Question

Y=3x^2+4 ,x=-3, x=2 ,ось Ox. Найти: Sф

Answer 1

$S=\int\limits^2_{-3} {(3x^2+4)} \, dx =x^3+4x|^2_{-3}=8+8-(-27-12)=16+39=55$

Answer 2

To find the area (Sф) between the curve Y=3x^2+4 and the x-axis (ось Ox) between x=-3 and x=2, we can use the definite integral:

Sф = ∫[-3,2] (3x^2+4)dx

To evaluate this integral, we can use the power rule of integration:

∫x^n dx = (x^(n+1))/(n+1) + C, where C is the constant of integration.

Using this rule, we have:

Sф = ∫[-3,2] (3x^2+4)dx = [(x^3)/1 + 4x]_(-3,2) (evaluating the integral using the power rule) = [(2^3)/1 + 4(2)] - [(-3^3)/1 + 4(-3)] (substituting x=2 and x=-3) = [8 + 8] - [-27 - 12] = 16 + 39 = 55

Therefore, the area between the curve Y=3x^2+4 and the x-axis between x=-3 and x=2 is 55 square units.