hlp plz asap

(a)

Because the curve C in Figure 3 crosses the x-axis at points A and B, the solutions of the equation are A and B:

y = x(x + 4)(x - 2)

0 = x(x + 4)(x - 2) ...

now solve for x -> x₁ = 0, x₂ = -4 and x₃ = 2

We already know A or B are not zero,

so A = -4 (because A is negative) and B = 2 (because B is positive).

(b)

y = x(x + 4)(x - 2)    | distribute

y = x(x² - 2x + 4x - 8)

y = x(x² + 2x - 8)      | distribute

y = x³ + 2x² - 8x

now use integration

\(\int (x^3+2x^2-8x)dx\)

now do it yourself.