Question

N•16
P.s полностью
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Answer 1

Ответ:

Объяснение:

$\displaystyle 2) \; f(x)=8x^2-5x^4\\\\\int\limits {(8x^2-5x^4)} \, dx =8*\frac{x^3}{3} -5*\frac{x^5}{5}=\frac{8}{3}x^3-x^5+C$

$\displaystyle 4)\;f(x)=\frac{5}{x^4}+\frac{3}{x^5}=5x^{-4}+3x^{-5}\\\\\int\limits {(5x^{-4}+3x^{-5})} \, dx =5*\frac{x^{-3}}{-3} +3*\frac{x^{-4}}{-4}=-\frac{5}{3x^3}-\frac{3}{4x^4} +C$

$\displaystyle 5)\;f(x)=\sqrt[3]{x} +3\sqrt[3]{x} =4\sqrt[3]{x} =4x^{\frac{1}{3} }\\\\\int\limits {(4x^{\frac{1}{3} })} \, dx =4*\frac{x^{\frac{4}{3} }}{\frac{4}{3} } =3x\sqrt[3]{x} +C$

$\displaystyle 6)\;f(x)=7\sqrt[3]{x} -5\sqrt{x} =7x^{\frac{1}{3} }-5x^{\frac{1}{2} }\\\\\int\limits {(7x^{\frac{1}{3} }-5x^{\frac{1}{2} })} \, dx =7*\frac{x^{\frac{4}{3} }}{\frac{4}{3} } -5*\frac{x^{\frac{3}{2} }}{\frac{3}{2} } =\frac{21x\sqrt[3]{x} }{4}-\frac{10x\sqrt{x} }{3}+C$

$\displaystyle 7)\;f(x)=5x^4+4x^3-2x^2\\\\\int\limits {(5x^4+4x^3-2x^2)} \, dx =5*\frac{x^5}{5} +4*\frac{x^4}{4}-2*\frac{x^3}{3}=x^5+x^4-\frac{2}{3}x^2+C$