the 2nd FT of Calculus is just the chain rule in reverse.
If F(x) = ∫f(x) dx
∫[u,v] f(t) dt = F(u)-F(v)
so, taking derivatives,
F'(x) = dF/dv * dv/dx - dF/du * du/dx
= f(v) v' - f(u) u'
So, for this problem,
g(x) = ∫[2x,3x] f(u) du
g'(x) = f(3x)*3 - f(2x)*2
= 3((3x)^2-5)/((3x)^2+5) - 2((2x)^2-5)/((2x)^2+5)
and you can simplify that in a couple of ways. But that's just algebra...
Find the derivative of the function
g(x)=S(u^2-5)/(u^2+5)du
g'(x)=?
S=integration symbol
b=3x
a=2x
1 answer