Question
$\int_{\,0}^{\,3} {|2 - x|dx} =$
$ = \int_0^2 {(2 - x)} \,dx + \int_2^3 { - (2 - x)\,dx} $
$ = \int_0^2 {(2 - x)} \,dx - \int_2^3 {\,(2 - x)\,dx} = \left[ {2x - \frac{{{x^2}}}{2}} \right]_0^2 - \left[ {2x - \frac{{{x^2}}}{2}} \right]_2^3$
$\Rightarrow$ $I = [4-2] - \left[ 6- \frac{{9}}{{2}} - (4-2) \right] $
$ = 2 - \left[ {4 - \frac{9}{2}} \right]$$ = \frac{5}{2}$.
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$\cos x d y=y(\sin x-y) d x, 0< x< \frac{\pi}{2}$ का हल है