Вычислите интегралы : (на фото)

[latex]\\ \int_{-2}^{3}x^2{\mathrm dx}={1\over3}x^3|_{-2}^{3}={1\over3}(27+8)={35\over3}\\\\ \int_{-2\pi}^{\pi}\sin{2x}{\mathrm dx}={1\over2}\int_{-2\pi}^{\pi}\sin{2x}{\mathrm d{2x}}=-{1\over2}\cos{2x}|_{-2\pi}^{\pi}=-{1\over2}(\cos{2\pi}-\cos{(-4\pi)}=0\\ \\ \int_{-2}^{1}\left ( -3x^2-4x+2 \right ){\mathrm dx}=-x^3|_{-2}^{1}-2x^2|_{-2}^{1}+2x|_{-2}^{1}=-9+6+6=3\\\\ \int_{\pi\over8}^{\pi\over4}{1\over \sin^2{2x}}{\mathrm dx}={1\over2}\int_{\pi\over8}^{\pi\over4}{1\over \sin^2{2x}}{\mathrm d{2x}}=-{1\over2}ctg{2x}|_{\pi\over8}^{\pi\over4}=-{1\over2}\left ( ctg{\pi\over2}-ctg{\pi\over4} \right )={1\over2}\\\\[/latex] [latex]\\ \int_{1}^{16}x^{0,25}{\mathrm dx}={x^{1,25}\over1,25}|_{1}^{16}={4\over5}\left ( \sqrt[4]{16^5} -\sqrt[4]{1} \right )={4\over5}\left ( \sqrt[4]{({2^{5})}^4} -\sqrt[4]{1} \right )={4\over5}(2^5-1)={4\over5}\cdot31=24,8\\[/latex]