Предмет: Алгебра, автор: 098102nik

Обчислити визначений інтеграл. Обов'язково написати, чому = первісна на верхній та нижній межах інтегрування

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Автор ответа: Аноним

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Автор ответа: NNNLLL54

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Интегрирование степенной функции : $\displaystyle \int x^{n}\, dx=\frac{x^{n+1}}{n+1}+C$ .

$\displaystyle 1)\ \ \int\limits_8^{27}\, \sqrt[3]{x}\, dx= \int\limits_8^{27}\, x^{\frac{1}{3}}\, dx=\frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}\, \Big|_8^{27}=\frac{3\sqrt[3]{x^4}}{4}\, \Big|_8^{27}=\frac{3}{4}\cdot \Big(\sqrt[3]{27^4}-\sqrt[3]{8^4}\Big)=\\\\\\=\frac{3}{4}\cdot \Big(27\sqrt[3]{27}-8\sqrt[3]{8}\Big)=\frac{3}{4}\cdot \Big(27\cdot 3-8\cdot 2\Big)=\frac{3}{4}\cdot \Big(81-16\Big)=\frac{3}{4}\cdot 65=48,75$

$2)\ \ \displaystyle \int\limits^1_{-2}\, (x^2+x)^2\, dx= \int\limits^1_{-2}\, (x^4+2x^3+x^2)\, dx=\Big(\frac{x^5}{5}+\frac{2x^4}{4}+\frac{x^3}{3}\Big)\, \Big|_{-2}^1=\\\\\\=\frac{1}{5}+\frac{1}{2}+\frac{1}{3}-\Big(-\frac{32}{5}+\frac{16}{2}-\frac{8}{3}\Big)=\Big(\frac{1}{5}+\frac{32}{5}\Big)+\Big(\frac{1}{2}-\frac{16}{2}\Big)+\Big(\frac{1}{3}+\frac{8}{3}\Big)=\\\\\\=\frac{33}{5}-\frac{15}{2}+\frac{9}{3}=6,6-7,5+3=2,1$

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