Question

N19
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Answer 1

Объяснение:

$1)\ f(x)=sin(4x)\ \ \ \ A(\frac{\pi }{7};7)\\\int\limits {sin(4x)} \, dx = \int\limits {sin(4x)} \, \frac{d(4x)}{4} = \frac{1}{4}*\int\limits {sin(4x)} \, d(4x)= -cos(4x)+C.\\7=-cos\frac{4*\pi }{4}+C\\7=-cos\pi+C\\C=7+cos\pi =7+(-1)=6\ \ \ \ \Rightarrow\\F(x)=-cos(4x)+6.$

$2)\ f(x)=cos(5x)\ \ \ \ \ A(\frac{\pi }{4} ;4).\\\int\limits {cos(5x)} \, dx=\frac{1}{5}*sin(5x)+C\\4=\frac{sin\frac{5*\pi }{4} }{5} +C\\C=4-\frac{-\frac{\sqrt{2} }{2} }{5} =4+\frac{\sqrt{2} }{10}=\frac{40+\sqrt{2} }{10} .\\F(x)=\frac{sin(5x)}{5} +\frac{40+\sqrt{2} }{10} .$

$3)\ f(x)=3x^2+\frac{2}{x+2} \ \ \ \ \ A(-1;0).\\\int\limits {(3x^2+\frac{2}{\sqrt{x+2} })} \, dx=\int\limits {3x^2} \, dx+\int\limits {\frac{2}{\sqrt{x+2} } } \, dx =3*\frac{x^3}{3} +2*\int\limits {\frac{1}{\sqrt{x+2} } } \, dx =\\2*\int\limits {\frac{1}{\sqrt{x+2} } } \, dx .\\u=x+2\ \ \ \ x=u-2 dx=du\ \ \ \ \Rightarrow\\2*\int\limits {\frac{1}{\sqrt{u} } } \, du =2*\int\limits {u^{-\frac{1}{2}} \, du =2*2*\sqrt{u} +C= 4\sqrt{x+2}+C.\ \ \ \ \Rightarrow$

$y=x^3+4\sqrt{x+2} +C\\0=(-1)^3+4*\sqrt{-1+2}+C\\C=0-(-1+4*\sqrt{1}) =-(-1+4)=-3. \ \ \ \ \Rightarrow\\F(x)=x^3+4\sqrt{x+2} -3.$

$4)\ f(x)=4x^3-\frac{1}{2\sqrt{x-1} } \ \ \ \ A(2;0).\\\int\limits {(4x^3-\frac{1}{2\sqrt{x-1} } )} \, dx =\frac{4*x^4}{4} +\frac{1}{2}\int\limits {\frac{1}{\sqrt{x-1} } } \, dx=\\\frac{1}{2}\int\limits {\frac{1}{\sqrt{x-1} } } \, dx .\\u=x-1\ \ \ \ x=u+1 \ \ \ \ dx=du.\ \ \ \ \ \ \Rightarrow\\\frac{1}{2} \int\limits {\frac{1}{\sqrt{u} } } \, du=\frac{2}{2}*\sqrt{u}=\sqrt{x-1}.$

$y=x^4-\sqrt{x-1} +C.\\0=2^4-\sqrt{2-1}+C \\C=0-(16-\sqrt{1}) =-(16-1)=-15.\ \ \ \ \Rightarrow\\F(x)=x^4-\sqrt{x-1}-15.$