2.2

238

$\int_{ }^{ }e^{2x}=\frac{1}{2}e^{2x}+C$

$H\left(0\right)=\frac{1}{2}e^{2\cdot0}+C$
$\frac{1}{2}e^{2\cdot0}+C=-1$

$C=-\frac{3}{2}$

243

$\int_{ }^{ }-6\sin3x\ dx=\int_{ }^{ }-2\cdot3\cdot\sin3x\ dx=-2\int_{ }^{ }3\cdot\sin3x\ dx=-2\cdot\left(-\cos3x\right)+C=2\cos3x+C$

$G\left(\frac{\pi}{2}{,}1\right)=2\cos\left(3\cdot\frac{\pi}{2}\right)+C$

$2\cos\left(3\cdot\frac{\pi}{2}\right)+C=1$
$2\cdot0+C=1$

$C=1$

$G\left(x\right)=2\cos3x+1$

$2\cos3x+1=2$
$2\cos3x=1$
$\cos3x=\frac{1}{2}$

$3x=\pm\frac{\pi}{3}+n\cdot2\pi\ \ \ \ \ \left|\right|:3$

$x=\pm\frac{\pi}{9}+n\cdot\frac{2\pi}{3}{,}\ n\in\mathbb{Z}$

250

a)

Koska 2,5 tunti on 150 min, on jäljellä n. 6 eliötä/min

b)

$g'\left(x\right)=54{,}06\cdot e^{-0{,}01x}$
$g\left(x\right)=\int_{ }^{ }54{,}06\cdot e^{-0{,}01x}dx=-3624{,}53e^{-0{,}01x}+C$

Koska tunti on 60 min

$g\left(0\right)=-3624{,}53e^{-0{,}01\cdot60}+C=500$
$C=4124.53$

$g\left(60\right)=2135.3457606281...\approx2135$

257
$\int_{ }^{ }\tan xdx=\int_{ }^{ }\left(\frac{\sin x}{\cos x}\right)dx=\int_{ }^{ }\left(\sin x\cdot\left(\cos x\right)^{-1}\right)dx=-\int_{ }^{ }-\sin x\left(\cos x\right)^{-1}dx$

$s\left(x\right)=\cos x{,}\ s'\left(x\right)=-\sin x{,}\ U=\ln\left|x\right|$

$=-\ln\left|\cos x\right|+C$

Koska $\cos x>0{,}\ -\frac{\pi}{2}<x<\frac{\pi}{2}$
$=-\ln\left(\cos x\right)+C$