since ln(g(b)/g(a) = ln(g(b)) - ln(g(a))
g'(x) = 1/x
Integral of 1/x dx = ln(x)
The integral from a to b of f(x)/g(x) equals ln[(g(b)/g(a)]. Find g'(x), and explain your reasoning.
3 answers
I still do not understand how you found g'(x) to be 1/x. Could you please elaborate a little more on the reasoning behind this?
Good question. My reasoning is bogus. I was thinking ln(b)-ln(a)
We know that ∫dg/g = ln(g)
dg = g' dx
so, dg/dx = g' = f
how's that?
For example
∫(2x+3)/(x^2 + 3x - 7) dx
let g = x^2 + 3x - 7
dg = (2x+3) dx
and we have
∫dg/g = ln(g)
evaluate at a and b to get
ln(g(b)) - ln(g(a)) = ln[g(b)/g(a)]
We know that ∫dg/g = ln(g)
dg = g' dx
so, dg/dx = g' = f
how's that?
For example
∫(2x+3)/(x^2 + 3x - 7) dx
let g = x^2 + 3x - 7
dg = (2x+3) dx
and we have
∫dg/g = ln(g)
evaluate at a and b to get
ln(g(b)) - ln(g(a)) = ln[g(b)/g(a)]