Question

Помогите с номером 4
#4
1)
2)
распишите подробно

Answer 1

$x^2+x-6=0 \\\ (x+3)(x-2)=0 \\\ x_1=-3; \ x_2=2 \\\ S= |\int\limits^2_{-3} {(x^2+x-6)} \, dx|=|( \frac{x^3}{3} + \frac{x^2}{2} -6x)^2_{-3}|= \\\ =|( \frac{2^3}{3} + \frac{2^2}{2} -6\cdot2)-( \frac{(-3)^3}{3} + \frac{(-3)^2}{2} -6\cdot(-3))|= \\\ =|( \frac{8}{3} +2 -12)-( -9 + \frac{9}{2} +18)|= |\frac{8}{3} -10-9- \frac{9}{2} |=|-20 \frac{5}{6}|=20 \frac{5}{6}$

$x^2+1=10 \\\ x^2=9 \\\ x=\pm3 \\\ S= \int\limits^3_{-3} {(10-(x^2+1))} \, dx = \int\limits^3_{-3} {(10-x^2-1)} \, dx = \int\limits^3_{-3} {(9-x^2)} \, dx = \\\ =(9x- \frac{x^3}{3} )|^3_{-3} =(9\cdot3- \frac{3^3}{3} )-(9\cdot(-3)- \frac{(-3)^3}{3} )= \\\ =27- 9-(-27+ 9 )=27- 9+27- 9=18+18=36$