R2 2014 vår LØSNING

DEL 1

Oppgave 1

a) $\displaystyle f(x) = \sin(3x)$

$\displaystyle f'(x) = 3\cos(3x)$

b) $\displaystyle g(x) = e^{2x} \cdot \cos x$

$\displaystyle g'(x) = 2e^{2x} \cdot \cos x + e^{2x} \cdot (-\sin x) = e^{2x} (2\cos x - \sin x)$

Oppgave 2

a) $\displaystyle \int 2x \cdot \sin (x^2) \, \mathrm{d}x$

La $\displaystyle u = x^2$

$\displaystyle \begin{align*} & \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = 2x \\ & \Rightarrow \mathrm{d}u = 2x \space \mathrm{d}x \end{align*}$

$\displaystyle \int 2x \cdot \sin (x^2) \, \mathrm{d}x = \int \sin u \, \mathrm{d}u = -\cos u + C = -\cos (x^2) + C$

b) $\displaystyle \int_1^{e} x \cdot \ln x \, \mathrm{d}x$

La $\displaystyle u = \ln x$ og $\displaystyle v' = x$:

$\displaystyle \begin{align*} \int_1^{e} x \cdot \ln x \, \mathrm{d}x & = \left[ \ln x \cdot \frac{1}{2} x^2 - \int \frac{1}{x} \cdot \frac{1}{2} x^2 \right]_1^{e} \\ & = \left[ \frac{1}{2} x^2 \cdot \ln x - \frac{1}{2} \int x \, \mathrm{d}x \right]_1^{e} \\ & = \left[ \frac{1}{2} x^2 \cdot \ln x - \frac{1}{2} \cdot \frac{1}{2} x^2 \right]_1^{e} \\ & = \frac{1}{2} \left[ x^2 \cdot \ln x - \frac{1}{2} x^2 \right]_1^{e} \\ & = \frac{1}{2} \left( (e^2 \cdot \ln e - \frac{1}{2} \cdot e^2) - (1^2 \cdot \ln1 - \frac{1}{2} \cdot 1^2) \right) \\ & = \frac{1}{2} \left( (e^2 - \frac{1}{2} \cdot e^2) - (0 - \frac{1}{2}) \right) \\ & = \frac{1}{2} \left( \frac{e^2}{2} + \frac{1}{2} \right) \\ & = \frac{1}{2} \cdot \frac{e^2 + 1}{2} \\ & = \frac{e^2 + 1}{4} \end{align*}$