Question

Помогите пожалуйстааа​

Answer 1

$1.\;F'(x)=3x^2-2=f(x)\\\\2.\;f(x)=3x^2+1\\F(x)=x^2+x+C\\M(1;\;-2):\\1^2+1+C=-2\\2+C=-2\\C=-4\\F(x)=x^2+x-4\\\\3.\;a)\;\int\limits_{-1}^23x^2dx=\left.x^3\right|_{-1}^2=2^3-(-1)^3=8+1=9\\\\b)\;\int\limits_1^2\frac1{x^3}dx=\left.-\frac1{2x^2}\right|_1^2=-\frac1{2\cdot2^2}-\left(-\frac1{2\cdot1^2}\right)=-\frac18+\frac12=\frac38\\\\c)\;\int\limits_{\frac\pi2}^\pi\sin xdx=\left.-\cos x\right|_{\frac\pi2}^\pi=(-\cos\pi)-(-\cos\frac\pi2)=1$

$4.\\1)\;16-x^2=0\\(4-x)(4+x)=0\\x_1=-4,\;x_2=4\\S=\int\limits_{-4}^{4}(16-x^2)dx=\left.(16x-\frac{x^3}3)\right|_{-4}^4=(64-\frac{64}3)-(-64+\frac{64}3)=\frac{128}3+\frac{128}3=\\=\frac{256}3=85\frac13\\\\2)\;S=\int\limits_0^3(x^2+2)dx=\left.(\frac{x^3}3+2x)\right|_0^3=(\frac{3^3}3+2\cdot3)-0=9+6=15$