Question

Помогите решить! Нужно срочно! Хотя бы 3 из 5 нужно решить!

Answer 1

$Large 1.\ y=6-x,, y=x^2\ 6-x=x^2,, x^2+x-6=0\ x_1=-3, x_2=2\ int_{-3}^{2}(6-x-x^2)mathrm dx=6x|_{-3}^{2}-{1over2}x^2|_{-3}^{2}-{1over3}x^3|_{-3}^{2}=6(2+3)-{1over2}(4-9)-{1over3}(8+27)=30+{5over2}-{35over3}=30+{15-70over6}=30-{55over6}={180-55over6}={125over6}$
$Large 2.\ picdotint_{0}^{6}(2x^2-12x)^2mathrm dx=picdotint_{0}^{6}(4x^4-48x^3+144x^2)mathrm dx=picdot({4over5}x^5|_{0}^{6}-12x^4|_{0}^{6}+48x^3|_{0}^{6})=picdot({4over5}(6^5-0)-12(6^4-0)+48(6^3-0))=picdot6^3(48-72+28,8)=picdot6^3cdot4,8=1036,8pi$

$Large 5.\ 5int_{OA}sin{y}: mathrm dl=*\ OA: {xover1}={yover -3}\ OA:y=-3x\ y'= -3\ *=5int_{0}^{1}sin{(-3x)}sqrt{1+(-3)^2}: mathrm dx=-5sqrt{10}int_{0}^{1}sin{(3x)}: mathrm dx=-{5sqrt{10}over3}int_{0}^{1}sin{(3x)}: mathrm d(3x)={5sqrt{10}over3}cdotcos{(3x)}|_{0}^{1}={5sqrt{10}over3}(cos{3}-1)$