Question

№65. Это одно задание

Answer 1

$65.\; \; S_1=\int\limits^0_{-1}(x+1-(x+1)^3)\, dx=\Big (\frac{(x+1)^2}{2}-\frac{(x+1)^4}{4}\Big )\Big |_{-1}^0= \\\\=\frac{1}{2}-\frac{1}{4}-0=\frac{1}{4}\\\\S_2=2\cdot S_1=\frac{1}{2}\\\\1.\; \; S=\int\limits^{-1}_{-3}(1-(x^2+4x+4))\, dx=\int\limits^{-1}_{-3}(1-(x+2)^2)\, dx=\\\\=\Big (x-\frac{(x+2)^3}{3}\Big )\Big |_{-3}^{-1}=-1-\frac{1}{3}-(-3+\frac{1}{3})=2-\frac{2}{3}=1\frac{1}{3}$

$2.\; \; S=\int\limits^{1}_{-1}(2-(1-x^3))\, dx=\int\limits^1_{-1}(1+x^3)\, dx=\Big (x+\frac{x^4}{4}\Big )\Big |_{-1}^1=\\\\=1+\frac{1}{4}-(-1+\frac{1}{4})=2\\\\3.\; \; S=\int\limits^{1}_{-2}(-x^2+4-(x+2))\, dx=\int\limits^1_{-2}(-x^2-x+2)\, dx=(-\frac{x^3}{3}-\frac{x^2}{2}+2x)\Big |_{-2}^1=\\\\=(-\frac{1}{3}-\frac{1}{2}+2)-(\frac{8}{3}-2-4)=\frac{27}{6}\\\\4.\; \; S=\int\limits^1_0(1+2x-x^2)-(x^2+1))\, dx=\int\limits^1_0(-2x^2+2x)\, dx=\\\\=(-\frac{2x^3}{3}+x^2)\Big |_0^1=-\frac{2}{3}+1= \frac{1}{3}$