Question

ребят, срочно, срочно, срочно! АЛГБЕРА 11 КЛАСС ​

Answer 1

Ответ:

$\boxed{\ \displaystyle \ \int\limits{x^{n}} \, dx=\frac{x^{n+1}}{n+1}+C\ \ ,\ \ \ \ \int C\cdot f(x)\, dx=C\cdot \int \, f(x)\, dx\ }$

$1a)\ \ f(x)=2x^2-3x+4\ \ ,\ \ A(2;8)\\\\F(x)=\dfrac{2x^3}{3}-\dfrac{3x^2}{2}+4x+C\\\\A(2;8):\ \ 8=\dfrac{2\cdot 8}{3}-\dfrac{3\cdot 4}{2}+4\cdot 2+C\ \ ,\ \ 8=\dfrac{16}{3}-6+8+C\ \ ,\\\\C=6-\dfrac{16}{3}=\dfrac{2}{3}\\\\F(x)\Big|_{A}=\dfrac{2x^3}{3}-\dfrac{3x^2}{2}+4x+\dfrac{2}{3}$

$b)\ \ f(x)=(1-2x)^2-cosx\ \ ,\ \ \ B(0;5)\\\\F(x)=-\dfrac{(1-2x)^3}{6}-sinx+C\\\\B(0;5):\ \ 5=-\dfrac{1^3}{6}-0+C\ \ ,\ \ C=5+\dfrac{1}{6}=\dfrac{31}{6}\\\\F(x)=-\dfrac{(1-2x)^3}{6}-sinx+\dfrac{31}{6}$

$2)\ \ \displaystyle \int\limits^3_1\, x^3\, dx=\dfrac{x^4}{4}\Big|_1^3=\dfrac{1}{4}\cdot (3^4-1^4)=\dfrac{80}{4}=20\\\\\\\int\limits^{\pi /4}_0\, sin2x\, dx=-\dfrac{1}{2}\, cos2x\Big|_0^{\pi /4}=-\dfrac{1}{2}\cdot (0-1)=\dfrac{1}{2}\\\\\\\int\limits^{\pi /3}_{\pi /6}\, \dfrac{dx}{cos^2x}=tgx\Big|_{\pi /6}^{\pi /3}=\sqrt3-\dfrac{\sqrt3}{3}=\dfrac{3-1}{\sqrt3}=\dfrac{2}{\sqrt3}=\frac{2\sqrt3}{3}$