Question

Пожалуйста помогите решить. Вопрос жизни и смерти

Answer 1

Ответ:

$18\dfrac{2}{3}-2\cos 2; \quad \dfrac{1}{3}-\sqrt{3} \ ; \quad \dfrac{1}{3};$

$\dfrac{1}{3};$

Объяснение:

$\displaystyle 1. \ a) \ \int\limits_{0}^{2}(x^{2}+2\sin x+7)dx=\int\limits_{0}^{2}x^{2}dx+\int\limits_{0}^{2}2\sin xdx+\int\limits_{0}^{2}7dx=\dfrac{x^{2+1}}{2+1} \bigg |_{0}^{2}-2\cos x \bigg |_{0}^{2}+$

$+7x \bigg |_{0}^{2}=\dfrac{2^{3}}{3}-\dfrac{0^{3}}{3}-2(\cos 2-\cos 0)+7 \cdot 2-7\cdot 0=\dfrac{8}{3}-2(\cos 2-1)+14=$

$=14+2\dfrac{2}{3}-2\cos 2+2=18\dfrac{2}{3}-2\cos 2;$

$b) \ (3-2\cos x)'=3'-(2\cos x)'=0-2 \cdot (\cos x)'=-2 \cdot (-\sin x)=2\sin x;$

$\sin x=\dfrac{1}{2} \cdot 2\sin x;$

$\displaystyle \int\limits_{\tfrac{3\pi}{2}}^{2\pi}\sqrt{3-2\cos x} \cdot \sin xdx=\dfrac{1}{2}\int\limits_{\tfrac{3\pi}{2}}^{2\pi}\sqrt{3-2\cos x} \ d(3-2\cos x)=$

$\displaystyle =\dfrac{1}{2}\int\limits_{\tfrac{3\pi}{2}}^{2\pi}(3-2\cos x)^{\tfrac{1}{2}}d(3-2\cos x)=\dfrac{1}{2} \cdot \dfrac{(3-2\cos x)^{\tfrac{1}{2}+1}}{\dfrac{1}{2}+1} \bigg |_{\tfrac{3\pi}{2}}^{2\pi}=$

$\displaystyle =\dfrac{1}{2} \cdot \dfrac{2}{3} \cdot (3-2\cos x)^{\tfrac{3}{2}} \bigg |_{\tfrac{3\pi}{2}}^{2\pi} = \dfrac{1}{3} \cdot \sqrt{(3-2\cos x)^{3}} \bigg |_{\tfrac{3\pi}{2}}^{2\pi}=\dfrac{1}{3} \cdot \bigg (\sqrt{(3-2\cos 2\pi)^{3}}-$

$-\sqrt{ \bigg (3-2\cos \dfrac{3\pi}{2} \bigg )^{3}} \ \bigg )=\dfrac{1}{3} \cdot (\sqrt{(3-2 \cdot 1)^{3}}-\sqrt{(3-2 \cdot 0)^{3}})=\dfrac{1}{3} \cdot (1-3\sqrt{3})=$

$=\dfrac{1}{3}-\sqrt{3} \ ;$

$c) \ (3x^{2}+1)'=(3x^{2})'+1'=3 \cdot (x^{2})'+0=3 \cdot 2 \cdot x^{2-1}=6x;$

$x=\dfrac{1}{6} \cdot 6x;$

$\displaystyle \int\limits_{\sqrt{5}}^{2\sqrt{2}}\dfrac{xdx}{\sqrt{3x^{2}+1}}=\dfrac{1}{6}\int\limits_{\sqrt{5}}^{2\sqrt{2}}\dfrac{d(3x^{2}+1)}{(3x^{2}+1)^{\tfrac{1}{2}}}=\dfrac{1}{6}\int\limits_{\sqrt{5}}^{2\sqrt{2}}(3x^{2}+1)^{-\tfrac{1}{2}}d(3x^{2}+1)=$

$=\dfrac{1}{6} \cdot \dfrac{(3x^{2}+1)^{-\tfrac{1}{2}+1}}{-\dfrac{1}{2}+1} \bigg |_{\sqrt{5}}^{2\sqrt{2}}=\dfrac{1}{6} \cdot 2 \cdot (3x^{2}+1)^{\tfrac{1}{2}} \bigg |_{\sqrt{5}}^{2\sqrt{2}}=\dfrac{1}{3} \cdot \sqrt{3x^{2}+1}\bigg |_{\sqrt{5}}^{2\sqrt{2}}=$

$=\dfrac{1}{3} \cdot (\sqrt{3 \cdot (2\sqrt{2})^{2}+1}-\sqrt{3 \cdot (\sqrt{5})^{2}+1})=\dfrac{1}{3} \cdot (\sqrt{25}-\sqrt{16})=\dfrac{1}{3} \cdot (5-4)=\dfrac{1}{3};$

2. График в приложении.

$\sqrt{x}=x^{2}; \quad x=x^{4}; \quad x(x^{3}-1)=0 \Rightarrow x=0 \ \vee \ x=1;$

$\displaystyle \int\limits_{0}^{1}(\sqrt{x}-x^{2})dx=\int\limits_{0}^{1}\sqrt{x} \ dx-\int\limits_{0}^{1}x^{2}dx=\dfrac{x^{\tfrac{1}{2}+1}}{\dfrac{1}{2}+1} \ \Bigg |_{0}^{1}-\dfrac{x^{2+1}}{2+1} \ \Bigg |_{0}^{1}=\dfrac{2}{3}x^{\tfrac{3}{2}} \bigg |_{0}^{1}-\dfrac{1}{3}x^{3} \bigg |_{0}^{1}=$

$=\dfrac{2}{3}\sqrt{x^{3}} \ \bigg |_{0}^{1}-\dfrac{1}{3} \cdot (1^{3}-0^{3})=\dfrac{2}{3}(\sqrt{1^{3}}-\sqrt{0^{3}})-\dfrac{1}{3}=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3};$