If the entire function g(x) is an integral with upper and lower limits, that makes it a definite integral. A definite integral always has a constant value.
That would make g(x) a constant value. So what would its derivative be?
(This might be incorrect if my interpretation of the question is inaccurate)
Find g'(x)
G(x)=9tcos(t^10)dt; intergral from [0 to sqrt(x)]
3 answers
using the 2nd FTC,
g'(x) = 9√x cos(x^5) * 1/(2√x)
If G(x) = ∫[a,u(x)] f(t) dt
G'(x) = f(u(x)) * u'(x)
This is just the chain rule, seen from the other side.
g'(x) = 9√x cos(x^5) * 1/(2√x)
If G(x) = ∫[a,u(x)] f(t) dt
G'(x) = f(u(x)) * u'(x)
This is just the chain rule, seen from the other side.
Note that
if F'(x) = f(x)
and G(x) = ∫[u(x),v(x)] f(t) dt
= F(v)-F(u)
so
G'(x) = f(v)*v' - f(u)*u'
If, as in your problem, the lower limit is a constant a, then F(a) is a constant, so F'(a) = 0 (that means that u is a constant, so u'=0 in the more general formula)
if F'(x) = f(x)
and G(x) = ∫[u(x),v(x)] f(t) dt
= F(v)-F(u)
so
G'(x) = f(v)*v' - f(u)*u'
If, as in your problem, the lower limit is a constant a, then F(a) is a constant, so F'(a) = 0 (that means that u is a constant, so u'=0 in the more general formula)
2 answers