Question

N19
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Answer 1

Объяснение:

5)

$\displaystyle f(x)=cos^2\;3x+sin^2\;3x+\frac{1}{4}sin\;4x=1+\frac{1}{4}sin\;4x;\;\;\;\;\;A(\frac{\pi }{8};\;\frac{\pi }{8} })$

$\displaystyle F(x)=\int\limits {(1+\frac{1}{4}sin\;4x) } \, dx=\int\limits {} \, dx +\frac{1}{4}\int\limits {sin\;4x} \, dx =\\\\=x+\frac{1}{4}* (-\frac{1}{4})*cos4x =x-\frac{1}{16}cos\;4x+C$

Подставим координаты точки А и найдем С:

$\displaystyle \frac{\pi }{8}=\frac{\pi }{8}-\frac{1}{16}cos\;\frac{\pi }{2}+C\\\\ C=0$

⇒

$\displaystyle F(x)=x-\frac{1}{16}cos4x$

6)

$\displaystyle f(x)=tg\;x*ctgx-2cos\;\frac{x}{2} =1-2cos\;\frac{x}{2},\;\;\;\;\;A(2\pi ;\;2\pi )$

$\displaystyle F(x)=\int\limits {(1-2cos\;\frac{x}{2} } )\, dx =\int\limits \, dx -2\int\limits {cos\;\frac{x}{2} } \, dx =\\\\=x-2*\frac{1}{\frac{1}{2} }sin\;\frac{x}{2} =x-4sin\;\frac{x}{2} +C$

Найдем С:

$\displaystyle 2\pi =2\pi -4sin\pi +C\\\\C=0$

$\displaystyle F(x)=x-4sin\;\frac{x}{2}$

7)

$\displaystyle f(x)=\frac{2}{\sqrt{5-2x} } +4x=2(5-2x)^{-\frac{1}{2}}+4x;\;\;\;\;\;A(2;\;6)\\\\F(x)=\int\limits {(2(5-2x)^{-\frac{1}{2} }+4x)} \, dx =2\int\limits {(5-2x)^{-\frac{1}{2} }} \, dx +4\int\limits {x} \, dx =\\\\=2*\frac{(5-2x)^{\frac{1}{2} }}{-2*\frac{1}{2} } +4*\frac{x^2}{2} =-2\sqrt{5-2x} +2x^2+C$

Найдем С:

$\displaystyle 6=-2\sqrt{5-2*2}+2*2^2+C\\\\C=6+2-8\\C=0$

⇒

$F(x)=-2\sqrt{5-2x}+2x^2$

8)

$\displaystyle f(x)=6x^2-\frac{1}{2\sqrt{2-x} }=6x^2-\frac{1}{2}(2-x)^{-\frac{1}{2} } ;\;\;\;\;\;A(-2;\;4)$

$\displaystyle \int\limits {(6x^2-\frac{1}{2} (2-x)^{-\frac{1}{2} })} \, dx =6\int\limits {x^2} \, dx -\frac{1}{2}\int\limits {(2-x)^{-\frac{1}{2} }} \, dx =\\\\=6*\frac{x^3}{3}-\frac{1}{2}*\frac{1}{-1}*\frac{(2-x)^{\frac{1}{2} }}{\frac{1}{2} } =2x^3+(2-x)^{\frac{1}{2} } =2x^3+\sqrt{2-x}+C$

Найдем С:

$\displaystyle 4=2*(-2)^3+\sqrt{2+2}+C\\\\4=-16+2+C\\\\C=18$

⇒

$\displaystyle F(x)=2x^3+\sqrt{2-x}+18$