the quesiton from ap calculus

Question

Suppose the integral from 2 to 8 of g of x, dx equals 13 , and the integral from 6 to 8 of g of x, dx equals negative 3 , find the value of 2 plus the integral from 2 to 6 of g of x, dx

Answer 1

∫₂⁸g(x)dx = 13; ∫₆⁸g(x)dx = - 3;

∫₂⁸g(x)dx = ∫₂⁶g(x)dx + ∫₆⁸g(x)dx; or 13 = ∫₂⁶g(x)dx - 3;

So, ∫₂⁶g(x)dx = 13 + 3 = 16 and ∫₂⁶g(x)dx + 2 = 16 + 2 = 18