Consider the function f(x)=(6/x^3)-42x^5, x>0. Let F(x) be the antiderivative of f(x) with F(1)=0. Then F(x)=

Answer 1

F(x) = ∫f(x) dx

F(x) = ∫[6/x³ - 42x⁵]dx

F(x) = ∫[6x^-3 - 42x⁵]dx

F(x) = -(6/2)x^-2 - (42/6)x⁶ + C where C is an arbitrary constant of integration

F(x) = -3/x² - 7x⁶ + C

We can determine a value for C by knowing that F(1) = 0.

F(1) = -3/1 - 7(1) + C = 0, or -10 + C = 0, or C = 10

Therefore, the antiderivative is

F(x) = -3/x² - 7x⁶ + 10