Question

допоможітьььь! будь-ласка ​

Answer 1

Пошаговое объяснение:

$\displaystyle\\f(x)=\frac{3}{2\sqrt{3x+4} } -2x\\\\F(x)=\int(\frac{3}{2\sqrt{3x+4} } -2x)dx=\int\frac{3}{2\sqrt{3x+4} } dx-\int(2x)dx.\\\\\\\int\frac{3}{2\sqrt{3x+4} } dx=\left [ {{u=3x+4\ \ \ \ x=\frac{u-4}{3} } \atop {dx=\frac{du}{3} }} \right]=\int\frac{3}{2\sqrt{u} }*\frac{du}{3} =\frac{1}{2} \int\frac{du}{\sqrt{u} } =\\\\\\=\frac{2\sqrt{u} }{2}+C =\sqrt{u}+C=\sqrt{3x+4} +C.\\\\\\\int(2x)dx=2*\frac{x^2}{2}+C=x^2+C.\ \ \ \ \ \ \Rightarrow\\\\ F(x)=\sqrt{3x+4} -x^2+C$$A(7;-2)\ \ \ \ x=7\ \ \ \ y=-2\\\\\sqrt{3*7+4} -7^2+C=-2\\\\\sqrt{21+4}-49+C=-2\\\\\sqrt{25} +C=47\\\\5+C=47\\\\C=42.\ \ \ \ \ \ \Rightarrow\\\\F(x)=\sqrt{3x+4}-x^2+42$