If F′(x)=x3−9 and F(1)=1, what is F(x)?

Answer 1

We simply find the anti-derivative of f'(x)

f'(x) = x³-9

The anti-derivative: x⁽ⁿ⁺¹⁾/(n+1)

f(x) = x⁴/4 - 9x + c

since f(1) = 1 we can determine the value of c

f(1) = (1)⁴/4 - 9(1) + c = 1

1/4 - 9 + c = 1

c = 10 - 1/4 = 9 3/4 = 39/4

So f(x) = x⁴/4 - 9x + 39/4

checking, take the derivative

f'(x) = 4(x³)/4 - 9 = x³ - 9

Answer 2

I assume that is x^3 rather than 3x...

dF/dx = x³ - 9

Int(dF) =Int((x³ - 9)dx) or

F = x⁴/4 - 9x +C (Solve for C so that F(1) = 1

Take care