Consider the function f(x) = 8/x^2 - 4/x^5

Question

Consider the function f(x) = 8/x^2 - 4/x^5

Let F(x) be the antiderivative of f(x) with F(1)=0.

Then F(x) =

Answer 1

See the video.

Answer 2

Hello Hey.

Let's try this and see if it works .....

(Back when Newton invented calculus, I told him to call it an "indefinite integral" I hate the term antiderivative.)

F(x) = ∫ f(x) dx = ∫ (8/x² - 4/x⁵ ) dx = ∫ (8*x^-2 - 4*x^-5 ) dx

F(x) = 8*(-2)*x^-1 - 4*(-5)*x^-4 + C = -16*x^-1 + 20*x^-4 + C

We are also given that F(1) = 0, so

F(1) = 0 = -16*(1)^-1 + 20*(1)^-4 + C = -16 + 20 + C = 4 + C

Solving for C gives C= -4

Therefore the answer is F(x) = 20*x^-4 -16*x^-1 -4 = 4*x^-1 (5*x^-3 -4 -x)

Agree???