1. S = -(10/3)
2. S = (7/3)-ln(2)
Объяснение:
1. y = -x²-1
[2]S[1](-x²-1)dx
Вычислим неопределенный интеграл:
S(-x²-1)dx = S(-x²)dx + S(-1)dx = -S(x²)dx - S(1)dx
(-)S(x²)dx = (-)(x³/3)
(-)S(1)dx = (-)x
-[2]S[1](x²)dx = -(8/3)+(1/3) = -(7/3)
-[2]S[1](1)dx = -2+1 = -1
-(7/3)+1 = -(10/3)
2. y = x²
y = 1/x
S = S(x²-(1/x))dx
S = S(x²)dx - S(1/x)dx
[2]S[1](x²)dx = (8/3)-(1/3) = 7/3
[2]S[1](1/x)dx = ln(2)-ln(1) = ln(2)
↓
S = (7/3)-ln(2)
