Question

алгебра задание 1,2 ​просьба не знаете не пишите

Answer 1

Ответ:

$\displaystyle 1)\ \ F(x)=\int (2x^2-5x)\, dx=\dfrac{2x^3}{3}-\dfrac{5x^2}{2}+C\ ;\\\\M(-1;4\, ):\ \ \frac{-2}{3}-\frac{5}{2}+C=4\ \ ,\ \ \frac{-4-15}{6}+C=4\ \ ,\ \ -\frac{19}{6}+C=4\ ,\\\\C=4+\frac{19}{6}=\frac{43}{6}\\\\F(x)\Big|_{M}=\frac{2x^3}{3}-\frac{5x^2}{2}+\frac{43}{6}$

$\displaystyle 2)\ \ \int (x^5+3)^5\cdot x^4\, dx=\Big[\ u=x^5+3\ ,\ du=5x^4\, dx\ \Big]=\\\\=\frac{1}{5}\int u^5\, du=\frac{1}{5}\cdot \frac{u^6}{6}+C=\frac{1}{30}\cdot (x^5+3)^6+C$

$\displaystyle 3)\ \ \int (x+5)\, cos\frac{x}{5}\, dx=\Big[\ u=x+5\ ,\ du=dx\ ,\ dv=cos\frac{x}{5}\, dx\ ,\ v=5sin\frac{x}{5}\ \Big]=\\\\\\=uv-\int v\, du=5(x+5)sin\frac{x}{5}-5\int sin\frac{x}{5}\, dx=5(x+5)sin\frac{x}{5}+25cos\frac{x}{5}+C\\\\\\\\\star \ \ \int cos(kx+b)\, dx=\frac{1}{k}\cdot sin(kx+b)+C\ ;\\\\{}\ \ \ \int sin(kx+b)\, dx=-\frac{1}{k}\cdot cos(kx+b)+C\ \ \star$