Oxford - přijímací zkoušky 1. část

Tak tady nevím?

\(  \int_0^1 (\mathrm{e}^{x}-x)(\mathrm{e}^{x}+x)\,\mathrm{d}x = \int_0^1 (\mathrm{e}^{2x}-x^2)\,\mathrm{d}x  =\int_0^1 (\mathrm{e}^{2x}\,\mathrm{d}x   -\int_0^1 x^2\,\mathrm{d}x  = \left.\frac{e^{2x}}{2}\right|_0^1 - \left.\frac{x^3}{3}\right|_0^1=\)

\(=\frac{e^2}{2} - \frac{1}{2}-\frac{1}{3}= \frac{2e^2-3}{6}\)

\( 1-4+9-16+25-...+99^2 - 100^2 = ? \)

\( 1-4+9-16+25-...+99^2 - 100^2 =  \sum_{n=1}^{99}{( n^2-(n+1)^2) }= \sum_{n=1}^{99}{( -2n-1) } = \)

\(= - \sum_{n=1}^{99}{(2n+1)}=-\frac{3+199}{2}\cdot100 = -5050 \)

\( S=\frac{nr^2}{2}\sin{\frac{2\pi}{n}}  \)

\( S=\frac{12}{2}\sin{\frac{2\pi}{12}}=6\sin\frac{\pi}{6}=3  \)