Question

Помогите решить ......................................

Answer 1

Ответ:

$1)\ \ y=2x^3\ \,\ \ x=1\ \ ,\ \ kasatelnaya\ v\ A(-1;-2)\\\\kasatelnaya:\ y(-1)=-2\ \ ,\ \ \ y'(x)=6x^2\ \ ,\ \ \ y'(-1)=6\\\\y=-2+6(x+1)\ \ ,\ \ \ \underline{\ y=6x+4\ }\\\\\\\displaystyle S=\int\limits^1_{-1}\, \Big((6x+4)-2x^3\Big)\, dx=\frac{1}{6}\cdot \frac{(6x+4)^2}{2}\, \Big|_{-1}^1-\frac{2x^4}{4}\, \Big|_{-1}^1=\\\\\\=\frac{1}{12}\cdot \Big(10^2-(-2)^2\Big)-\frac{1}{2}\cdot \Big(1^4-(-1)^4\Big)=\frac{1}{12}\cdot 96-\frac{1}{2}\cdot 0=\frac{96}{12}=8$

$2)\ \ y=2x-x^2\ \ ,\ \ y=0\ \ ,\ \ kasatelnaya\ v\ A\Big(\ \dfrac{1}{2}\ ;\ \dfrac{3}{4}\ \Big)\\\\kasatelnaya:\ \ y\Big(\dfrac{1}{2}\Big)=\dfrac{3}{4}\ \ ,\ \ y'(x)=2-2x\ \ ,\ \ \ y'\Big(\dfrac{1}{2}\Big)=2-1=1\\\\y=\dfrac{3}{4}+1\cdot \Big(x-\dfrac{1}{2}\Big)\ \ ,\ \ \ \underline {y=x+\dfrac{1}{4}\ }$

$\displaystyle S=\int\limits^{\frac{1}{2}}_{-\frac{1}{4}}\, \Big(x+\frac{1}{4}\Big)\, dx-\int\limits_0^{\frac{1}{2}}\, (2x-x^2)\, dx=\Big(\frac{x^2}{2}+\frac{x}{4}\Big)\Big|_{-\frac{1}{4}}^{\frac{1}{2}}-\Big(x^2-\frac{x^3}{3}\Big)\Big|_0^{\frac{1}{2}}=\\\\\\=\frac{1}{8}+\frac{1}{8}-\Big(\frac{1}{32}-\frac{1}{16}\Big)-\Big(\frac{1}{4}-\frac{1}{24}\Big)=\frac{1}{4}+\frac{1}{32}-\frac{5}{24}=\frac{24+3-20}{96}=\frac{7}{96}$