dF/dx = f(x) then
dF/dx = f(x)/x^5 = 6x^-2 - 2
see what you can do with that.
Find a function f and a number a such that
x
2+∫ (f(t)/t^(5)) dt=6x^(−2)
a
f(x)=
a=
2 answers
actually, I kind of mangled that. If
F(x) = ∫[a,x] f(t)/t^5 dt
then
F'(x) = f(x)/x^5 = -12/x^3
so, f(x) = -12x^3
F(x) = 6/x^2
So now we have
∫[a,x] f(t)/t^5 dt + 2 = F(x) - F(a) = 6/x^2 - 6/a^2 + 2 = 6/x^2
so
6/a^2 = 2
a = ±√3
F(x) = ∫[a,x] f(t)/t^5 dt
then
F'(x) = f(x)/x^5 = -12/x^3
so, f(x) = -12x^3
F(x) = 6/x^2
So now we have
∫[a,x] f(t)/t^5 dt + 2 = F(x) - F(a) = 6/x^2 - 6/a^2 + 2 = 6/x^2
so
6/a^2 = 2
a = ±√3
1 answer