Suppose the integral from 2 to 10 of g of x, dx equals 10 and the integral from 8 to 10 of g of x, dx equals negative 6, find the value of the integral from 2 to 8 of one-half times g of x, dx .

Answer 1

∫[2,8] 1/2 g(x) dx = 1/2 ∫[2,8] g(x) dx
Now, assuming all integrals are of g(x), we have
∫[2,10] = ∫[2,8] + ∫[8,10]
10 = ∫[2,8] - 6
∫[2,8] = 16
1/2 ∫[2,8] = 8