle = \frac{2}{\sqrt{2}} - 1 - \sin n - \cos n + \frac{2}{\sqrt{2}}$

$\displaystyle = \frac{4}{\sqrt{2}} - 1 - \sin n - \cos n$

Applications of Intergration - Applications of Intergration - General

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