Yefim S. answered 01/30/21
Math Tutor with Experience
- ∫2∞dx/(x√x2 - 4) = lim ε→0∫2+ε3dx/(x√x2 - 4) + lim A→∞∫3Adx/(x√x2 - 4) = lim ε→0 1/2sec-1Ix/2I2 + ε3 +
lim A→∞1/2sec-1Ix/2I3A = lim ε →0[1/2sec-1I3/2I - 1/2sec-1I(2 + ε)/2] + lim A → ∞ [1/2sec-1IA/2I -1/2sec-1I3/2I] = lim A → ∞ 1/2sec-1IA/2I - lim ε → 0 1/2sec-1I(2 + ε)/2! = 1/2sec-1(∞) - 1/2sec-1(1) = 1/2·π/2 - 1/2·0 = π/4.
2.x - coordinates of points of intersection: x3 = 4x; x = 0, x = ±2.
So, area A = ∫-20(x3- 4x)dx + ∫02(4x - x3)dx = (x4/4 - 2x2)-20 + (2x2 - x4/4)02 = 0 - 4 + 8 + 8 - 4 - 0 = 8