Question

40 балов тому кто решит правильно!

Answer 1

$\displaystyle\\\\1)S=\int\limits^6_{-1} {11-x-x^2+6x-5} \, dx =\int\limits^6_{-1} {6+5x-x^2} \, dx=(6x+\frac{5x^2}{2}-\frac{x^3}{3})\mid^6_{-1}=\\\\\\ =6*6+\frac{5*6^2}{2}-\frac{6^3}{3}-(6*(-1)+\frac{5*(-1)^2}{2}-\frac{(-1)^3}{3})=\frac{343}{6}\\\\\\ 2)S=\int\limits^3_{-1} {\sqrt{x+1}} \, dx+\int\limits^7_3 {\sqrt{7-x}} \, dx \\\\\\\ \int\limits^3_{-1} {\sqrt{x+1}} \, dx=(\frac{2}{3}*(x+1)\sqrt{x+1})\mid^3_{-1}=\frac{2(3+1)\sqrt{3+1}}{3}-\frac{2(-1+1)\sqrt{-1+1}}{3}=\\\\\\=\frac{16}{3}$

$\displaystyle \int\limits^7_3 {\sqrt{7-x}} \, dx=\int\limits^3_{-1} {\sqrt{x+1}} \, dx =\frac{16}{3}\\\\\\ S=\frac{16}{3}+\frac{16}{3}=\frac{32}{2}\\\\\\ 3)S=\int\limits^2_{-2} {4-x^2} \, dx -\int\limits^2_0 {4-x^2-(x-2)^2} \, dx \\\\\\\int\limits^2_{-2} {4-x^2} \, dx=(4x-\frac{x^3}{3})\mid^2_{-2}=4*2-\frac{2^3}{3}-(4*(-2)-\frac{(-2)^3}{3})=\frac{32}{3}\\\\\\ \int\limits^2_0 {4-x^2-(x-2)^2} \, dx=\int\limits^2_0 {-2x^2+4x} \, dx=(-\frac{2x^3}{3}+2x^2)\mid^2_0=$

$\displaystyle =-\frac{2*2^3}{3}+2*2^2-(-\frac{2*0}{3}+2*0)=\frac{8}{3}\\\\\\ S=\frac{32}{3}-\frac{8}{3}=8$